Answer:
The force of gravity is not the same as being on the earth. when your on the earth there no gravitational pull its all up to the air
Explanation:
No explanation
would it be m/s or kg?
Answer:
m.s
Explanation:
7. A particle of mass 3 kg is held in equilibrium by two light unextensible strings. One string is horizontal, as shown in Figure 7.30. The tension in the horizontal string is PN and the tension in the other string is N. Find a) the value of 0 b) the value of P.
The tension in the strings are 31.47 and 19.25 N respectively.
Mass of the block, m = 3 kg
From the figure, consider the vertical components,
T₁ sin45° + T₂ sin30° = mg
(T₁/√2) + (T₂/2) = 3 x 9.8 = 29.4
Also, consider the horizontal components,
T₁ cos45° = T₂ cos30°
T₁/√2 = T₂ x√3/2
T₁ = T₂ x √3/2 x √2
So,
T₁ = 0.612T₂
Applying in the first equation,
(T₁/√2) + (T₂/2) = 29.4
(0.612T₂/1.414) + 0.5T₂ = 29.4
0.434 T₂ + 0.5 T₂ = 29.4
0.934 T₂ = 29.4
Therefore, the tension,
T₂ = 29.4/0.934
T₂ = 31.47 N
So, the tension,
T₁ = 0.612 T₂
T₁ = 0.612 x 31.47
T₁ = 19.25 N
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Blocks A (mass 5.00 kg) and B (mass 6.50 kg) move on a frictionless, horizontal surface. Initially, block B is at rest and block A is moving toward it at 3.00 m/s. The blocks are equipped with ideal spring bumpers. The collision is head-on, so all motion before and after the collision is along a straight line.
(a) Find the maximum energy stored in the spring bumpers and the velocity of each block at that time.
(b) Find the velocity of each block after they have moved apart.
Answer: i believe its B Find the velocity of each block after they have moved apart sorry
Explanation: have a nice day buddy
A car is moving around a circular track of radius 25m with a speed of 30 m/s. What is the centripetal acceleration of the car? *
A 72 m/s2
B 36 m/s2
C 30 m/s2
D 6 m/s2
Answer:
I don't know how to do it the subject
The centripetal acceleration of the car when there is the radius and the speed is given so it should be considered as the option b. 36 [tex]m/s^2[/tex].
Calculation of the centripetal acceleration of the car:Since
The radius is 25 cm and the speed should be 30 m/s
Now the following formula should be used
Acceleration = speed^2/ radius
= 30^2 / 25
= 900 / 25
= 36 [tex]m/s^2[/tex]
Therefore, The centripetal acceleration of the car should be considered as the option b. 36 [tex]m/s^2[/tex].
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Light of wavelength 656 nm and 410 nm emitted from a hot gas of hydrogen atoms strikes a grating with 5300 lines per centimeter. a) Determine the angular deflection of both wavelengths in the 1st and 2nd order.
Answer:
[tex]20.32^{\circ}[/tex] and [tex]44.08^{\circ}[/tex]
[tex]12.56^{\circ}[/tex] and [tex]25.77^{\circ}[/tex]
Explanation:
[tex]\lambda[/tex] = Wavelength
[tex]\theta[/tex] = Angle
m = Order
Distance between grating is given by
[tex]d=\dfrac{1}{5300}\\\Rightarrow d=0.0001886\ \text{cm}[/tex]
[tex]\lambda=656\ \text{nm}[/tex]
We have the relation
[tex]d\sin\theta=m\lambda\\\Rightarrow \theta=\sin^{-1}\dfrac{m\lambda}{d}[/tex]
m = 1
[tex]\theta=\sin^{-1}\dfrac{1\times 656\times 10^{-9}}{0.0001886\times 10^{-2}}\\\Rightarrow \theta=20.35^{\circ}[/tex]
m = 2
[tex]\theta=\sin^{-1}\dfrac{2\times 656\times 10^{-9}}{0.0001886\times 10^{-2}}\\\Rightarrow \theta=44.08^{\circ}[/tex]
The first and second order angular deflection is [tex]20.32^{\circ}[/tex] and [tex]44.08^{\circ}[/tex]
[tex]\lambda=410\ \text{nm}[/tex]
m = 1
[tex]\theta=\sin^{-1}\dfrac{1\times 410\times 10^{-9}}{0.0001886\times 10^{-2}}\\\Rightarrow \theta=12.56^{\circ}[/tex]
m = 2
[tex]\theta=\sin^{-1}\dfrac{2\times 410\times 10^{-9}}{0.0001886\times 10^{-2}}\\\Rightarrow \theta=25.77^{\circ}[/tex]
The first and second order angular deflection is [tex]12.56^{\circ}[/tex] and [tex]25.77^{\circ}[/tex].
