Answer:Newton’s law also states that larger bodies with heavier masses exert more gravitational pull, which is why those who walked on the much smaller moon experienced a sense of weightlessness, as it had a smaller gravitational pull. To help explain his theories of gravity and motion, Newton helped create a new, specialized form of mathematics.
Explanation:I'm very stupid so I don't think this is the right answer
Doug rubs a piece of fur on a hard rubber rod, giving the rod a negative charge. Which of the following statements best describes what happens in this process?A) Protons are removed from the rod.B) Electrons are added to the rod.C) The fur is also charged negatively.D) The fur is left neutral.
Answer:
B is correct. Electrons are added to the rod.
Explanation:
Because the fur lose electrons to the rode and because positively charged while the rod because negative
What is the car's average velocity (in m/s) in the interval between t = 1.0 s
to t = 1.5 s?
Answer:
1.4 m/s
Explanation:
From the question given above, we obtained the following data:
Initial Displacement (d1) = 0.9 m
Final Displacement (d2) = 1.6 m
Initial time (t1) = 1.5 secs
Final time (t2) = 2 secs
Velocity (v) =..?
The velocity of an object can be defined as the rate of change of the displacement of the object with time. Mathematically, it can be expressed as follow:
Velocity = change of displacement /time
v = Δd / Δt
Thus, with the above formula, we can obtain the velocity of the car as follow:
Initial Displacement (d1) = 0.9 m
Final Displacement (d2) = 1.6 m
Change in displacement (Δd) = d2 – d1 = 1.6 – 0.9
= 0.7 m
Initial time (t1) = 1.5 secs
Final time (t2) = 2 secs
Change in time (Δt) = t2 – t1
= 2 – 1.5
= 0.5 s
Velocity (v) =..?
v = Δd / Δt
v = 0.7/0.5
v = 1.4 m/s
Therefore, the velocity of the car is 1.4 m/s
Explain the Law of Conservation of Momentum
A snowboarder on a slope starts from rest and reaches a speed of 8.9 m/s after 9.5 s.
How far does the snowboarder travel in this time?
Explanation:
Given that,
Initial speed, u = 0
Final speed, v = 8.9 m/s
Time, t = 9.5 s
Let a is the acceleration of the snowboarder. It can be calculated as follows :
[tex]a=\dfrac{v-u}{t}\\\\a=\dfrac{8.9-0}{9.5}\\\\a=0.936\ m/s^2[/tex]
The distance traveled by the snowboarder in this time is calculated using third equation of motion as follows :
[tex]v^2=u^2=2as\\\\s=\dfrac{v^2}{2a}\\\\s=\dfrac{(8.9)^2}{2\times 0.936}\\\\s=42.31\ m[/tex]
So, 42.31 m of distance is traveled by snowboarder in this time.
If you dropped a rock 4.9 m in 1 s how far would it fall in 3 s? express your answer in meters to three significant figures
Answer:
14.7 m
Explanation:
It would fall 14.7 m in 3 s . You multiply 4.9 by 3.
The quantity represented by vi is a function of time (i.e., is not constant).A. TrueB. False
Answer:
The answer is false
Work done of frictional force from instant
Answer:
[tex]-100\ J[/tex]Step-by-step explanation:
1. Find acceleration:
[tex]m=2\ kg[/tex] [tex]F=-5\ N[/tex] [tex]a=\frac{F}{m}[/tex] (Newton's second law)[tex]a=\frac{-5}{2} =-2.5\ \frac{m}{s^{2}}[/tex]2. Find distance traveled:
[tex]v_0=10\ \frac{m}{s}[/tex] [tex]v=0[/tex][tex]a=-2.5\ \frac{m}{s^{2} }[/tex] [tex]v^2-v_0^2=2ad[/tex] (Kinematic equation)[tex]-100=-5d[/tex] [tex]d=20\ m[/tex]3. Find work done by friction:
[tex]W=Fd[/tex] (Work formula when angle between Force and Displacement vectors are 0°) [tex]W=-5\times20=-100\ J[/tex]The period of a simple pendulum, defined as the time necessary for one complete oscillation, is measured in time units and is given by T=2πlg where l is the length of the pendulum and g is the acceleration due to gravity, in units of length divided by time squared. Show that this equation is dimensionally consistent.
