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polet [3.4K]
3 years ago
15

If a water wave vibrates up and down three times each second and the distance between wave crests is 2 meters, what are:

Physics
2 answers:
finlep [7]3 years ago
4 0

a). 3 Hz

b). 2 meters

c). 6 m/s

Helga [31]3 years ago
3 0

Answer:

im here for help

Explanation:

help

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Why might a scientist repeat an experiment if she did not make a mistake in the first one?
iren2701 [21]

Answer:

B

Explanation:

4 0
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A delivery drone uniformly slowed down from 45 m/s to 23 m/s in a straight line over a 30-minute duration within a 16-kilometer
Neporo4naja [7]
I think it is letter A
7 0
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the shock absorbers in a car act as a big spring with k= 21900 N/m. when a 92.5 kg person gets in, how far does the spring stret
r-ruslan [8.4K]

Answer: 0.04139m

Explanation:

First, we need to calculate the weight of the man which will be:

Weight = mass × acceleration due to gravity

Weight = mg

Weight = 92.5 × 9.8

Weight = 906.5N

Then, we calculate the force which will be:

F = kx

mg = kx

x = mg/k

x = 906.5/21900

x = 0.04139m.

The spring stretched for 0.04139m.

4 0
3 years ago
7. Explain how changes in temperature affect the particle motion of a substance. (3 points)
Morgarella [4.7K]

Answer:

When heat is added to a substance, the molecules and atoms vibrate faster. As atoms vibrate faster, the space between atoms increases. The motion and spacing of theparticles determines the state of matter of thesubstance. The end result of increased molecular motion is that the object expands and takes up more space.

6 0
3 years ago
An object is attached to a vertical spring and slowly lowered to its equilibrium position. This stretches the spring by an amoun
trasher [3.6K]

Answer:

2d

Explanation:

For any instance equivalent force acting on the body is

mg-kd= m\frac{d}{dt}\frac{dx}{dt}

Where

m is the mass of the object

k is the force constant of the spring

d is the extension in the spring

and

d/dt(dx/dt)=  is the acceleration of the object

solving the above equation we get

x= Asin\omega t +d

where

\omega= \sqrt{\frac{k}{m} } = \frac{2\pi}{T}

A is the amplitude of oscillation from the mean position.

k= spring constant , T= time period

Here  we are assuming  that at t=T/4

x= 0   since, no extension in the spring

then

A=- d

Hence

x=- d sin wt + d

now, x is maximum when sin wt=- 1

Therefore,

x(maximum)=2d

7 0
3 years ago
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