5) Which statement about leaders is true?
A) Leaders always make the right decisions.
B) A leader keeps the team focused on achieving its goals.
C) A leader's opinion counts more than the opinions of the other team members.
D) all of the above.
Answer: D) All of the above
Explanation:
Which factor affects the color of the star?
Luminosity
Temperature
Apparent Magnitude
None of the aobve
Answer:
temperature
Explanation:
if you look at a hertzsprung-russel diagram. you can understand how I got that answer
Make a poem about waves with 12 Lines and 3 Stanzas.
In a ocean full of storms
A new wave was born
Deep into that darkness flooding
Suddenly, I heard some pummeling
By the grave I saw the winds
And the sun just shined
Answer:
a friendly face that comes with waves,
the waves of all the memorial days,
and with these days we smile with pride,
as for the waves we used to ride,
given up the day has passed,
how it went away like an hour glass,
as if we knew the world was right,
just like the waves, oh so bright,
the time has come the days have passed,
the waves ashore the waves alast,
as if the friendly face was right,
the waves that rode, oh goodnight.
If an athlete runs the triathlon of 10 km in 2 hours, what is her average speed in kilometers per hour?
Answer: 5 km per hour
Explanation:
if in 10 km there is 2 hours, then 10 divided by 2 is 5.
Suppose a diode consists of a cylindrical cathode with a radius of 6.200×10−2 cm , mounted coaxially within a cylindrical anode with a radius of 0.5580 cm . The potential difference between the anode and cathode is 400 V . An electron leaves the surface of the cathode with zero initial speed (vinitial=0). Find its speed vfinal when it strikes the anode.
Answer:
The final speed will be "[tex]1.185\times 10^7 \ m/sec[/tex]".
Explanation:
The given values are:
Potential difference,
Δv = 400 v
Radius,
r = 0.5580 cm
As we know,
⇒ [tex]W=e \Delta v[/tex]
and,
⇒ [tex]\frac{1}{2}mv^2=e \Delta v[/tex]
then,
⇒ [tex]v=\sqrt{\frac{2e \Delta v}{m} }[/tex]
On substituting the values, we get
⇒ [tex]=\sqrt{\frac{2\times 1.6\times 10^{-19}\times 400}{9.11\times 10^{-31}} }[/tex]
⇒ [tex]=\sqrt{\frac{1.6\times 10^{-19}\times 800}{9.11\times 10^{-31}}}[/tex]
⇒ [tex]=1.185\times 10^7 \ m/sec[/tex]
the atom of an element x has 21protrons and 23neutrons. What is the
(a) Electron number
(b) Mass number
(c) Neutron number
A 4.0 kg block is moving at 5.0 m/s along a horizontal frictionless surface toward and ideal spring that is attached to a wall , After the block collides with the spring, the spring is compressed a maximum distance of 0.68m . what is the speed of the block when the spring is compressed to only one-half of the maximum distance?
A 4.0 kg block is moving at 5.0 m/s along a horizontal frictionless surface toward an ideal spring that is attached to a wall, the maximum speed of the block when the spring is compressed to one-half of the maximum distance is 4.33 m/s
From the conservation of energy; the kinetic energy of the mass is equal to the work done on the spring.
i.e.
