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just olya [345]
3 years ago
11

When a spring is compressed, the energy changes from kinetic to potential. Which best describes what is causing this change?

Physics
2 answers:
vredina [299]3 years ago
8 0

Answer:

gravitational energy

Explanation:

Lelu [443]3 years ago
3 0

Answer:

work

Explanation:

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Absorption occurs when <span>all of the energy from light waves is transferred to a medium! </span>
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The vibrations along a transverse wave move in a direction _________.
GrogVix [38]

Answer: perpendicular to it oscillations.

Explanation: A transverse wave is a wave whose oscillations is perpendicular to the direction of the wave.

By perpendicular, we mean that the wave is oscillating on the vertical axis (y) of a Cartesian plane and the vibration is along the horizontal axis (x) of the plane.

Examples of transverse waves includes wave in a string, water wave and light.

Let us take a wave in a string for example, you tie one end of a string to a fixed point and the other end is free with you holding it.

If you move the rope vertically ( that's up and down) you will notice a kind of wave traveling away from you ( horizontally) to the fixed point.

Since the oscillations is perpendicular to the direction of wave, it is a transverse wave

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3 years ago
A merry-go-round with a rotational inertia of 600 kg m2 and a radius of 3. 0 m is initially at rest. A 20 kg boy approaches the
Margaret [11]

Hi there!

\boxed{\omega = 0.38 rad/sec}

We can use the conservation of angular momentum to solve.

\large\boxed{L_i = L_f}

Recall the equation for angular momentum:

L = I\omega

We can begin by writing out the scenario as a conservation of angular momentum:

I_m\omega_m + I_b\omega_b = \omega_f(I_m + I_b)

I_m = moment of inertia of the merry-go-round (kgm²)

\omega_m = angular velocity of merry go round (rad/sec)

\omega_f = final angular velocity of COMBINED objects (rad/sec)

I_b = moment of inertia of boy (kgm²)

\omega_b= angular velocity of the boy (rad/sec)

The only value not explicitly given is the moment of inertia of the boy.

Since he stands along the edge of the merry go round:

I = MR^2

We are given that he jumps on the merry-go-round at a speed of 5 m/s. Use the following relation:

\omega = \frac{v}{r}

L_b = MR^2(\frac{v}{R}) = MRv

Plug in the given values:

L_b = (20)(3)(5) = 300 kgm^2/s

Now, we must solve for the boy's moment of inertia:

I = MR^2\\I = 20(3^2) = 180 kgm^2

Use the above equation for conservation of momentum:

600(0) + 300 = \omega_f(180 + 600)\\\\300 = 780\omega_f\\\\\omega = \boxed{0.38 rad/sec}

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2 years ago
Why do plane mirrors and convex mirrors form only virtual images?
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A golf ball is struck with a five iron on level ground. it lands 100.0 m away 4.60 s later. what was the magnitude and direction
finlep [7]

consider the motion in x-direction

v_{ox} = initial velocity in x-direction = ?

X = horizontal distance traveled = 100 m

a_{x} = acceleration along x-direction = 0 m/s²

t = time of travel = 4.60 sec

Using the equation

X = v_{ox} t + (0.5) a_{x} t²

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v_{ox} = 21.7 m/s


consider the motion along y-direction

v_{oy} = initial velocity in y-direction = ?

Y = vertical displacement  = 0 m

a_{y} = acceleration along x-direction = - 9.8 m/s²

t = time of travel = 4.60 sec

Using the equation

Y = v_{oy} t + (0.5) a_{y} t²

0 = v_{oy} (4.60) + (0.5) (- 9.8) (4.60)²

v_{oy} = 22.54 m/s

initial velocity is given as

v_{o} = sqrt((v_{ox})² + (v_{oy})²)

v_{o} = sqrt((21.7)² + (22.54)²) = 31.3 m/s

direction: θ = tan⁻¹(22.54/21.7) = 46.12 deg

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3 years ago
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