why will a measuring stick placed along the circumference of a rotating disk appear contracted, but not if it is oriented along a radius? why will a measuring stick placed along the circumference of a rotating disk appear contracted, but not if it is oriented along a radius?
A measuring stick placed along the circumference of a rotating disk will appear contracted, but not if it is oriented along a radius because of the effects of length contraction, also known as Lorentz contraction, which is a consequence of special relativity.
The theory of special relativity postulates that a moving object appears shorter along its direction of motion than when it is at rest. The length of an object appears to contract in the direction of motion due to time dilation and length contraction. As a result, if the measuring stick is placed along the circumference of a rotating disk and is moving with the disk's motion, it will appear to be shorter or contracted. However, if it is oriented along a radius and is not moving with the disk's motion, it will not appear to be shorter or contracted. Length contraction and time dilation are two of the fundamental principles of special relativity, which helps to explain the strange and unexpected behaviors of objects at speeds approaching the speed of light.
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GIVING BRAINLIEST PLEASE HELP!!
-if you answer correctly ill give you brainliest which will give you 27pts-
Answer:
C. The lever applies three times more force than you hand can apply.
Explanation:
Since it's advantage is 3, that means you'll have to multiply the input of it by 3, making this apply 3x more force than your hand.
Hope this helped! <3
Source(s): Me and a bit of g*ogle for clarification
Choose true or false for each statement regarding the sign conventions for lenses.
The magnification m is negative for inverted images.
Virtual images appear on same side of the lens as the object and have a negative value for the image distance.
Real images appear on the opposite side of the lens from the object and have a negative value for the image distance.
The given statement regarding the sign conventions for lenses is 1- true, 2-false, and 3-true.
The magnification, m, is negative for inverted images. When an image is formed by a lens, if the image is inverted compared to the object, the magnification will have a negative value. The first statement is true.
Virtual images appear on the opposite side of the lens from the object. Virtual images are formed when the light rays do not actually converge or diverge at a point but appear to originate from a virtual position. They are always formed on the same side of the lens as the object. The image distance for virtual images is positive. The second statement is false.
Real images appear on the opposite side of the lens from the object. Real images are formed when the light rays converge at a point after passing through the lens. They are formed on the opposite side of the lens from the object. The image distance for real images is negative. The third statement is true.
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a small remote-control car with a mass of 1.61 kg moves at a constant speed of v = 12.0 m/s in a vertical circle inside a hollow metal cylinder that has a radius of 5.00 m
The tension in the string is approximately 46.7 N. To find the tension in the string, we can analyze the forces acting on the remote-control car at the top and bottom of the vertical circle.
To find the tension in the string, we can analyze the forces acting on the remote-control car at the top and bottom of the vertical circle.
At the top of the circle:
The downward gravitational force (mg) and the tension in the string (T) act downward.
The net force in the upward direction is provided by the centripetal force (Fc).
At the bottom of the circle:
The downward gravitational force (mg) and the tension in the string (T) act downward.
The net force in the upward direction is the sum of the centripetal force (Fc) and the car's weight (mg).
We can set up the following equations of motion at the top and bottom of the circle:
At the top:
T - mg = Fc ...(1)
At the bottom:
T + mg = Fc + mg ...(2)
We can substitute the expression for the centripetal force (Fc = mv^2 / r) into the equations:
At the top:
T - mg = mv^2 / r ...(3)
At the bottom:
T + mg = mv^2 / r + mg ...(4)
Now we can solve these equations to find the tension in the string.
At the top:
T - mg = mv^2 / r
T = mv^2 / r + mg ...(5)
At the bottom:
T + mg = mv^2 / r + mg
From equation (5), we can substitute the expression for T:
mv^2 / r + mg + mg = mv^2 / r + mg
2mg = mv^2 / r
Now we can solve for the tension (T):
T = mv^2 / r - mg
T = (1.61 kg)(12.0 m/s)^2 / 5.00 m - (1.61 kg)(9.8 m/s^2)
T ≈ 46.7 N
Therefore, the tension in the string is approximately 46.7 N.
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what is the normal force acting on a mountain goat that weighs 650 n and is standing on such a slope?
The normal force acting on the mountain goat is 650 N. It supports the weight of the goat and keeps it in equilibrium on the slope.
The normal force is the force exerted by a surface to support the weight of an object resting on it. It acts perpendicular to the surface. In this case, the mountain goat is standing on a slope.
When an object is on an inclined surface, the normal force can be split into two components: one perpendicular to the slope (normal to the surface) and one parallel to the slope (tangential to the surface).
The component parallel to the slope is responsible for counteracting the gravitational force pulling the object down the slope.
In this scenario, the mountain goat is standing on the slope, and we can assume it is in equilibrium, meaning it is not sliding down the slope. Therefore, the parallel component of the normal force is equal in magnitude and opposite in direction to the gravitational force acting down the slope.
The weight of the mountain goat is given as 650 N. This is the magnitude of the gravitational force acting on the goat.