Answer:
Explanation:
T = 2π √l/g
The dimension for l = m
The dimension for g = m/s²
The dimension for 2π is nothing. Since it's a constant, it is dimensionless.
Now we proceed ahead. Since we are not using the 2π, for the sake of this proving, our formula will temporarily be written as
T = √l/g
Inputting the dimensions, we have
T = √(m) / (m/s²)
T = √(m * s²/m)
T = √s²
T = s
Since the unit of period itself is in s, we can adjudge that the equation is dimensionally constant.
Refer to the data. How many males are taller than 175 cm and approximately what percentage of the total is that?
Answer:
Males over 175 cm = 10 Percentage of total = 50%Explanation:
Males measuring;
161 - 165 = 2
166 - 170 = 1
171 - 175 = 7
176 - 180 = 1
181 - 185 = 6
186 - 190 = 3
Males taller than 175 cm = 1 + 6 + 3 = 10
Total number of Males = 2 + 1 + 7 + 1 + 6 + 3 = 20
Percentage of total over 175 cm
= 10 / 20
= 50%
What are some superheroes that resemble neurotransmitter functions. Dopamine: Acetylcholine: Endorphins: GABA: Glutamates: Norepinephrine: Serotonin:
Answer:
Serotonin
___________
An object has the acceleration graph shown in (Figure 1). Its velocity at t=0s is vx=2.0m/s. Draw the object's velocity graph for the values of t that are separated by the step Δt = 2s.
Answer:
Explanation:
We may notice that change in velocity can be obtained by calculating areas between acceleration lines and horizontal axis ("Time"). Mathematically, we know that:
[tex]v_{b}-v_{a} = \int\limits^{t_{b}}_{t_{a}} {a(t)} \, dt[/tex]
[tex]v_{b} = v_{a}+ \int\limits^{t_{b}}_{t_{a}} {a(t)} \, dt[/tex]
Where:
[tex]v_{a}[/tex], [tex]v_{b}[/tex] - Initial and final velocities, measured in meters per second.
[tex]t_{a}[/tex], [tex]t_{b}[/tex] - Initial and final times, measured in seconds.
[tex]a(t)[/tex] - Acceleration, measured in meters per square second.
Acceleration is the slope of velocity, as we know that each line is an horizontal one, then, velocity curves are lines with slopes different of zero. There are three region where velocities should be found:
Region I (t = 0 s to t = 4 s)
[tex]v_{4} = 2\,\frac{m}{s} +\int\limits^{4\,s}_{0\,s} {\left(-2\,\frac{m}{s^{2}} \right)} \, dt[/tex]
[tex]v_{4} = 2\,\frac{m}{s}+\left(-2\,\frac{m}{s^{2}} \right) \cdot (4\,s-0\,s)[/tex]
[tex]v_{4} = -6\,\frac{m}{s}[/tex]
Region II (t = 4 s to t = 6 s)
[tex]v_{6} = -6\,\frac{m}{s} +\int\limits^{6\,s}_{4\,s} {\left(1\,\frac{m}{s^{2}} \right)} \, dt[/tex]
[tex]v_{6} = -6\,\frac{m}{s}+\left(1\,\frac{m}{s^{2}} \right) \cdot (6\,s-4\,s)[/tex]
[tex]v_{6} = -4\,\frac{m}{s}[/tex]
Region III (t = 6 s to t = 10 s)
[tex]v_{10} = -4\,\frac{m}{s} +\int\limits^{10\,s}_{6\,s} {\left(2\,\frac{m}{s^{2}} \right)} \, dt[/tex]
[tex]v_{10} = -4\,\frac{m}{s}+\left(2\,\frac{m}{s^{2}} \right) \cdot (10\,s-6\,s)[/tex]
[tex]v_{10} = 4\,\frac{m}{s}[/tex]
Finally, we draw the object's velocity graph as follows. Graphic is attached below.
