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3 years ago

Trevor places a mass on a spring. When he releases it, the mass and spring move in simple harmonic motion.

Physics

2 answers:

777dan777 [17]3 years ago

8 0

The correct answer is D He could pull the mass down farther.

Amplitude, is the maximum displacement or distance moved by a point on a vibrating body or wave measured from its equilibrium position. So to increase the amplitude the spring should be pulled down further which can increase the amplitude.

Elis [28]3 years ago

8 0

The answer is D, He could pull the mass down further.

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A coil of wire with 60 turns lies on the plane of the page, and a uniform magnetic field points into the page with a strength of

baherus [9]

Answer:

0.85V

Explanation:

The emf is calculated by using the Lenz's Law

$\epsilon=-\frac{\Delta\Phi_B}{\Delta t}$

But for this case we have that the magnetic field is constant. Hence we have

$\epsilon=-\frac{B(A_{2}-A_{1})}{t_2-t_1}=-\frac{(1.10T)(0.270m^{2}-0.89m^{2})}{0.8s}\\\\\epsilon=0.85V$

where we have taken that the intial time is t1=0

I hope this is useful for you

regards

4 0

3 years ago

Two loudspeakers emit sound waves along the x-axis. A listener in front of both speakers hears a maximum sound intensity when sp

yKpoI14uk [10]

Answer

given,

difference between the two consecutive maximum

λ = 0.870 - 0.540

λ = 0.33 m

speed of sound = 340 m/s

b) frequency of the sound

v = f x λ

340 = f x 0.33

$f =\dfrac{340}{0.33}$

f = 1030.3 Hz

a) phase difference

the expression of phase difference is given by

$\phi = \dfrac{2\pi}{\lambda}\ \delta$

$\delta = \Delta x - \lambda$

$\delta = 0.540 - 0.33$

$\delta = 0.21\ m$

now,

$\phi = \dfrac{2\pi}{\lambda}\ \times 0.21$

$\phi = \dfrac{2\pi}{0.33}\ \times 0.21$

$\phi = 3.99 rad$

8 0

3 years ago

Sophia is planning on going down an 8-m water slide. Her weight is 50 N. She knows that she has gravitational potential energy (

RideAnS [48]

Answer:

Explanation:

graph would be a straight line from (0, 0) to (400, 8)

Plot points are

PE = mgh

50(0) = 0 J

50(2) = 100 J

50(4) = 200 J

50(6) = 300 J

50(8) = 400 J

4 0

3 years ago

The earth's radius is 6.37×106m; it rotates once every 24 hours. What is the earth's angular speed? What is the speed of a point

Zarrin [17]

Answer:

a) w = 7.27 * 10^-5 rad/s

b) v1 = 463.1 m/s

c) v1 = 440.433 m/s

Explanation:

Given:-

- The radius of the earth, R = 6.37 * 10 ^6 m

- The time period for 1 revolution T = 24 hrs

Find:

What is the earth's angular speed?

What is the speed of a point on the equator?

What is the speed of a point on the earth's surface located at 1/5 of the length of the arc between the equator and the pole, measured from equator?

Solution:

- The angular speed w of the earth can be related with the Time period T of the earth revolution by:

w = 2π / T

w = 2π / 24*3600

w = 7.27 * 10^-5 rad/s

- The speed of the point on the equator v1 can be determined from the linear and rotational motion kinematic relation.

v1 = R*w

v1 = (6.37 * 10 ^6)*(7.27 * 10^-5)

v1 = 463.1 m/s

- The angle θ subtended by a point on earth's surface 1/5 th between the equator and the pole wrt equator is.

π/2 ........... s

x ............ 1/5 s

x = π/2*5 = 18°

- The radius of the earth R' at point where θ = 18° from the equator is:

R' = R*cos(18)

R' = (6.37 * 10 ^6)*cos(18)

R' = 6058230.0088 m

- The speed of the point where θ = 18° from the equator v2 can be determined from the linear and rotational motion kinematic relation.

v2 = R'*w

v2 = (6058230.0088)*(7.27 * 10^-5)

v2 = 440.433 m/s

5 0

3 years ago

In the waves lab you change blank to see how it would affect wave speed

GaryK [48]

where is the question in this ????

7 0

3 years ago