[tex]\mathbf{\dfrac{1}{2} mv^2 = \dfrac{1}{2}kx^2_{max}}[/tex]
Given that:
the mass of the block = 4.0 kg the speed at which it is moving = 5.0 m/scompression of the spring = 0.68 m∴
From the equation above, multiplying both sides with 2, we have:
[tex]\mathbf{mv^2 =kx^2_{max}}[/tex]
Making (k) the subject of the formula;
[tex]\mathbf{k = \dfrac{mv^2}{x^2_{max}}}[/tex]
[tex]\mathbf{k = \dfrac{4 \times 5^2}{0.68^2}}[/tex]
k = 216.26 N/m
However, when compressed to one-half of the maximum distance; the speed is computed as follows:
x = 0.68/2 = 0.34 m
∴
[tex]\mathbf{\dfrac{1}{2}mv_o^2 - \dfrac{1}{2}mv^2 = \dfrac{1}{2}kx^2}[/tex]
[tex]\mathbf{m(v_o^2 -v^2) =kx^2}[/tex]
[tex]\mathbf{(v_o^2 -v^2) =\dfrac{kx^2}{m}}[/tex]
[tex]\mathbf{(5^2 -v^2) =\dfrac{216.26 \times 0.34^2}{4.0}}[/tex]
25 - v² = 6.25
25 -6.25 = v²
v² = 18.75
[tex]\mathbf{ v= \sqrt{18.75 }}[/tex]
v = 4.33 m/s
Therefore, we can conclude that the speed of the block when the spring is compressed to only one-half of the maximum distance is 4.33 m/s
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A spiral staircase winds up to the top of a tower in an old castle. To measure the height of the tower, a rope is attached to the top of the tower and hung down the center of the staircase. However, nothing is available with which to measure the length of the rope. Therefore, at the bottom of the rope a small object is attached so as to form a simple pendulum that just clears the floor. The period of the pendulum is measured to be 6.82 s. What is the
Answer:
The answer is "[tex]11.55780\ m[/tex]"
Explanation:
Using formula:
[tex]= 2 \pi f= \frac{2\pi}{T} =\sqrt{\frac{g}{L}}[/tex]
L = length of pendulum.
[tex]= T =2 \pi \sqrt{\frac{L}{g}}[/tex]
Calculate the value for L:
[tex]L= g (\frac{T}{2 \pi})^2 \\\\[/tex]
[tex]= (9.80 \ \frac{m}{s^2}) (\frac{6.82 \ s}{2 \pi})^2\\\\= (9.80 \ \frac{m}{s^2}) (\frac{46.5124 \ s^2}{4 \times \pi^2})\\\\= (9.80 \ \frac{m}{s^2}) (\frac{46.5124\ s^2}{4 \times 9.8596 })\\\\= (9.80 \ \frac{m}{s^2}) (\frac{46.5124 \ s^2}{ 39.4384 })\\\\= \frac{455.82152}{39.4384} \ m\\\\=11.55780\ m[/tex]
The height of the tower = 11.55780 m
A 1.65-m-long wire having a mass of 0.100 kg is fixed at both ends. The tension in the wire is maintained at 16.0 N. (a) What are the frequencies of the first three allowed modes of vibration
Answer:
Explanation:
mass per unit length ρ = .100 / 1.65 = .0606 . kg /m
length of wire L = 1.65 m
For fundamental frequency , the expression is as follows
n = [tex]\frac{1}{2L} \sqrt{\frac{T}{m} }[/tex]
L = 1.65 , T = 16 n and m = .0606
n = [tex]\frac{1}{2\times 1.65} \sqrt{\frac{16}{.0606} }[/tex]
= 4.9 /s .
This is fundamental frequency .
other mode of vibration ( first three ) will be as follows
4.9 x 2 = 9.8 /s ,
4.9 x 3 = 14.7 /s .
If you live in Melbourne, Australia, the local magnetic field has a strength of about 4x10-5 T. The magnetic field vector is directed northward, making an angle of 30 deg above the horizontal. An electron in Melbourne is moving parallel to the ground, in the west direction, at a speed of 9x105 m/s. What are the magnitude and direction of the magnetic force on the electron
Answer:
[tex]5.76\times 10^{-18}\ \text{N}[/tex] perpendicular to the velocity and magnetic field
Explanation:
B = Magnetic field = [tex]4\times 10^{-5}\ \text{T}[/tex]
[tex]\theta[/tex] = Angle the magnetic field makes with the horizontal = [tex]30^{\circ}[/tex]
v = Velocity of electron = [tex]9\times 10^5\ \text{m/s}[/tex]
q = Charge of electron = [tex]1.6\times 10^{-19}\ \text{C}[/tex]
Magnetic force is given by
[tex]F=qvB\sin\theta\\\Rightarrow F=1.6\times 10^{-19}\times 9\times 10^5\times 4\times 10^{-5}\sin30^{\circ}\\\Rightarrow F=2.88\times 10^{-18}\ \text{N}[/tex]
The magnitude of the magnetic force is [tex]2.88\times 10^{-18}\ \text{N}[/tex] and the direction is perpendicular to the velocity and magnetic field.
