Why does gravitational energy increase away from earth

A mass is attached to a vertical spring, which then goes into oscillation. At the high point of the oscillation, the spring is i

andrew-mc [135]

Answer:

0.34 sec

Explanation:

Low point of spring ( length of stretched spring ) = 5.8 cm

midpoint of spring = 5.8 / 2 = 2.9 cm

Determine the oscillation period

at equilibrum condition

Kx = Mg

g= 9.8 m/s^2

x = 2.9 * 10^-2 m

k / m = 9.8 / ( 2.9 * 10^-2 ) = 337.93

note : w = $\sqrt{k/m}$ = $\sqrt{337.93}$ = 18.38 rad/sec

Period of oscillation = $2\pi / w$

= 0.34 sec

8 0

3 years ago

A 3-kg block, attached to a spring, executes simple harmonic motion according to x= 2cos(50t) where x is in meters and t is in s

saw5 [17]

Explanation:

w=50=√k/m=√k/3

SO

(50²)*3 is 7500N/m

4 0

3 years ago

You are traveling in a car toward a hill at a speed of 36.4 mph. The car's horn emits sound waves of frequency 231 Hz, which mov

Marina CMI [18]

Answer:

<em>a. The frequency with which the waves strike the hill is 242.61 Hz</em>

<em>b. The frequency of the reflected sound wave is 254.23 Hz</em>

<em>c. The beat frequency produced by the direct and reflected sound is </em>

Explanation:

Part A

The car is the source of our sound, and the frequency of the sound wave it emits is given as 231 Hz. The speed of sound given can be used to determine the other frequencies, as expressed below;

$f_{1} = f[\frac{v_{s} }{v_{s} -v} ]$ ..............................1

where $f_{1}$ is the frequency of the wave as it strikes the hill;

f is the frequency of the produced by the horn of the car = 231 Hz;

$v_{s}$ is the speed of sound = 340 m/s;

v is the speed of the car = 36.4 mph

Converting the speed of the car from mph to m/s we have ;

hint (1 mile = 1609 m, 1 hr = 3600 secs)

v = $36.4 mph *\frac{1609 m}{1 mile} *\frac{1 hr}{3600 secs}$

v = 16.27 m/s

Substituting into equation 1 we have

$f_{1}$ = $231 Hz (\frac{340 m/s}{340 m/s - 16.27 m/s})$

$f_{1}$ = 242.61 Hz.

Therefore, the frequency which the wave strikes the hill is 242.61 Hz.

Part B

At this point, the hill is the stationary point while the driver is the observer moving towards the hill that is stationary. The frequency of the sound waves reflecting the driver can be obtained using equation 2;

$f_{2} = f_{1} [\frac{v_{s}+v }{v_{s} } ]$

where $f_{2}$ is the frequency of the reflected sound;

$f_{1}$ is the frequency which the wave strikes the hill = 242.61 Hz;

$v_{s}$ is the speed of sound = 340 m/s;

v is the speed of the car = 16.27 m/s.

Substituting our values into equation 1 we have;

$f_{2} = 242.61 Hz [\frac{340 m/s+16.27 m/s }{340 m/s } ]$

$f_{2}$ = 254.23 Hz.

Therefore, the frequency of the reflected sound is 254.23 Hz.

Part C

The beat frequency is the change in frequency between the frequency of the direct sound and the reflected sound. This can be obtained as follows;

Δf = $f_{2}$ - $f_{1}$

The parameters as specified in Part A and B;

Δf = 254.23 Hz - 242.61 Hz

Δf = 11.62 Hz

Therefore the beat frequency produced by the direct and reflected sound is 11.62 Hz

3 0

3 years ago

Velocity can best be described by which of the following statements?

pogonyaev

Answer: Velocity can best be described as, the speed in a given direction.

Explanation: To find the answer, we need to know more about the Velocity of a body.

<h3>What is Velocity of a body?</h3>

Velocity is the rate of change of displacement.
It's a vector quantity and is measured in m/s.
It can be positive, negative or zero.
A body is said to be in uniform motion, then its velocity remains constant.
Change in velocity can be a change in speed.
The magnitude of velocity is less than or equal to speed.

Thus, we can conclude that, the option C is best describing velocity.

Learn more about velocity here:

brainly.com/question/28108466

#SPJ4

8 0

2 years ago

Read 2 more answers

Velocity and Time

Murrr4er [49]

Answer:

h=112.35

Explanation:

becuase it had to

3 0

3 years ago