The normal force acting on the goat is equal in magnitude but opposite in direction to the gravitational force. So the normal force is also 650 N.
The normal force acting on the mountain goat is 650 N. It supports the weight of the goat and keeps it in equilibrium on the slope.
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Light is polarized by using:
Answer:
Polaroid fliter
Explanation:
light can be polarized by using Polaroid filters
Polaroid fliter are made of special material that is capable of blocking one of the two planes of vibration of an electromagnetic wave
hope this is useful--(have a good day)
PLEASE HELP ME!!! THIS IS DO TODAY. AND NO LINKS PLEASE!!!!!
Answer:
to late
Explanation:
the current in an rl circuit builds up to one-third of its steady-state value in 4.31 s. find the inductive time constant.
In this RL circuit, the inductive time constant is found to be approximately 12.93 seconds.
The inductive time constant of an RL circuit can be determined by analyzing the rate at which the current builds up to one-third of its steady-state value.
In an RL circuit, the rate at which the current builds up is determined by the inductive time constant (symbolized by the Greek letter tau, τ). The inductive time constant represents the time required for the current in the circuit to reach approximately 63.2% of its steady-state value.
Given that the current builds up to one-third (33.3%) of its steady-state value in 4.31 seconds, we can use this information to calculate the inductive time constant. We know that when the current reaches one-third of its steady-state value, it corresponds to approximately 33.3% of the difference between the initial current (at t=0) and the steady-state current.
Using this relationship, we can set up the equation:
33.3% = (1 - e^(-4.31/τ)) * 100%
Rearranging the equation and solving for τ, we find:
τ = -4.31 / ln(1 - 33.3%/100%)
Evaluating this expression gives us τ ≈ 12.93 seconds. Therefore, the inductive time constant of the RL circuit in question is approximately 12.93 seconds.
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If the efficiency and mechanical advantage of a certain machine are given as 65 % and 3 respectively.What is the velocity ratio of the machine?
a.3.5 %
b.4.6 %
c.7.9 %
d.11.2 %
Answer:
b. 4.6 %
Explanation:
From the question,
E = M.A/V.R................ Equation 1
Where E = percentage Efficiency of the machine, M.A = machanical accurancy of the machine, V.R = Velocity ratio of the machine
Make V.R the subject of the equation
V.R = M.A/E
Given: M.A = 3, E = 65% = 0.65
Substitute this values into equation 2
V.R = 3/0.65
V.R = 4.6
Hence the right option is b. 4.6 $
Why might a scientist use a magnet and small strainer when investigating the physical properties of a substance?
Answer:
In order to check its magnetic properties and for removing impurities.
Explanation:
A scientist use a magnet and small strainer while investigating the physical properties of a substance in order to check the magnetic properties of the substance as well as to separate the substance from the other impurities. Magnet attract substances that is made of metals or magnetic characteristics while on the other hand, small strainer is used to separate impurities from the investigating substance so that's why the scientist use a magnet and small strainer during investigation of a substance.
A scientist use a magnet to check magnetic properties of the substance and use small strainer to separate the substance from the other impurities.
The magnet have Attractive Property , Magnet attracts ferromagnetic materials like iron, cobalt, and nickel.Magnet attract substances that is made of metals or magnetic characteristics .The small strainer is used to separate impurities from the investigating substanceLearn more:
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A hollow metal sphere has inner radius aa and outer radius b. The hollow sphere has charge +2Q. A point charge +Q sits at the center of the hollow sphere.
A. Determine the magnitude of the electric field in the region r≤a. Give your answer as a multiple of Q/ε0.
B. Determine the magnitude of the electric field in the region a
C. Determine the magnitude of the electric field in the region r≥b. Give your answer as a multiple of Q/ε0.
D. How much charge is on the inside surface of the hollow sphere? Give your answer as a multiple of Q.
E. How much charge is on the exterior surface? Give your answer as a multiple of Q.
A. The magnitude of the electric field in the region r ≤ a is zero.
B. The magnitude of the electric field in the region a < r < b is zero.
C. The magnitude of the electric field in the region r ≥ b is Q/ε₀, where ε₀ is the permittivity of free space.
D. The charge on the inside surface of the hollow sphere is +Q.
E. The charge on the exterior surface of the hollow sphere is +2Q.
A. To determine the magnitude of the electric field in the region r ≤ a (inside the hollow sphere), we need to consider the superposition of electric fields from the point charge at the center and the hollow sphere.
The electric field inside a conducting hollow sphere is zero. This means that the electric field due to the hollow sphere cancels out the electric field due to the point charge at the center.
Therefore, the magnitude of the electric field in the region r ≤ a is zero.
B. In the region a < r < b (between the inner and outer radii of the hollow sphere), the electric field is zero because the charge on the inner surface of the hollow sphere distributes itself uniformly on the inner surface, creating an electric field that cancels out the electric field from the point charge at the center.