The velocity of a body under constant acceleration increases steadily with time
Please find attached the required velocity graph for values of t that are separated by Δt = 2s
The reasons the attached graph is correct are given as follows:
At t = 0 secondsThe initial velocity of the object at t = 0 is v = 2.0 m/s
The first point on the graph is (0, 2.0)
From t = 0 s, to t = 4 sThe acceleration from t = 0, to t = 4, a₁ = -2 m/s²
The velocity at t = 4 s, v₂ = 2.0 + (-2)×4 = -6
Therefore, the next point on the graph is (4, -6)
From t = 4 s to t = 6 sFrom t = 4 to t = 6, the acceleration, a₂ = 1 m/s²
Therefore, v₃ = -6 + 1 × 2 = -4
The third point on the velocity graph is (6, -4)
From t = 6 s to t = 10 sFrom t = 6 s to t = 10 s, we have, the acceleration, a₃ = 2 m/s²
The velocity, v₄ = -4 + 4 × 2 = 4
Therefore, the fourth point on the velocity graph is (10, 4)
With the above points, the velocity graph can be plotted using MS Excel
Learn more about plotting velocity time graph, here:
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If an initially neutral insulator is charged on its left end, would you get an electrical shock by touching its right end?
Answer:
No there won't be.
Explanation:
An insulator does not conduct electricity readily. An insulator that is charged at one end will have the charges remain at that end, since it does not allow the free flow of charges through it. If you touch the other end of the insulator that is void of electric charges, there would be no charge transfer between you and the insulator, and hence no electric shock.
Name a statistical quantity that can be used to indicate an accuracy level.
Answer:
100%
Explanation:
The accuracy level can be measured by the percent of error which is defined as the average accepted 100%.
Accuracy is related to the proximity of the measurement. It is called the true value of the accuracy where as the precision is referred to the ability at the same time to be reproduced in measurement .
The instrument should be proper and produce the reliable and accurate measurement tools.
Thus here the statistical quantity that is used by the to indicate the accuracy level is purely 100%.
We can reasonably model a 75 W incandescent light bulb as a sphere 6.0 cm in diameter. Typically, only about 5 % of the energy goes to visible light: the rest foes largely to nonvisible infrared radiation.(a) What is the visible light intensity (in W/m^2) at the surface of the bulb?(b) What are the amplitudes of the electric and magnetic fields at this surface, for a sinusoidal wave with this intensity?
Answer:
a
[tex]I = 6637 \ W/m^2[/tex]
b
[tex]E_{max} = 500 \ N/m[/tex]
And
[tex]B_{max} = 1.67*10^{-6} \ T[/tex]
Explanation:
From the question we are told that
The power is [tex]P = 75 \ W[/tex]
The diameter is [tex]d = 6.0 \ cm = 0.06 \ m [/tex]
Generally the radius is mathematically represented as
[tex]r = \frac{d}{2}[/tex]
=> [tex]r = \frac{ 0.06}{2}[/tex]
=> [tex]r = 0.03 \ m[/tex]
Generally the area of the sphere is mathematically evaluated as
[tex]A = 4 \pi r^2[/tex]
=> [tex]A = 4 * 3.142 * (0.03)^2[/tex]
=> [tex]A = 0.0113 \ m^2[/tex]
Generally the total Intensity of the incandescent light bulb is mathematically represented as
[tex]I= \frac{P}{A}[/tex]
=> [tex]I = \frac{75}{ 0.0113}[/tex]
=> [tex]I = 6637 \ W/m^2[/tex]
Given that 5% of the energy goes to visible light
Then the intensity that goes visible light is
[tex]I_v = 0.05 * 6637[/tex]
[tex]I_v = 332 \ W/m^2[/tex]
The amplitude of the electric field at the surface is mathematically represented as
[tex]E_{max} = \sqrt{\frac{2 * I_v}{\epsilon_o * c } }[/tex]
=> [tex]E_{max} = \sqrt{\frac{2 * 332}{ 8.85*10^{-12} * 3.0*10^8} }[/tex]
=> [tex]E_{max} = 500 \ N/m[/tex]
The amplitude of the magnetic field at the surface is mathematically represented as
[tex]B_{max} = \frac{E_{max}}{c}[/tex]
=> [tex]B_{max} = \frac{ 500}{3.0*10^8}[/tex]
=> [tex]B_{max} = 1.67*10^{-6} \ T[/tex]
A sound wave is observed to travel through a liquid with a speed of 1400 m/s. The specific gravity of the liquid is 1.8.
