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Andreas93 [3]
3 years ago
12

Define solar system.Mention 3 types of energy.​

Physics
2 answers:
sveticcg [70]3 years ago
4 0

Answer:

The Solar System is the Sun and all the objects that orbit around it

The Sun is orbited by planets, asteroids, comets and other things. ...

It has strong gravity

There are many types of energies

three types are

kinetic energy

potential energy

gravitational energy

Gnom [1K]3 years ago
3 0

Answer:

solar panels: monocrystalline, polycrystalline, and thin-film. Each type has its own unique advantages and disadvantages, and the solar panel type best suited for your installation will depend on factors specific to your own property and desired system characteristics.

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I think D am not pretty show

8 0
3 years ago
A Ping-Pong ball moving East at a speed of 4 m/s collides with a stationary bowling ball. The Ping-Pong ball bounces back to the
In-s [12.5K]

Answer:

They experience the same magnitude impulse

Explanation:

We have a ping-pong ball colliding with a stationary bowling ball. According to the law of conservation of momentum, we have that the total momentum before and after the collision must be conserved:

p_i = p_f\\p_p + p_b = p'_p+p'_b

where

p_p is the initial momentum of the ping-poll ball

p_b is the initial momentum of the bowling ball (which is zero, since the ball is stationary)

p'_p is the final momentum of the ping-poll ball

p'_f is the final momentum of the bowling ball

We can re-arrange the equation as follows

p_p - p'_p = p_b'-p_b

or

-\Delta p_p = \Delta p_b

which means

|\Delta p_p | = |\Delta p_b| (1)

so the magnitude of the change in momentum of the ping-pong ball is equal to the magnitude of the change in momentum of the bowling ball.

However, we also know that the magnitude of the impulse on an object is equal to the change of momentum of the object:

I=\Delta p (2)

Therefore, (1)+(2) tells us that the ping-pong ball and the bowling ball experiences the same magnitude impulse:

|I_p| = |I_b|

4 0
4 years ago
A hollow sphere of radius 0.200 m, with rotational inertia I = 0.0484 kg·m2 about a line through its center of mass, rolls witho
d1i1m1o1n [39]

Answer:

Part a)

KE_r = 8 J

Part b)

v = 3.64 m/s

Part c)

KE_f = 12.7 J

Part d)

v = 2.9 m/s

Explanation:

As we know that moment of inertia of hollow sphere is given as

I = \frac{2}{3}mR^2

here we know that

I = 0.0484 kg m^2

R = 0.200 m

now we have

0.0484 = \frac{2}{3}m(0.200)^2

m = 1.815 kg

now we know that total Kinetic energy is given as

KE = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2

KE = \frac{1}{2}mv^2 + \frac{1}{2}I(\frac{v}{R})^2

20 = \frac{1}{2}(1.815)v^2 + \frac{1}{2}(0.0484)(\frac{v}{0.200})^2

20 = 1.5125 v^2

v = 3.64 m/s

Part a)

Now initial rotational kinetic energy is given as

KE_r = \frac{1}{2}I(\frac{v}{R})^2

KE_r = \frac{1}{2}(0.0484)(\frac{3.64}{0.200})^2

KE_r = 8 J

Part b)

speed of the sphere is given as

v = 3.64 m/s

Part c)

By energy conservation of the rolling sphere we can say

mgh = (KE_i) - KE_f

1.815(9.8)(0.900sin27.1) = 20- KE_f

7.30 = 20 - KE_f

KE_f = 12.7 J

Part d)

Now we know that

\frac{1}{2}mv^2 + \frac{1}{2}I(\frac{v}{r})^2 = 12.7

\frac{1}{2}(1.815) v^2 + \frac{1}{2}(0.0484)(\frac{v}{0.200})^2 = 12.7

1.5125 v^2 = 12.7

v = 2.9 m/s

8 0
3 years ago
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