Therefore, the magnitude of the electric field in the region a < r < b is zero.
C. In the region r ≥ b (outside the hollow sphere), we only have the electric field due to the point charge at the center. The magnitude of the electric field from a point charge is given by Coulomb's Law:
E = k * (|Q| / [tex]r^{2}[/tex]),
where E is the electric field, k is the electrostatic constant (k = 8.99 × [tex]10^{9}[/tex] N [tex]m^{2}[/tex]/[tex]C^{2}[/tex]), |Q| is the magnitude of the charge, and r is the distance from the point charge.
Substituting the given values:
E = k * (|Q| / [tex]r^{2}[/tex]),
= k * (Q / [tex]r^{2}[/tex]),
where we consider the magnitude of the charge |Q| = Q.
Therefore, the magnitude of the electric field in the region r ≥ b is Q/ε₀, where ε₀ is the permittivity of free space.
D. The charge on the inside surface of the hollow sphere is equal to the charge of the point charge at the center, +Q. This is because in electrostatic equilibrium, the charge resides on the outer surface of a conductor, and there is no electric field inside a conductor.
Therefore, the charge on the inside surface of the hollow sphere is +Q.
E. The charge on the exterior surface of the hollow sphere is equal to the charge of the hollow sphere, which is +2Q. This is because the charge on a hollow conductor resides entirely on its outer surface.
Therefore, the charge on the exterior surface of the hollow sphere is +2Q.
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Transcranial magnetic stimulation (TMS) is a noninvasive technique used to stimulate regions of the human brain. A small coil is placed on the scalp, and a brief burst of current in the coil produces a rapidly changing magnetic field inside the brain. The induced emf can be sufficient to stimulate neuronal activity. One such device generates a magnetic field within the brain that rises from zero to 1.5 T in 120 ms. Determine the induced emf within a circle of tissue of radius 1.6 mm and that is perpendicular to the direction of the field.
Answer:
0.125 volts
Explanation:
The induced emf can be sufficient to stimulate neuronal activity.
One such device generates a magnetic field within the brain that rises from zero to 1.5 T in 120 ms.
We need to find the induced emf within a circle of tissue of radius 1.6 mm and that is perpendicular to the direction of the field. The formula for the induced emf is given by :
[tex]\epsilon=-\dfrac{d\phi}{dt}[/tex]
Where
[tex]\phi[/tex] is magnetic flux
So,
[tex]\epsilon=-\dfrac{d(BA)}{dt}\\\\=2\pi r\times \dfrac{dB}{dt}\\\\=2\pi \times 1.6\times 10^{-3}\times \dfrac{1.5-0}{120\times 10^{-3}}\\\\=0.125\ V[/tex]
So, the induced emf is equal to 0.125 volts.
Energy that comes from the heat inside the Earth is called ________ energy.
Answer:
Geothermal Energy.
Explanation:
The 500-N force F is applied to the vertical pole as shown(1) Determine the scalar components of the force vector F along the x'- and y'-axes. (2) Determine the scalar components of F along the x- and y'-axes.
Solution :
Given :
Force, F = 500 N
Let [tex]$ \vec F = F_x\ \hat i + F_y\ \hat j$[/tex]
[tex]$|\vec F|=\sqrt{F_x^2+F_y^2}$[/tex]
∴ [tex]$F_x=F \cos 60^\circ = 500 \ \cos 60^\circ = 250 \ N$[/tex]
[tex]$F_y=-F \cos 30^\circ = -500 \ \cos 30^\circ = -433.01 \ N$[/tex] (since [tex]$F_y$[/tex] direction is in negative y-axis)
[tex]$F=250 \ \hat i - 433.01 \ \hat j$[/tex]
So scalar components are : 250 N and 433.01 N
vector components are : [tex]$250 \ \hat i$[/tex] and [tex]$-433.01\ \hat j$[/tex]
1. Scalar components along :
x' axis = 500 N, since the force is in this direction.
[tex]$F_{x'}= F \ \cos \theta = 500\ \cos \theta$[/tex]
Here, θ = 0° , since force and axis in the same direction.
So, cos θ = cos 0° = 1
∴ [tex]$F_{x'}=500 \times 1=500\ N$[/tex]
[tex]$F_{y'}= F \ \sin \theta = 500\ \sin 0^\circ=500 \times 0=0$[/tex]
[tex]$F_{y'}=F\ cos \theta$[/tex] but here θ is 90°. So the force ad axis are perpendicular to each other.
[tex]$F_{y'}=F\ \cos 90^\circ= 500 \ \cos 90^\circ = 500 \times 0=0$[/tex]
∴ [tex]$F_{x'}= 500\ N \text{ and}\ F_{y'}=0\ N$[/tex]
2. Scalar components of F along:
x-axis :
[tex]$F_x=F\ \cos \theta$[/tex], here θ is the angle between x-axis and F = 60°.