Determine the bulk modulus for this fluid.
Answer:
21
Explanation:
I think its 21 because its 21
Light of wavelength 600 nm illuminates a diffraction grating. The second order maximum is at an angle of 65 degrees. a) List your known variables. b) What is the spacing between slits in this grating? c) How many lines per millimeter does this grating have?
Answer:
A. Known variables include
Wavelength = 600nm
Theta= 65°
m= 2
B.
d= m x wavelength / sin theta
= 2 * 600*10^-9 /sin 65°
= 1.3*10^-6m
C.
N = 1/d
So N = 1/1.3*10^-6m
=76.9 lines per micro meter
Which element is most likely to be shiny? sulfur (S) boron (B) calcium (Ca) carbon (C)
Answer:
C.
Explanation:
The element that is most likely to be shiny is calcium. The correct option is 3.
Calcium (Ca) is the most likely to be sparkly of the elements shown. Calcium is a metallic element with a glossy or shining look when newly cut or polished.
It belongs to the alkaline earth metals group, which have metallic characteristics and a distinctive lustre.
Sulphur (S), boron (B), and carbon (C), on the other hand, are non-metallic elements with a dull look.
They come in a variety of forms, including powders, crystals, and amorphous solids, although they are not noted for their metallic lustre.
Thus, the answer is 3. calcium.
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Your question seems incomplete, the probable complete question is:
Which element is most likely to be shiny?
sulfur (S) boron (B) calcium (Ca) carbon (C)a 20 ft shipping container on a cargo ship has a mass of 24000 kg and a volume of 33.2m3. what is the density of the shipping container
Answer:
722.89
Explanation:
mass=24000kg
volume=33•2
density=?
now,
density=mass/volume
=24000/33•2
=722•89
density=722•89 kg/m^3
The acceleration of a particle is given by a(t)= -2.00 m/s^2 + (3 m/s^3)t. Required:a. Find the initial velocity vo such that the particle will have the same x-coordinate at t=4.00 s as it had at t=0. b. What will be the velocity at t=4.00 s ?
Answer:
Explanation:
a(t)= -2.00 m/s^2 + (3 m/s^3)t.
dv / dt = -2.00 m/s^2 + (3 m/s^3)t.
dv = (-2.00 m/s^2 + (3 m/s^3)t.)dt
v = - 2t + 3 t² / 2 + c , where c is a constant
for initial velocity t = 0
v0 = c
v = - 2t + 3 t² / 2 + v0
ds / dt = - 2t + 3 t² / 2 + v0
ds = (- 2t + 3 t² / 2 + v0)dt
s = - 2t²/2 + 3 t³/6 + vot + c₁
At t = 0
s = c₁
At t = 4
s = -16 + 32 + 4v0 + c₁
= 4v0 + c₁ + 16
Given
4v0 + c₁ + 16 = c₁
v0 = - 4 m /s
Putting this value in the equation of velocity
v = - 2t + 3 t² / 2 - 4
At t = 4
v = -8 + 24 - 4
= 12 m / s
What are the components of vector a
Answer:
Option (A)
Explanation:
From the picture attached,
Given : Vector A with magnitude 12 m.
To Find : Components of vector A.
There are two components of any vector.
1). Horizontal component
2). Vertical component
If vector A represents the velocity, horizontal component of a vector decides the horizontal motion and vertical component decides the vertical motion.
