All categories

3 years ago

If a 1200kg car travels 20 m/s and slows down for a stop, what is its momentum?

Physics

1 answer:

frosja888 [35]3 years ago

3 0

Answer:

24000kgm/s

Explanation:

Given parameters:

Mass of car = 1200kg

Velocity = 20m/s

Unknown:

Momentum = ?

Solution:

Momentum is the amount of motion a body possesses.

Momentum = mass x velocity

Now insert the parameters and solve;

Momentum = 1200 x 20 = 24000kgm/s

You might be interested in

Does light travel faster through water or glass?

Len [333]

Answer: The refractive index or n shows the relationship between c and the speed v when light travels through a material. The equation is n=c/v. In water, the refractive index is 1.3 and in glass, it is 1.5. Therefore, light travels faster through water, than it travels through glass.

Explanation:

4 0

4 years ago

A golden-colored cube is handed to you. The person wants you to buy it for $100, saying that is a gold nugget. You pull out your

kenny6666 [7]

Answer:

Explanation:

Volume of cube=side³

=2*2*2

=8 cm³

Mass of cube=40g

Its density=mass/volume

=5g/cm³

which differs the value given in the text hence it is not gold and u should not bye it...

Instead u should call the police for arresting the person lol...

6 0

3 years ago

Read 2 more answers

What piece of sports equipment is aerodynamic

Mashcka [7]

Aerobie. Frisbee. Discus. Javelin. I suppose an American football to some extent.
Pull! Clay pigeons. Arrows. Wingsuit. Kites. Hang gliders. Sails. sailboat keels/dagger boards. Water skis. Ski jumping skis. Boomerang. 
I'm excluding spheres and parachutes as bluff bodies even though aerodynamics often plays a big part in their motion.

6 0

3 years ago

Force F acts between a pair of charges, q1 and q2, separated by a distance d. For each of the statements, use the drop-down menu

lora16 [44]

The initial force between the two charges is given by:

$F=k \frac{q_1 q_2}{d^2}$

where k is the Coulomb's constant, q1 and q2 the two charges, d their separation. Let's analyze now the other situations:

1. F

In this case, q1 is halved, q2 is doubled, but the distance between the charges remains d.

So, we have:

$q_1' = \frac{q_1}{2}\\q_2' = 2 q_2\\d' = d$

So, the new force is:

$F'=k \frac{q_1' q_2'}{d'^2}= k \frac{(\frac{q_1}{2})(2q_2)}{d^2}=k \frac{q_1 q_2}{d^2}=F$

So the force has not changed.

2. F/4

In this case, q1 and q2 are unchanged. The distance between the charges is doubled to 2d.

So, we have:

$q_1' = q_1\\q_2' = q_2\\d' = 2d$

So, the new force is:

$F'=k \frac{q_1' q_2'}{d'^2}= k \frac{q_1 q_2)}{(2d)^2}=\frac{1}{4} k \frac{q_1 q_2}{d^2}=\frac{F}{4}$

So the force has decreased by a factor 4.

3. 6F

In this case, q1 is doubled and q2 is tripled. The distance between the charges remains d.

So, we have:

$q_1' = 2 q_1\\q_2' = 3 q_2\\d' = d$

So, the new force is:

$F'=k \frac{q_1' q_2'}{d'^2}= k \frac{(2 q_1)(3 q_2)}{d^2}=6 k \frac{q_1 q_2}{d^2}=6F$

So the force has increased by a factor 6.

8 0

3 years ago

Read 2 more answers

In midair in the international space station a 1 kg chunk of putty moving at 1 m/s collides with and sticks to a 5 kg chunk of p

Delicious77 [7]

We know that the momentum keeps constant in a inelastic collisions, so the product of mass and speed do not change:
m1 * v1 + m2 * v2 = m * v
1 * 1 + 5 * 0 = (1 + 5) * v
1 = 6 * v
v = 1/6 m/s
So the final speed of the 6 kg chunk will travel at 0.167 m/s

6 0

3 years ago

If a 1200kg car travels 20 m/s and slows down for a stop, what is its momentum?​

If a 1200kg car travels 20 m/s and slows down for a stop, what is its momentum?