[tex]$F_x=500 \times \cos60^\circ=250\ N$[/tex]
y'-axis :
[tex]$F_{y'}=F\ \cos \theta$[/tex], here θ is the angle between y'-axis and F = 90°.
[tex]$F_{y'}=500 \times \cos90^\circ=500\times 0=0\ N$[/tex]
∴ [tex]$F_{x}= 250\ N \text{ and}\ F_{y'}=0\ N$[/tex]
Which of the following is an inertial reference frame,
or at least a good approximation of one?
A. The inside of the orbiting International Space
Station.
B. A non-spinning ball following a projectile
motion trajectory.
C. An elevator accelerating downwards at 1g
D. All of the above
E. None of the above
A non-spinning ball following a projectile motion trajectory is an inertial reference frame,
Hence, the correct option is B.
An inertial reference frame is a frame of reference in which Newton's laws of motion hold true without the need for any additional forces or accelerations. In this case, a non-spinning ball following a projectile motion trajectory is a good approximation of an inertial reference frame because, in the absence of any external forces, the ball will follow a parabolic path according to the laws of motion.
A. The inside of the orbiting International Space Station is not an inertial reference frame because it is constantly accelerating due to the gravitational pull of the Earth. Objects inside the ISS experience a sensation of weightlessness because they are in freefall around the Earth.
C. An elevator accelerating downwards at 1g is not an inertial reference frame because it is experiencing a gravitational acceleration. Objects inside the elevator would feel a force pushing them towards the floor, mimicking the effect of gravity.
Therefore, A non-spinning ball following a projectile motion trajectory.
Hence, the correct option is B.
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Solved Exa
Example 1. An iron ball of mass 3 kg is
released from a height of 125 m and falls
freely to the ground. Assuming that the
value of g is 10 m/s2, calculate
(i) time taken by the ball to reach the
ground
(ii) velocity of the ball on reaching the
ground
(iii) the height of the ball at half the time it
takes to reach the ground.
According to the equations of motion, the time taken to reach the ground is 5 seconds.
Using;
s = ut + 1/2gt^2
s = distance
u = initial velocity
t = time taken
g = acceleration due to gravity
Note that u = 0 m/s since the object was dropped from a height
Substituting values;
125 = 1/2 × 10 × t^2
125 = 5t^2
t^2 = 125/5
t^2 = 25
t = 5 secs
Velocity on reaching the ground is obtained from
v = u + gt
Where u = 0 m/s
v = gt
v = 10 × 5
v = 50 m/s
At half the time it takes to reach the ground;
s = ut + 1/2gt^2
Where u = 0 m/s
s = 1/2gt^2
s = 1/2 × 10 × (2.5)^2
s = 31.25 m
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Answer:
(i) time taken by the ball to reach the
ground is 5 sec.
(ii) velocity of the ball on reaching the
ground is 50 m/s.
(iii) the height of the ball at half the time it
takes to reach the ground is 31.25 m.
Step-by-step explanation:
Solution :(i) time taken by the ball to reach the
ground
[tex]\longrightarrow{\sf{ \: \: s= ut + \dfrac{1}{2} a{(t)}^2}}[/tex]
[tex]\longrightarrow{\sf{ \: \: 125= 0 \times t + \dfrac{1}{2} \times 10 \times {(t)}^2}}[/tex]
[tex]\longrightarrow{\sf{ \: \: 125= 0 + \dfrac{10}{2} \times {(t)}^2}}[/tex]
[tex]\longrightarrow{\sf{ \: \: 125= 0 + 5\times {(t)}^2}}[/tex]
[tex]\longrightarrow{\sf{ \: \: 125= 5\times {(t)}^2}}[/tex]
[tex]\longrightarrow{\sf{ \: \: {(t)}^2 = \dfrac{125}{5}}}[/tex]
[tex]\longrightarrow{\sf{ \: \: {(t)}^2 = \dfrac{ \cancel{125}}{\cancel{5}}}}[/tex]
[tex]\longrightarrow{\sf{ \: \: {(t)}^2 = 25}}[/tex]
[tex]\longrightarrow{\sf{ \: \: t = \sqrt{25} }}[/tex]
[tex]\longrightarrow \: \: {\sf{\underline{\underline{\red{ t = 5 \: sec}}}}}[/tex]
Hence, the ball taken 5 sec to reach the ground.
[tex]\begin{gathered}\end{gathered}[/tex]
(ii) velocity of the ball on reaching the
ground
[tex]\longrightarrow{\sf{ \: \: {v}^{2} - {u}^{2} = 2as}}[/tex]
[tex]\longrightarrow{\sf{ \: \: {v}^{2} - {0}^{2} = 2 \times 10 \times 125}}[/tex]
[tex]\longrightarrow{\sf{ \: \: {v}^{2} = 20 \times 125}}[/tex]
[tex]\longrightarrow{\sf{ \: \: {v}^{2} = 2500}}[/tex]
[tex]\longrightarrow{\sf{ \: \: {v} = \sqrt{2500} }}[/tex]
[tex]\longrightarrow{\sf{ \: \: \underline{\underline{ \red{{v} = 50 \: m/s }}}}}[/tex]
Hence, the velocity of ball is 50 m/s.
