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3 years ago

Which object has the most gravitional potential energy?

Physics

1 answer:

KengaRu [80]3 years ago

7 0

When an object is above the earth surface gravitational potential energy

i hope this helped :)

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How would I graph the data in the table on the graph? ********WORTH 50 POINTS********WILL MARK BRAINLIEST**********

Natali5045456 [20]

Answer:

Save the file and try to put stuff over the file. (kinda off like a picture)

3 0

3 years ago

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A uniform electric field is directed parallel to the +y axis. If a positive test charge begins at the origin and moves upward al

lukranit [14]

Answer: option 1 : the electric potential will decrease with an increase in y

Explanation: The electric potential (V) is related to distance (in this case y) by the formulae below

V = kq/y

Where k = 1/4πε0

Where V = electric potential,

k = electric constant = 9×10^9,

y = distance of potential relative to a reference point, ε0 = permittivity of free space

q = magnitude of electronic charge = 1.609×10^-19 c

From the formulae, we can see that q and k are constants, only potential (V) and distance (y) are variables.

We have that

V = k/y

We see the potential(V) is inversely proportional to distance (y).

This implies that an increase in distance results to a decreasing potential and a decrease in distance results to an increase in potential.

This fact makes option 1 the correct answer

3 0

3 years ago

The organelle responsible for breaking down food into energy is the

kompoz [17]

C.) Mitochondria is the answer...

7 0

3 years ago

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A 4-kg toy car with a speed of 5 m/s collides head-on with a stationary 1-kg car. After the collision, the cars are locked toget

mihalych1998 [28]

Kinetic energy lost in collision is 10 J.

<u>Explanation:</u>

Given,

Mass, $m_{1}$ = 4 kg

Speed, $v_{1}$ = 5 m/s

$m_{2}$ = 1 kg

$v_{2}$ = 0

Speed after collision = 4 m/s

Kinetic energy lost, K×E = ?

During collision, momentum is conserved.

Before collision, the kinetic energy is

$\frac{1}{2} m1 (v1)^2 + \frac{1}{2} m2(v2)^2$

By plugging in the values we get,

$KE = \frac{1}{2} * 4 * (5)^2 + \frac{1}{2} * 1 * (0)^2\\\\KE = \frac{1}{2} * 4 * 25 + 0\\\\$

K×E = 50 J

Therefore, kinetic energy before collision is 50 J

Kinetic energy after collision:

$KE = \frac{1}{2} (4 + 1) * (4)^2 + KE(lost)$

$KE = 40J + KE(lost)$

Since,

Initial Kinetic energy = Final kinetic energy

50 J = 40 J + K×E(lost)

K×E(lost) = 50 J - 40 J

K×E(lost) = 10 J

Therefore, kinetic energy lost in collision is 10 J.

4 0

4 years ago

A ray of light passes from air into a block of clear plastic. How does the angle of incidence in the air compare to the angle of

andre [41]

Answer:

The angle of incidence is greater than the angle of refraction

Explanation:

Refraction occurs when a light wave passes through the boundary between two mediums.

When a ray of light is refracted, it changes speed and direction, according to Snell's Law:

$n_1 sin \theta_1 = n_2 sin \theta_2$

where :

$n_1$ is the index of refraction of the 1st medium

$n_2$ is the index of refraction of the 2nd medium

$\theta_1$ is the angle of incidence (the angle between the incident ray and the normal to the boundary)

$\theta_2$ is the angle of refraction (the angle between the refracted ray and the normal to the boundary)

In this problem, we have a ray of light passing from air into clear plastic. We have:

$n_1=1.00$ (index of refraction of air)

$n_2=1.50$ approx. (index of refraction in clear plastic)

Snell's Law can be rewritten as

$sin \theta_2 =\frac{n_1}{n_2}sin \theta_1$

And since $n_2>n_1$ , we have

$\frac{n_1}{n_2}$

And so

$\theta_2$

Which means that

The angle of incidence is greater than the angle of refraction

6 0

3 years ago