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cestrela7 [59]
3 years ago
14

How do very small objects behave?

Physics
2 answers:
nataly862011 [7]3 years ago
7 0
A. very small objects behave like like particles.
In-s [12.5K]3 years ago
7 0

Answer:

A) PARTICLES

Explanation:

VERY SMALL OBJECTS ARE BEHAVE LIKE A PARTICLES.

<h3>HOPE IT HELP.....❤❤</h3>
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A 5 kg ball is sitting on the floor without moving. What is the momentum of the ball?
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Answer:

The ball has no momentum

Explanation:

The given parameters are;

The mass of the ball = 5 kg

The velocity of the ball = 0 (The ball is sitting on the floor without moving)

The momentum of the ball = The mass of the ball × Velocity of the ball

Therefore, the momentum of the ball = 5 kg × 0 m/s = 0

The momentum of the ball is zero, the ball has no momentum.

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Radiation can carry thermal energy.
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Far out in space, a 100,000-kg rocket and a 200,000-kg rocket are docked at opposite ends of a motionless 90m-long tunnel. The t
postnew [5]

Answer:

a) X_{cm} = 60m

b) I = 5.4*10^8Kg*m^2

c) T= 4.5*10^6 N*m

d) a = 8.3*10^{-3}rad/s^2

e) W= 0.25 rad/s

Explanation:

a) we know that:

X_{cm} = \frac{m_1x_1+m_2x_2}{m1+m2}

where X_cm is the ubicaton of the center of mass, m_1 the mass of the first rocket, x_1 its distances with the rocket 1, m_2 the mass of the second rocket and x_2 its distance with the rocket 1. So, replacing values, we get:

X_{cm} = \frac{(100,000kg)(0)+(200,000kg)(90m)}{100,000kg+200,000kg}

X_{cm} = 60m

So, the center of mass is at 60m from the rocket 1.

b) we know that:

I = M_1R_1^2 +M_2R_2^2

where I is the moment of inertia, M_1 is the mass of the rocket 1, R_1 its distance from the center of mass, M_2 the mass of the second rocket and R_2 the distance between the rocket 2 and the center of mass. So, replacing values, we get:

I = (100,000kg)(60m)^2 +(200,000kg)(30m)^2

I = 5.4*10^8Kg*m^2

c) We know that:

T = Fr

where T is the net torque, F is the force and r is the distance between the rocket and the radius. Then:

FR_1+FR_2 = T

Replacing values, we get:

50,000N(60m)+50,000N(30m) = T

T= 4.5*10^6 N*m

d) We know that:

T = Ia

where T is the net torque, I the moment of inertia and a is the angular aceleration. So, replacing values, we get:

4.5*10^6Nm = 5.4*10^8Kg*m^2a

solving for a:

a = 8.3*10^{-3}rad/s^2

e) Finally, using:

W = at

where W is the angular velocity, a is the angular aceleration and t is the time.

Then, replacing values. we get:

W = (8.3*10^{-3}rad/s^2)(30s)

W = 0.25 rad/s

3 0
2 years ago
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