What is wrong with the following momentum value: 25 kg*m/s
Answer:
25N/s or 25kg*m/s^2
Explanation:
It is written wrong because the unit of momentum is kgm/s^2 or N/s
Pls help ASAP
Imagine that Maritans launch a rocket toward the Earth at a great speed. While the
rocket is traveling toward us, it will appear
than it actually is.
O more blue
darker
larger
more red
Answer:
The rocket will appear larger than it actually is
Look at the diagram showing the different wavelengths in sunlight.
A diagram showing the human eye and visible light. Visible light is broken down by color with wavelength in nanometers. Red is 700, orange is 600, yellow is 580, green is 550, blue is 475, indigo is 450, violet is 400.
Which has a wavelength of 350 nanometers?
red light
violet light
infrared light
ultraviolet light
Answer:
ultraviolet light
plz mark me as brainliest.
Answer:
Ultra violet
Explanation:
True or False
Microscopic organisms grow on the rocks near a volcano if the rocks are cooled to 120 degrees Celsius or less.
Explanation:
it's true
Kiss me if I'm wrong. But dinosaurs still exist right?
Answer:
True
Definetely true
A car weighing 1,500kg possesses 20 000 units of momentum. What would be the car's new momentum if it's velocity was tripled
Answer:
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Explanation:
6. The rate at which velocity changes is called
O speed
O direction
O acceleration
O displacement
A fisherman notices that his boat is moving up and down periodically without any horizontal motion, owing to waves on the surface of the water. It takes a time of 2.60 s for the boat to travel from its highest point to its lowest, a total distance of 0.630 m . The fisherman sees that the wave crests are spaced a horizontal distance of 5.70 m apart.
Required:
a. How fast are the waves traveling?
b. What is the amplitude of each wave?
c. If the total vertical distance traveled by the boat were 0.30 m but the other data remained the same, how would the answers to parts (a) and (b) be affected?
Answer:
a) v = 1.1 m/s
b) A = 0.315 m
c) v = 1.1 m/s A= 0.15 m
Explanation:
a)
In any travelling wave, there exists a fixed relationship between the propagation speed, the wavelength and the frequency, as follows:[tex]v = \lambda * f (1)[/tex]
If the wave crests are spaced a horizontal distance of 5.7 m apart, this means that the wavelength of the wave is just the same, i.e., 5.70 m.Regarding the frequency, we know that the frequency is just the inverse of the period, i.e., the time needed to complete one oscillation.If it takes a time of 2.60 s to go from the highest point to the lowest, the time needed to complete an oscillation (the period T) will be just double of this time:⇒ T = 2.60 s * 2 = 5.20 s (2)Since we have now T, we can find the frequency f as follows:[tex]f = \frac{1}{T} = \frac{1}{5.20s} = 0.19 Hz (3)[/tex]
Replacing f and λ in (1) we get:[tex]v = \lambda * f = 5.70 m * 0.19 Hz = 1.10 m/s (4)[/tex]
b)
The amplitude of the wave is just the amount that the water aparts from its equilibrium level, which is just the half of the distance between its highest point and the lowest one, as follows:[tex]A = \frac{0.630m}{2} = 0.315 m (5)[/tex]
c)
Part a) will not be affected by the new amplitude, because we have showed that the speed is independent of the amplitude, so v can be written as follows:v = 1.10 m/s (6)
Part b) will change , due to the amplitude changes. If the total vertical distance traveled by the boat is 0.30 m, by the same token as explained in b), the new amplitude will be just half of this, as follows:[tex]A = \frac{0.30m}{2} = 0.15 m (7)[/tex]
which particle have a mass of 1 u
Answer:
Explanation:
proton
A student using a stopwatch finds that the time for 10 complete orbits of a ball on the end of a string is 25 seconds. The period of the orbiting ball is
Answer:
T = 2.5 s
Explanation:
Given that,
Number of complete orbits = 10
Time, t = 25 seconds
We need to find the period of the orbiting ball. Let it is T. We know that number of oscillations per unit time is called frequency and the reciprocal of frequency is called period of the ball.
