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3 years ago

What is the momentum of a lab cart with a mass of 0.60 [kg] and a speed of 2.2 [m/s]?

Physics

1 answer:

pentagon [3]3 years ago

8 0

Answer:

1.32kgm/s

Explanation:

use the formula: p=mv

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What state of matter takes the shape and volume of its container, like the steam coming from a pot A) gas B) liquid Eliminate C)

Gnoma [55]

Ummm gas takes the shape and volume of its container, but I don't know about plasma. Solids definitely don't, and liquids don't change their volumes. So I would say gas. :)

6 0

3 years ago

Read 2 more answers

Given a force of 88 N and an acceleration of 4 m/s2, what is the mass?

WARRIOR [948]

Newton's second law of motion says:

Force = (mass) x (acceleration)

88 = (mass) x (4)

Divide each side by 4 :

Mass = 22 kg.

7 0

4 years ago

A body of mass 450g changes it speed from 5ms¹ to 25ms¹. what is the work done by the body?

Andrej [43]

Answer:

135J

Explanation:

So we know ΔKinetic Energy= ΔWork

Kinetic energy=1/2mv²

So Kf-Ki=ΔK

ΔK=1/2*0.45(25²-5²)=135J

135J=ΔWork

3 0

3 years ago

Before colliding, the momentum of block A is -100 kg*m/, and block B is -150 kg*m/s. After, block A has a momentum -200 kg*m/s.

rjkz [21]

Answer:

Momentum of block B after collision = $-50\ kg\ ms^{-1}$

Explanation:

Given

Before collision:

Momentum of block A = $p_{A1}$ = $-100\ kg\ ms^{-1}$

Momentum of block B = $p_{B1}$ = $-150\ kg\ ms^{-1}$

After collision:

Momentum of block A = $p_{A2}$ = $-200\ kg\ ms^{-1}$

Applying law of conservation of momentum to find momentum of block B after collision $p_{B2}$ .

$p_{A1}+p_{B1}=p_{A2}+p_{B2}$

Plugging in the given values and simplifying.

$-100-150=-200+p_{B2}$

$-250=-200+p_{B2}$

Adding 200 to both sides.

$200-250=-200+p_{B2}+200$

$-50=p_{B2}$

∴ $p_{B2}=-50\ kg\ ms^{-1}$

Momentum of block B after collision = $-50\ kg\ ms^{-1}$

6 0

3 years ago

A uniform sphere with mass 28 kg and radius 0.38 m is rotating at constant angular velocity about a stationary axis that lies al

elixir [45]

To solve this problem it is necessary to apply the concepts related to the equations of description of the rotational movement, also include there Inertia and Rotational Kinetic Energy.

From the definition for a sphere the moment of inertia can be defined as

$I = \frac{2}{5} mr^2$

Where,

m = mass

r = radius

At the same time the rotational kinetic energy is described as

$KE = \frac{1}{2} I\omega^2$

Where,

I = Moment of Inertia

$\omega =$ Angular velocity

From the description of the angular movement we know that the product of the angular velocity and the radius is equivalent to the linear / tangential velocity:

$V = r \omega$