To find these components we draw a right triangle OBA as shown in the figure,
From ΔOBA,
Sin(37)° = [tex]\frac{\text{AB}}{\text{OA}}[/tex]
= [tex]\frac{{A_x}}{{A}}[/tex]
[tex]A_x=A\text{Sin}(37)[/tex]
= 12Sin(37)°
= 7.22 m
[tex]A_y=A\text{Cos}(37)[/tex]
= 12Cos(37)°
= 9.58 m
Therefore, Option (1) is the correct option.
A carpenter on the roof of a building accidentally drops herhammer. As the hammer falls it passes two windows of equal height,a) Is the increase in seed of the hammer as it drops past window 1greater than, less than or equal to the increase in speed as itdrops past window 2? b) Choose the best explanation from thefollowing:
I. The greater seed at window 2 results in a greater increase inspeed.
II. Constant acceleration means the hammer speeds up the sameamount for each window.
III. The hammer spends more time dropping past window 1.
Answer:
Explanation:
a) the speed increment of the hammer as it drops past the first window, is greater than that of the speed of the hammer as it drops past the second window. This can also be translated as saying that the hammer spent more time at the second window.
b) III
The best answer would be answer III, The hammer spends more time dropping past window 1, which I had already included in my explanation in (a) above.
A bowling ball of mass 3 kg is dropped from the top of a tall building. It safely lands on the ground 3.5 seconds later. Neglecting air friction, what is the height of the building in meters? (Give the answer without a unit and round it to the nearest whole number)
Answer:
The height of the building is 60 m.
Explanation:
Given;
mass of the mass of the ball, m = 3 kg
time of motion, t = 3.5 s
The velocity of the ball is given by;
v = u + gt
where;
u is the initial velocity of the ball = 0
v = 0 + 9.8 x 3.5
v = 34.3 m/s
When the ball hits the ground, energy is conserved;
mgh = ¹/₂mv²
gh = ¹/₂v²
h = (0.5 v²) / g
h = (0.5 x 34.3²) / (9.8)
h = 60.025 m
h = 60 m
Therefore, the height of the building is 60 m.
The height of the building is 60 m.
calculation of building height:The velocity of the ball should be provided by
v = u + gt
here,
u is the initial velocity of the ball = 0
v = 0 + 9.8 x 3.5
v = 34.3 m/s
Now
When the ball hits the ground, energy is conserved;
mgh = ¹/₂mv²
gh = ¹/₂v²
h = (0.5 v²) / g
h = (0.5 x 34.3²) / (9.8)
h = 60.025 m
h = 60 m
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Suppose we have Cl-, Na+, and Ca2+ ions in an aqueous solution (with dielectric constant κ = 80.4—we will discuss this later). Consider the situation where a sodium ion is between a chlorine ion and a calcium ion as shown:
(Cl-) (Na+) (Ca2+)
If the sodium ion is 1.50 nm from the chlorine ion and 3.00 nm from the calcium ion, find the electric force on the Na+ ion.
Answer:
a) 1.19 x 10^7 N/C
b) 2 x 10^-12 N
Explanation:
field due to Cl on (9.0xE¡)(1.6x10-19 C) (1.5Å—10-9m) -6.4x10 N/C field due to Ca+3 ion, 2(90x10 (1.6x10-19c) ((4.5-1.5)Å—10-9 m)' magnitude of net field at given point without dielectric E E+E 6.4x108 +3.2x108-9.6x10 N/C magnitude of net field at given point with di electric K 80.4 the force on sodium ion at this point, F-Edq= (1.19å—107 N/C)(1.6å—10-19C)= 2.0Å—10-12 N
What is the car's average velocity (in m/s) in the interval between t = 0.5 s
to t = 2 s?
Answer:
[tex]1.0\; \rm m \cdot s^{-1}[/tex].
Explanation:
Consider a time period of duration [tex]\Delta t[/tex]. Let change in the position of an object during that period of time be denoted as [tex]\Delta x[/tex]. The average velocity of that object during that period would be:
[tex]\displaystyle \bar{v} = \frac{\Delta x}{\Delta t}[/tex].
