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OLga [1]
3 years ago
13

An object has 22.4 kg•m/s of

Physics
2 answers:
Bad White [126]3 years ago
5 0

Answer:

22.4\:\mathrm{kg\cdot m/s}

Explanation:

Momentum is given by p=mv.

Let p_1 be the initial momentum of the object:

p_1=m\cdot v=22.4

If we doubled the mass, the new mass would be 2\cdot m, and if we halved the velocity, the new velocity would be \frac{1}{2}\cdot v.

Therefore, the final momentum would be:

p_f=2\cdot m\cdot \frac{1}{2}\cdot v=m\cdot v=\fbox{$22.4\:\mathrm{kg\cdot m/s}$}..

SSSSS [86.1K]3 years ago
4 0

Answer:

the answer is 22.4 kg*m/s

Explanation:

Such a bizarre unit!

Acellus is cool. I receive and share knowledge almost every day.

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6 0
3 years ago
1.when viable light shines on a black object it get warm because.
Lera25 [3.4K]

Answer:

1. Black absorbs heat and does not bounce the light off. 2. White bounces the light off reducing heat

4 0
3 years ago
Read 2 more answers
What is resistivity?​
Rom4ik [11]

Answer:

Resistivity, electrical resistance of a conductor of unit cross-sectional area and unit length.

Explanation:

A characteristic property of each material, resistivity is useful in comparing various materials on the basis of their ability to conduct electric currents. High resistivity designates poor conductors.

3 0
3 years ago
A ball of mass 0.120 kg is dropped from rest from a height of 1.25 m. It rebounds from the floor to reach a height of 0.600 m. W
Karo-lina-s [1.5K]
Before we find impulse, we need to find the initial and final momentum of the ball.

To find the momentum of the ball before it hit the floor, we need to figure out its final velocity using kinematics.

Values we know:
acceleration(a) - 9.81m/s^2 [down]
initial velocity(vi) - 0m/s
distance(d) - 1.25m [down]

This equation can be used to find final velocity:

Vf^2 = Vi^2 + 2ad

Vf^2 = (0)^2 + (2)(-9.81)(-1.25)

Vf^2 = 24.525

Vf = 4.95m/s [down]

Now we need to find the velocity the ball leaves the floor at using the same kinematics concept.

What we know:
a = 9.81m/s^2 [down]
d = 0.600m [up]
vf = 0m/s

Vf^2 = Vi^2 + 2ad

0^2 = Vi^2 + 2(-9.81)(0.6)

0 = Vi^2 + -11.772

Vi^2 = 11.772

Vi = 3.43m/s [up]

Now to find impulse given to the ball by the floor we find the change in momentum.

Impulse = Momentum final - momentum initial

Impulse = (0.120)(3.43) - (0.120)(-4.95)

Impulse = 1.01kgm/s [up]
8 0
3 years ago
Read 2 more answers
A 5.30 g bullet moving at 963 m/s strikes a 610 g wooden block at rest on a frictionless surface. The bullet emerges, traveling
Tems11 [23]

Answer:

a) V_{wf} = 4.67m/s

b) V = 8.29 m/s

Explanation:

Givens:

The bullet is 5.30g moving at 963m/s and its speed reduced to 426m/s. The wooden block is 610g.

a) From conservation of linear momentum

Pi = Pf

m_{b}V{b_{i} }  + V_{wi}  = m_{w} V_{wf} + m_{b}V_{bf}

where m_{b},V{b_{i} are the mass and the initial velocity of the bullet, m_{w} and V_{wi} are the mass and the initial velocity of the wooden block, and V_{wf} and V_{bf} are the final velocities of the wooden block and the bullet

The wooden block is initial at rest (V_{wi} = 0) this yields

m_{b}V{b_{i} }  = m_{w} V_{wf} + m_{b}V_{bf}

By solving for V_{wf} adn substitute the givens

V_{wf} = \frac{m_{b}V_{bi} - m_{b} V_{bf}    }{m_{W} }

= \frac{5.3(g)(963(m/s)-426(m/s) }{610(g)}

V_{wf} = 4.67m/s

b) The center of mass speed is defined as

V = \frac{m_{b} }{m_{b}+m_{w} } V_{bi}

substituting:

V = \frac{5.3(g)}{5.3(g)+610(g)} X 963(m/s)

V = 8.29 m/s

7 0
3 years ago
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