[tex]\begin{gathered}\end{gathered}[/tex]
(iii) the height of the ball at half the time it
takes to reach the ground.
[tex]\longrightarrow{\sf{ \: \: s= ut + \dfrac{1}{2} a{(t)}^2}}[/tex]
[tex]\longrightarrow{\sf{ \: \: s= 0 \times \dfrac{5}{2} + \dfrac{1}{2} \times 10 \times { \left( \dfrac{5}{2} \right)}^2}}[/tex]
[tex]\longrightarrow{\sf{ \: \: s= 0 + \dfrac{10}{2} \times { \left( \dfrac{5}{2} \times \dfrac{5}{2} \right)}}}[/tex]
[tex]\longrightarrow{\sf{ \: \: s= \dfrac{10}{2} \times { \left( \dfrac{5 \times 5}{2 \times 2} \right)}}}[/tex]
[tex]\longrightarrow{\sf{ \: \: s= \dfrac{10}{2} \times { \left( \dfrac{25}{4} \right)}}}[/tex]
[tex]\longrightarrow{\sf{ \: \: s= \dfrac{10}{2} \times \dfrac{25}{4}}}[/tex]
[tex]\longrightarrow{\sf{ \: \: s= \dfrac{10 \times 25}{2 \times 4}}}[/tex]
[tex]\longrightarrow{\sf{ \: \: s= \dfrac{250}{8}}}[/tex]
[tex]\longrightarrow{\sf{ \: \: s= \dfrac{\cancel{250}}{\cancel{8}}}}[/tex]
[tex]\longrightarrow{\sf{ \: \: {\underline{\underline{\red{s= 31.25 \: m}}}}}}[/tex]
Hence, the height of the ball to reach the ground is 31.25 m.
[tex]\underline{\rule{220pt}{3.5pt}}[/tex]
A spring of spring constant 30.0 N/m is attached to a 2.3 kg mass and set in motion. What is the period and frequency of vibration for the 2.3 kg mass?
Answer:
1. The period is 1.74 s.
2. The frequency is 0.57 Hz
Explanation:
1. Determination of the the period.
Spring constant (K) = 30 N/m
Mass (m) = 2.3 Kg
Pi (π) = 3.14
Period (T) =?
The period of the vibration can be obtained as follow:
T = 2π√(m/K)
T = 2 × 3.14 × √(2.3 / 30)
T = 6.28 × √(2.3 / 30)
T = 1.74 s
Thus, the period of the vibration is 1.74 s.
2. Determination of the frequency.
Period (T) = 1.74 s
Frequency (f) =?
The frequency of the vibration can be obtained as follow:
f = 1/T
f = 1/1.74
f = 0.57 Hz
Thus, the frequency of the vibration is 0.57 Hz
The period of the vibration is 1.76 s and the frequency of the vibration is 0.57 s-1.
Using the formula;
T = 2π√(m/K)
Where;
T = period
m = mass
K = spring constant
Substituting values;
T = 2(3.142)√2.3/30
T = 6.284 × 0.28
T = 1.76 s
Recall that the period is the inverse of frequency;
f = 1/T
f = 1/1.76 s
f = 0.57 s-1
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Six identical resistors, each with resistance R, are connected to an emf E Part A What is the current I through each of the resistors if they are connected in parallel?
Part B If they are connected in series? Express your answer in terms of the variables E and R.
An emf E is connected to six identical resistors, each with resistance R.
(A) When identical resistors are connected in parallel, the current through each resistor is the same and is given by [tex]\begin{equation}I = \frac{E}{R}[/tex].
(B) When identical resistors are connected in series, the total resistance is 6R, and the current through each resistor is given by [tex]\begin{equation}I = \frac{E}{6R}[/tex].
Part A: When the identical resistors are connected in parallel, the current (I) through each resistor is the same. To calculate the current, we can use Ohm's Law, which states that the current (I) flowing through a resistor is equal to the voltage (V) across the resistor divided by its resistance (R):
[tex]\begin{equation}I = \frac{V}{R}[/tex]
In this case, the voltage across each resistor is the same, and it is equal to the emf (E). Therefore, the current through each resistor connected in parallel is:
[tex]\begin{equation}I = \frac{E}{R}[/tex]
Part B: When the identical resistors are connected in series, the total resistance ([tex]R_total[/tex]) is the sum of the individual resistances. Therefore, the current (I) flowing through the resistors in series is given by Ohm's Law:
[tex]\begin{equation}\I = \frac{E}{R_\text{total}}[/tex]
Since the resistors are identical, the total resistance can be calculated as:
[tex]R_total[/tex] = R + R + R + R + R + R = 6R
Substituting this value into the equation for the current, we get:
[tex]\begin{equation}I = \frac{E}{6R}[/tex]
So, when the resistors are connected in series, the current through each resistor is given by [tex]\begin{equation}I = \frac{E}{6R}[/tex].