For the toy car in this question, the time interval has a duration of [tex]2.0 \; \rm s - 0.5\; \rm s = 1.5\; \rm s[/tex]. That is: [tex]\Delta t = 1.5\; \rm s[/tex]. (One decimal place, two significant figures.)
On the other hand, what would be the change in the position of this toy car during that [tex]1.5\; \rm s[/tex]?
Note, that from readings on the snapshot in the diagram:
The position of the toy car was [tex]0.1\; \rm m[/tex] at [tex]t = 0.5\; \rm s[/tex] (the beginning of this [tex]1.5[/tex]-second time period.)The position of the toy car was [tex]1.6\; \rm m[/tex] at [tex]t = 2.0\; \rm s[/tex] (the end of this [tex]1.5[/tex]-second time period.)Therefore, the change to the position of this toy car over that time period would be [tex]\Delta x = 1.6\; \rm m - 0.1\; \rm m = 1.5\; \rm m[/tex]. (One decimal place, two significant figures.)
The average velocity of this car over this period of time would thus be:
[tex]\displaystyle \bar{v} = \frac{\Delta x}{\Delta t} = \frac{1.5\; \rm m}{1.5\; \rm s} = 1.0\; \rm m \cdot s^{-1}[/tex]. (Two significant figures.)
A diet decreases a person's mass by 6 %. Exercise creates muscle and reduces fat, thus increasing the person's density by 1 %. Determine the percent change in the person's volume.
Answer:
-5%
Explanation:
We know that volume is mass/density
So to find percentage
Let first mass be = M1
Second be 0.94m( 1-6%)
Second density be=0.99
First density be d
So finlq volume v2= 0.94m/0.99p=
V2= (0.94/0.99)V1
V2= 0.95v1
So volume decrease will be 1-0.95= 0.05
= -5%
A system of 1470 particles, each of which is either an electron or a proton has a net charge of -5.456 x 10^-17 C. (a) How many electrons are in this system? (b) What is the mass of this system?
Answer:
a
[tex]N_e = 906 \ electrons[/tex]
b
[tex]M = 9.43 *10^{-25} \ kg[/tex]
Explanation:
From the question we are told that
The total number of particles is n = 1470 particles
The the total amount of charge on the particles is [tex]Q = - 5.456*10^{-17} \ C[/tex]
Generally this total number of particles can be mathematically represented as
[tex]n = N_p + N_e[/tex]
Where [tex]N_p \ and \ N_e[/tex] represent the number of proton and electron respectively
=> [tex]N_p = 1470 - N_e[/tex]
Also the total charge of these particles can be mathematically represented as
[tex]Q = (N_p - N_e ) e[/tex]
Here e is the charge on a single electron or proton with value
The negative sign is due to the fact the electrons are negative signed
[tex]e = 1.60 *10^{-19} \ C[/tex]
[tex]-5.456*10^{-17} = ( 1470 -2N_e )* 1.60 *10^{-19} [/tex]
[tex]N_e = 906[/tex]
Thus
[tex]N_p = 1470 - 906[/tex]
[tex]N_p = 564[/tex]
Generally the mass of the system is mathematically represented as
[tex]M = N_e * M_e + N_p * M_p[/tex]
Here [tex]M_e \ and \ M_p \ are \ mass \ of \ electron \ and \ proton \ with \ values \\ M_e = 9.1 *10^{-31} \ kg \ and \ M_p = 1.67 *10^{-27} \ kg \ respectively[/tex]
So
[tex]M = 906 * 9.1 *10^{-31} + 564 * 1.67 *10^{-27}[/tex]
[tex]M = 9.43 *10^{-25} \ kg[/tex]
A student throws a heavy red ball horizontally from a balcony of a tall building with an initial speed v0. At the same time, a second student drops a lighter blue ball from the same balcony. Neglecting air resistance, which statement is true?
