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3 years ago

What is the momentum of and 200 kg car traveling south at 22 m/sec?

Physics

1 answer:

motikmotik3 years ago

5 0

The momentum of the car is 4.4x10^3 kg•m/sec

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(NEED HELP PLEASE) A physics student goes to the roof of the school, 24.15 m above the ground, and drops a pumpkin straight down

slavikrds [6]

Answer:

t = 2.2 s

Explanation:

Given that,

Height of the roof, h = 24.15 m

The initial velocity of the pumpkin, u = 0

We need to find the time taken for the pumpkin to hit the ground. Let the time be t. Using second equation of kinematics to find it as follows :

$h=ut+\dfrac{1}{2}at^2$

Here, u = 0 and a = g

$h=\dfrac{1}{2}gt^2\\\\t=\sqrt{\dfrac{2h}{g}} \\\\t=\sqrt{\dfrac{2\times 24.15}{9.8}} \\\\t=2.22\ s$

So, it will take 2.22 s for the pumpkin to hit the ground.

7 0

3 years ago

PLZZZ HELPPP ASAP I really need help as soon as possible

julsineya [31]

Answer:

Friction

Explanation:

As the toy cars rolls away, more friction is created. The more friction there is, the more friction on surface rubs against another which creates friction which in-term slows it down. Hope this helps.

4 0

2 years ago

Air passing over an airplane's wing travels ____________________, and therefore exerts ____________________ pressure than air tr

Aleksandr [31]

c. faster, less

is the answer

6 0

3 years ago

Read 2 more answers

NEED HELP FAST Are the circuits in a car parallel or series? how do you know?

guajiro [1.7K]

The answer is parallel

If the circuits in a car were series, they would go out at the same time.

I hope this helps! :3

3 0

3 years ago

The coefficient of the restitution of an object is defined as the ratio of its outgoing to incoming speed when the object collid

IgorLugansk [536]

Answer:

48.16 %

Explanation:

coefficient of restitution = 0.72

let the incoming speed be = u

let the outgoing speed be = v

kinetic energy = 0.5 x mass x $x velocity^{2}$

incoming kinetic energy = 0.5 x m x $x u^{2}$

coefficient of restitution = $\frac{v}{u}$

0.72 = $\frac{v}{u}$

v = 0.72u

therefore the outgoing kinetic energy = 0.5 x m x $(0.72u)^{2}$

outgoing kinetic energy = 0.5 x m x $0.5184 x u^{2}$

outgoing kinetic energy = 0.5184 (0.5 x m x $x u^{2}$ )

recall that 0.5 x m x $x u^{2}$ is our incoming kinetic energy, therefore

outgoing kinetic energy = 0.5184 x (incoming kinetic energy)

from the above we can see that the outgoing kinetic energy is 51.84 % of the incoming kinetic energy.

The energy lost would be 100 - 51.84 = 48.16 %

5 0

3 years ago