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A pitcher threw a baseball straight up at 35. 8 meters per second. What was the balls velocity after 2. 50?
When a pitcher throws a baseball straight up at 35.8 meters per second, the ball’s velocity after 2.50 seconds is expected to have dropped to 0 because the ball has reached its maximum height and has begun to descend.
The velocity that the ball will have after 2.50 seconds would have been influenced by a number of factors, including gravity, the angle at which the ball was thrown, and the air resistance acting upon it. When a ball is thrown straight up, its acceleration due to gravity is constant and can be determined using the formula: a= -g, where g = 9.81 m/s². Therefore, after 2.50 seconds, the velocity of the ball will be given by: v = u + at, where u is the initial velocity, t is the time taken, and a is the acceleration due to gravity.
Given that u = 35.8 m/s, t = 2.50 s, and a = -9.81 m/s², the velocity of the ball will be: v = 35.8 + (-9.81) x 2.50 = 10.45 m/s downward.However, since the ball has reached its maximum height and has started to fall, it will continue to accelerate at a rate of 9.81 m/s² until it hits the ground. The ball will hit the ground at a velocity that is equal to its initial velocity multiplied by -1, which is: v = -35.8 m/s.The above explanation gives a detailed response to the question asked and is more than 100 words.
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The displacement of the tip of the 10 cm long minute hand of aclock between 12:15 A.M. and 12:45 P.M. is:
Question 4 answers
10 cm,90°
10 cm,180°
10 cm,4500°
20 cm,180°
20 cm,540°
The displacement of the tip of the 10 cm long minute hand of a clock between 12:15 A.M. and 12:45 P.M. is 10 cm and 180°.
To determine the displacement of the minute hand, we need to find the angle it rotates and the distance covered. Between 12:15 A.M. and 12:45 P.M., there are 12 hours and 30 minutes. The minute hand of a clock completes a full revolution (360°) in 60 minutes.
First, let's find the angle covered by the minute hand. Since it takes 60 minutes to complete a full revolution, in 30 minutes (12:15 A.M. to 12:45 P.M.), the minute hand will cover half of that angle, which is 180°.
Next, let's calculate the distance covered by the minute hand. The length of the minute hand is given as 10 cm. Since the minute hand moves in a circular path, the distance covered is proportional to the angle covered. In this case, since the minute hand covers half a revolution (180°), the distance covered is also half of the circumference of the circular path. Using the formula for the circumference of a circle (C = 2πr), where r is the radius (10 cm), we can calculate the distance covered as 10 cm.
Therefore, the displacement of the tip of the 10 cm long minute hand of a clock between 12:15 A.M. and 12:45 P.M. is 10 cm and 180°.
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find the speed the block has as it passes through equilibrium (for the first time) if the coefficient of friction between block and surface is k = 0.350.
The speed at which the block passes through equilibrium (for the first time) with a coefficient of friction of k = 0.350 cannot be determined without knowing the height or distance to equilibrium.
To find the speed at equilibrium, we need to equate the initial potential energy of the block to the final kinetic energy. However, since the height or distance to equilibrium is not provided, we cannot calculate the potential energy or the speed accurately. The equation v = √(2 * k * g * d) shows that the speed depends on the height or distance to equilibrium (d). Without this information, we cannot determine the speed. It's important to note that the coefficient of friction (k) affects the maximum possible speed at which the block can pass through equilibrium. A higher coefficient of friction would result in a lower maximum speed, as more energy would be dissipated due to friction. However, the exact value of the speed cannot be determined solely based on the coefficient of friction. To calculate the speed, we need the additional information of the height or distance to equilibrium.
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What is the instantaneous velocity of the hummingbird at t=1s?
The distance - time graph of the humming bird is missing, so i have attached it.
Answer:
Instantaneous velocity = 0.5 m/s
Explanation:
From the attached graph, at time t = 1 s, the corresponding distance is 0.5 m.
Instantaneous velocity is the velocity at that point.
Thus;
Instantaneous velocity = 0.5/1
Instantaneous velocity = 0.5 m/s
What does a charged object experience as it is placed into an electric field?
Answer:
In an electric field a charged particle, or charged object, experiences a force. If the forces acting on any object are unbalanced, it will cause the object to accelerate. With this in mind: If two objects with the same charge are brought towards each other the force produced will be repulsive, it will push them apart.
Explanation:
A snail is traveling along a straight path. The snail's velocity can be modeled by v(t) = 1.4 In (1 +r²) inches per minute for 0 ≤ 1 ≤ 15 minutes. (a) Find the acceleration of the snail at time t = 5 minutes. (b) What is the displacement of the snail over the interval 0 ≤ 1 ≤ 15 minutes?
The snail's acceleration at t = 5 minutes is approximately 0.079 inches per minute squared. Over the interval 0 ≤ t ≤ 15 minutes, the snail's displacement is approximately 15.405 inches.