Answer:
The correct statement must be: both balls hit the floor at the same time
Explanation:
This is a kinematics exercise. The ball thrown horizontally does not have vertical speed and the ball that is released does not have vertical speed, therefore both take the same time to reach the ground, if we neglect the air resistance
The correct statement must be: both balls hit the floor at the same time
The portion of a uniform violin string that vibrates is from the "nut" to the "bridge" at the end of the finger board, and has length LA and mass m. The string tension F can only be increased or decreased by tightening or loosening the tuning pegs above the nut. Say that a string is tuned to produce a note with fundamental frequency ft.
a) Then, to play a different note with fundamental frequency fp, the violinist uses her finger to push the string against the fingerboard, reducing the length of the vibrating part of the string to L2 (The string now vibrates from her finger to the bridge.)
i) When she makes this change, determine if each of the following quantities increase, decrease, or remain unchanged: (1) string tension, (2) mass density, (3) wave speed, and (4) wavelength. Explain each.
ii) Is fy greater or less than fx ? Explain.
iii) Find an equation for Lg in terms of all or some of the given parameters. Simplify.
b) Find an equation for F in terms of the given parameters. c) Calculate numerical values for L, and F if m= 2.00 g, LA = 60.0 cm, fa = 440 Hz (A note), fB = 494 Hz (B note). -Bridge
Answer:
A)i) 1. constant, 2. constant, 3. constant, 4. decrease
ii) frecuency increase
iii) L = n /2f √T/μ
B) L_b = 0.534 m
Explanation:
We can approximate the violin string as a system of a fixed string at its two ends, therefore we have a node at each end and a maximum in the central part for a fundamental vibration,
λ = 2L / n
where n is an integer
The wavelength and frequency are related
v = λ f
and the speed of the wave is given by
v = √T /μ
with these expressions we can analyze the questions
A)
i) In this case the woman decreases the length of the rope L = L₂
therefore the wavelength changes
λ₂ = 2 (L₂) / n
as L₂ <L₀ the wavelength is
λ₂ < λ₀
The tension of the string is given by the force of the plug as it has not moved, the tension must not change and the density of the string is a constant that does not depend on the length of the string, therefore the speed of the string wave in the string should not change.
ii) how we analyze if the speed of the wave does not change
v = λ f
as the wavelength decreases, the frequency must increase so that the speed remains constant
fy> fx
iii) It is asked to find the length of the chord
let's use the initial equations
λ = 2L / n
v = λ f
v = 2L / n f
v = √ T /μ
we substitute
2 L / n f = √ T /μ
L = n /2f √T/μ
this is the length the string should be for each resonance
b) in this part they ask to calculate the frequency
f = n / 2L √ T /μ
the linear density is
μ = m / L
μ = 2.00 10⁻³ / 60.0 10⁻²
μ = 3.33 10⁻³ kg / m
we assume that the length is adequate to produce a fundamental frequency in each case
f_{a} = 440Hz
λ = 2La / n
λ = 2 0.60 / 1
λ = 1.20 m
v = λ f
v = 1.20 440
v = 528 m / s
v² = T /μ
T = v² μ
T = 528² 3.33 10⁻³
T = 9.28 10² N
Let's find the length of the chord for fb
f_{b} = 494 hz
L_b = 1 /(2 494) √(9.28 10² / 3.33 10⁻³)
L_b = 0.534 m
A cyclist traveling in a straight line has the velocity of +5 m/s for 10 sec. She then accelerates at a rate of 0.5 m/s2 over another 10 sec. How far does the cyclist travel during those 20 sec?
Answer: 125
Explanation: use the equation d=vt to find the first distance, 50. Then use the equation
X=Vot + 1/2at to find the second distance. Then add both together.
Vo=5
V=5
T=10
A=5
A boat is pulled into the dock by a rope attached to the bow of the boat and passing through a pulley on the dock that is 1 m higher than the bow of the boat.
If the rope is pulled in at a rate of 1.4 m/s, how fast is the boat approaching the dock when it is 9 m from the dock?
Answer:
√82/9 m/s
Explanation:
Let's say x²+1²= y²(y length of rope)
Using Pythagoras
With repect to t diff.
So 2x dx / dt= 2y dy / dt
But dy/ dt = 1.4m/s
X = 9 so from Pythagoras y = √82
So 9dx/dt = √82
So the boat will approved de dock at √82/9 m/s