So,
[tex]T=\dfrac{t}{n}\\\\T=\dfrac{25}{10}\\\\T=2.5\ s[/tex]
So, the period of the orbiting ball is equal to 2.5 seconds.
Which particle needs to be added to this equation to show that the total numbers of neutrons and protons are not changed by the reaction? MARKLING BRAINLIEST 70 points must be correct!
Answer:
C.
Explanation:
Answer:A
Explanation:ap3x
A 5.0-kg box is pulled by a horizontal force F applied to the top of the box. When the box meets a low doorstep, it begins to rotate around it. The box is 60 cmcm wide and 70 cmcm high. What minimum magnitude of the force F is needed to cause this movement
Answer:
the required minimum magnitude of the force F is 21 N
Explanation:
Given the data in the question,
m = 5 kg
width = 60 cm
height = 80 cm
Let force is F represent in the image below,
so when the block about to rotate normal shifted to edge of cube
mg(w/2) = Fh
F = mg(w/2) / h
we know that g = 9.8 m/s²
we substitute
F = (5 × 9.8 ( 60/2)) / 70
F = (5 × 9.8 × 30 ) / 70
F = 1470 / 70
F = 21 N
Therefore, the required minimum magnitude of the force F is 21 N
The minimum magnitude of the force needed to cause the movement is 34.3 N.
Given:
Mass of the box (m) = 5.0 kg
Width of the box (w) = 60 cm = 0.6 m
Height of the box (h) = 70 cm = 0.7 m
The torque acting is:
τ_weight = (mg) × (h÷2)
The total torque is zero. Therefore, the force is:
F × (h÷2) + (mg) × (h÷2) = 0
F = -(mg) × (h÷2) ÷ (h÷2)
F = -(5.0 × 9.8 ) × (0.7 m ÷ 2) / (0.7 m ÷ 2)
F = -34.3 N
|F| = 34.3 N
Hence, the minimum magnitude of the force F needed to cause the movement is 34.3 N.
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The Brazilian rain forest is an area with significant biodiversity. As the rain forest is replaced with agricultural land, it is reasonable to predict a reduction in -
Answers-
A: consumption of solar energy.
B: sustainability over time.
C: precipitation levels.
D: average daily temperature.
Answer:
Bb
Explanation:
The magnitude of the force can be determined as?
Answer:
the mass of the object multiplied by the acceleration of the object
Explanation:
N2L states that F = ma (force equals mass times acceleration).
A copper wire of resistivity 2.6 × 10-8 Ω m, has a cross sectional area of 35 × 10-4 cm2
. Calculate
the length of this wire required to make a 10 Ω coil.
Answer:
the length of the wire is 134.62 m.
Explanation:
Given;
resistivity of the copper wire, ρ = 2.6 x 10⁻⁸ Ωm
cross-sectional area of the wire, A = 35 x 10⁻⁴ cm² = ( 35 x 10⁻⁴) x 10⁻⁴ m²
resistance of the wire, R = 10Ω
The length of the wire is calculated as follows;
[tex]R = \frac{\rho L}{A} \\\\L = \frac{RA}{\rho} \\\\L= \frac{10 \times (35\times 10^{-4}) \times 10^{-4}}{2.6 \times 10^{-8}} \\\\L = 134.62 \ m[/tex]
Therefore, the length of the wire is 134.62 m.
Why is it harder to breathe on a
mountain?
A. The air pressure is so high the lungs can't expand.
B. The air is denser and oxygen can't flow easily into the
lungs.
C. The denser oxygen molecules sink below the
surrounding air.
D. The air is less dense so there are fewer oxygen
molecules.