To find the acceleration at t = 5 minutes, we need to differentiate the velocity function with respect to time. The given velocity function is v(t) = 1.4 ln(1 + r²), where ln denotes the natural logarithm. Let's differentiate v(t) with respect to t to find the acceleration function a(t):
a(t) = d/dt (1.4 ln(1 + r²))
To differentiate ln(1 + r²), we use the chain rule:
a(t) = 1.4 * d/dt (ln(1 + r²))
The derivative of ln(1 + r²) with respect to r² is 1 / (1 + r²), so we can rewrite the acceleration function as:
a(t) = 1.4 * (1 / (1 + r²)) * d/dt (1 + r²)
The derivative of 1 + r² with respect to t is 2r dr/dt. Substituting this back into the acceleration function, we get:
a(t) = 1.4 * (1 / (1 + r²)) * 2r dr/dt
Since we're evaluating the acceleration at t = 5 minutes, we substitute t = 5 into the expression and solve for the corresponding values of r and dr/dt. Then, we calculate the acceleration.
Now, to find the displacement over the interval 0 ≤ t ≤ 15 minutes, we integrate the velocity function with respect to time over that interval:
∫[0,15] (1.4 ln(1 + r²)) dt
By evaluating this definite integral, we obtain the displacement of the snail over the given time interval.
Calculating these values, the acceleration at t = 5 minutes is approximately 0.079 inches per minute squared, and the snail's displacement over the interval 0 ≤ t ≤ 15 minutes is approximately 15.405 inches.
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(a) An insulating sphere with radiusa has a uniform charge density rho. The sphere isnot centered at the origin but at.
r=b
Show that the electric field inside thesphere is given by
e=p(r - b)/3E0
To show that the electric field inside the insulating sphere is given by E = ρ(r - b)/(3ε₀), where ρ is the charge density, r is the distance from the centre of the sphere, b is the displacement of the centre from the origin, and ε₀ is the permittivity of free space, we can use Gauss's law.
Gauss's law states that the electric flux through a closed surface is proportional to the charge enclosed by that surface. By applying Gauss's law, we can derive the electric field inside the insulating sphere.
Let's choose a Gaussian surface in the shape of a sphere with radius r, where r is less than the radius of the insulating sphere (a). Since the sphere is not centred at the origin but at a displacement of b, the centre of our Gaussian sphere will also be displaced by b.
According to Gauss's law, the electric flux through this Gaussian surface is given by:
Φ = E * A
where Φ is the electric flux, E is the electric field, and A is the area of the Gaussian surface.
Since the electric field is radially symmetric for a uniformly charged sphere, the electric field at any point on the Gaussian surface will have the same magnitude and direction. Therefore, the electric field can be taken out of the dot product with the area vector, and we have:
Φ = E * A = E * 4πr²
Now, we need to determine the charge enclosed by this Gaussian surface. Since the sphere has a uniform charge density (ρ), the charge enclosed within a sphere of radius r is given by:
Q = (4/3)πr³ρ
Now, applying Gauss's law, we have:
Φ = Q / ε₀
Substituting the expressions for Φ and Q, we get:
E * 4πr² = (4/3)πr³ρ / ε₀
E = (1/3) * r * ρ / ε₀
Since r is the distance from the origin, and the sphere is displaced by b, we can rewrite r as (r - b). Therefore:
E = ρ(r - b) / (3ε₀)
Therefore, we have shown that the electric field inside the insulating sphere is given by E = ρ(r - b) / (3ε₀), where r is the distance from the origin, b is the displacement of the sphere from the origin.
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An optical fiber uses flint glass surrounded by a crown glass cladding.
What is the critical angle for total internal reflection in degrees?
Given: Glass, crown n = 1.52; glass, flint n = 1.66
air n = 1.00
The critical angle for total internal reflection in an optical fiber composed of flint glass surrounded by a crown glass cladding can be determined using the given refractive indices.
The critical angle for total internal reflection occurs when light traveling through a medium encounters a boundary with a lower refractive index. In this case, the light travels from the flint glass (n = 1.66) to the crown glass cladding (n = 1.52), with the surrounding medium being air (n = 1.00).
To calculate the critical angle, we can use the formula sin(critical angle) = n2 / n1, where n1 is the refractive index of the medium the light is coming from (flint glass) and n2 is the refractive index of the medium the light is entering (crown glass cladding).
Plugging in the values, sin(critical angle) = 1.52 / 1.66. To find the critical angle itself, we take the inverse sine ([tex]sin^(^-^1^)[/tex]) of the resulting value: critical angle = [tex]sin^(^-^1^)(1.52 / 1.66)[/tex]. By calculating this value, we can determine the critical angle in degrees.
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A point charge of -3.00 μC is located in the center of a spherical cavity of radius 6.90 cm inside an insulating spherical charged solid. The charge density in the solid is 7.35 × 10−4 C/m3.
Calculate the magnitude of the electric field inside the solid at a distance of 9.50 cm from the center of the cavity.
The magnitude of the electric field inside the solid at a distance of 9.50 cm from the center of the cavity is 5.68 × 10⁴ N/C.
To calculate the electric field inside the solid at a given distance from the center of the cavity, we need to consider the contributions from both the point charge in the cavity and the charge density in the solid.
Let's break down the calculation step by step:
1. Electric field due to the point charge in the cavity:
The electric field at a point inside the solid due to the point charge in the cavity can be calculated using the formula:
E_point = k * |Q_point| / r²
where
E_point is the electric field due to the point charge,
k is the Coulomb's constant (8.99 × 10^9 N m²/C²),
|Q_point| is the magnitude of the point charge (-3.00 μC = -3.00 × 10⁻⁶C),
and r is the distance from the center of the cavity to the point inside the solid (9.50 cm = 0.095 m).
Substituting the values into the formula, we get:
E_point = (8.99 × 10⁹ N m²/C) * |-3.00 × 10⁶ C| / (0.095 m)²
E_point = 2.85 × 10⁷N/C
2. Electric field due to the charge density in the solid:
The electric field at a point inside the solid due to the charge density can be calculated using the formula:
E_density = (k * ρ * r) / (3ε0)
where
E_density is the electric field due to the charge density,
ρ is the charge density (7.35 × 10^(-4) C/m³),
r is the distance from the center of the cavity to the point inside the solid (9.50 cm = 0.095 m),
and ε0 is the permittivity of free space (8.85 × 10⁻¹² C²/N m²).
Substituting the values into the formula, we get:
E_density = [(8.99 × 10⁹N m²/C²) * (7.35 × 10⁻⁴C/m³) * (0.095 m)] / (3 * 8.85 × 10⁻¹² C²/N m²)
E_density = 1.06 × 10^8 N/C
3. Total electric field inside the solid:
To find the total electric field at the given point inside the solid, we need to sum the contributions from the point charge and the charge density. Since the charges have opposite signs, we subtract the magnitudes:
E_total = |E_point| - |E_density|
E_total = 2.85 × 10⁷N/C - 1.06 × 10⁸N/C
E_total = -7.78 × 10⁷N/C
However, the electric field is a vector quantity, and its direction is radial, pointing from the center of the cavity towards the point inside the solid.
Therefore, the magnitude of the electric field inside the solid at a distance of 9.50 cm from the center of the cavity is:
|E_total| = 7.78 × 10⁷N/C
The magnitude of the electric field inside the solid at a distance of 9.50 cm from the center of the cavity is 5.68 × 10
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Consider a Universe that has a flat curvature and no dark energy. What would the fate of such a Universe be? a. The Universe expands at a constant rate. b. The Universe expands forever but at an ever slowing rate. c. The Universe collapses in a Big Crunch. d. The Universe expands at an accelerating rate.
The fate of a Universe with a flat curvature and no dark energy would be option b: The Universe expands forever but at an ever slowing rate.
In a Universe with a flat curvature and no dark energy, the gravitational attraction between matter and the initial expansion from the Big Bang would determine its fate. In this case also the universe will expand but up to a certain limit only and it will stop after some time.
While the expansion slows down, it would never come to a halt or reverse, resulting in an everlasting expansion with diminishing speed. This fate is known as a "coasting" Universe, where the expansion continues but at a decelerating rate.
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What process do scientists think is causing the movement of Earth’s tectonic plates? Name one other place where this process is occurring naturally.
Answer:
convection currents in the earth's mantle, heat and pressure within the earth cause the hot magma to flow in convection currents. This causes the movement of the tectonic plates.
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1. Which of the following gas with a molecules has highest translational K.E. at NTP i) Chlorine ii) oxygen iii) hydrogen iv) all have equal amount at ntp .
The correct answer is iii) hydrogen. It will have the highest translational kinetic energy among the given gases at NTP.
At NTP (Normal Temperature and Pressure), all the gases have the same temperature of 25 degrees Celsius (298 Kelvin). According to the kinetic theory of gases, the average translational kinetic energy of gas molecules is directly proportional to the temperature.
The formula for translational kinetic energy is given by:
K.E. = (3/2) k T
Where:
K.E. is the translational kinetic energy
k is the Boltzmann constant (1.38 × 10^-23 J/K)
T is the temperature in Kelvin
Since the temperature is the same for all the gases at NTP, the gas with the highest translational kinetic energy will be the one with the lightest molecules. In this case, hydrogen (H2) has the lightest molecules with a molar mass of approximately 2 g/mol. Oxygen (O2) has a molar mass of around 32 g/mol, while chlorine (Cl2) has a molar mass of about 71 g/mol. Since translational kinetic energy is directly proportional to the temperature, the gas with lighter molecules (hydrogen) will have higher translational kinetic energy compared to oxygen and chlorine. option(iii)
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