Ask question

Ask question

All categories

2 years ago

9

For a movie stunt, an empty truck with a mass of 2000 kg goes 10 m/s and runs into a stopped car of mass 1000 kg. The truck then

keeps moving and pushes the car along with it.
If there are no other forces acting on this system, which best describes the results of the collision?

The collision is perfectly elastic.
Kinetic energy stays the same.
The momentum after the collision is greater than the momentum before the collision.
The speed of the combined vehicles is less than the initial speed of the truck.

1 answer:

gizmo_the_mogwai [7]2 years ago

7 0

Answer:

D. The speed of the combined vehicles is less than the initial speed of the truck.

Explanation:

Ik this is late

You might be interested in

In 1656, the Burgmeister (mayor) of the town of Magdeburg, Germany, Otto Von Guericke, carried out a dramatic demonstration of t

tino4ka555 [31]

The force required to pull the two hemispheres is 46622.72N

<h3>Calculation and Parameters</h3>

( Note: 1 millibar=100 N/m2. One atmosphere is 1013 millibar = 1.013×105 N/m2 ]

The contact area between the hemispheres is (pi x 0.400^2) = 0.5024m^2.

Pressure difference = (940 - 12)

= 928 millibars.

(928 x 100)

= 92,800N/m^2.

Therefore, the required force to pull the two hemispheres is

(92800 x 0.5024)

= 46622.72N.

Read more about force here:

brainly.com/question/15300777

#SPJ1

6 0

2 years ago

Two difference between thrust and upthrust .

velikii [3]

Upthrust is the upward force exerted by fluids on the surface of an object immersed in fluids. Thrust:- It is the force acting perpendicular to the surface. Upthrust:- It is the upward force exerted by the fluid on the surface of an object immersed in liquid.

I hope it's help you

6 0

2 years ago

An air balloon is moving upward at a constant speed of 3 m/s. Suddenly a passenger realizes that she left her camera on the grou

frosja888 [35]

Answer:t=0.3253 s

Explanation:

Given

speed of balloon is $u=3\ m/s$

speed of camera $u_1=20\ m/s$

Initial separation between camera and balloon is $d_o=5\ m$

Suppose after t sec of throw camera reach balloon then,

distance travel by balloon is

$s=ut$

$s=3\times t$

and distance travel by camera to reach balloon is

$s_1=ut+\frac{1}{2}at^2$

$s_1=20\times t-\frac{1}{2}gt^2$

Now

$\Rightarrow s_1=5+s$

$\Rightarrow 20\times t-\frac{1}{2}gt^2 =5+3t$

$\Rightarrow 5t^2-17t+5=0$

$\Rightarrow t=\dfrac{17\pm \sqrt{17^2-4(5)(5)}}{2\times 5}$

$\Rightarrow t=\dfrac{17\pm 13.747}{10}$

$\Rightarrow t=0.3253\ s\ \text{and}\ t=3.07\ s$

There are two times when camera reaches the same level as balloon and the smaller time is associated with with the first one .

(b)When passenger catches the camera time is $t=0.3253\ s$

velocity is given by

$v=u+at$

$v=20-10\times 0.3253$

$v=16.747\ m/s$

and position of camera is same as of balloon so

Position is $=5+3\times 0.3253$

$=5.975\approx 6\ m$

8 0

3 years ago

A bomb of mass 6kg initially at rest explodes into two fragments of masses 4kg and2kg respectively. If the greater mass moves wi

Katen [24]

Answer:

v = 10 [m/s]

Explanation:

The largest mass is that of 4 [kg], in this way the momentum can be calculated by means of the product of the mass by velocity.

$P=m*v\\$

where:

P = momentum [kg*m/s]

m = mass = 4 [kg]

v = velocity = 5 [m/s]

Now the momentum:

$P=4*5\\P=20[kg*m/s]$

This same momentum is equal for the other mass, in this way we can find the velocity.

$P=m*v\\20=2*v\\v=10[m/s]$

7 0

3 years ago

You are driving your motorcycle in a circle of radius 75 m on wet pavement. what is the fastest you can go before you lose tract

stira [4]

On driving your motorcycle in a circle of radius 75 m on wet pavement, the fastest you can go before you lose traction, assuming the coefficient of static friction is 0.20 is 147m/s

Friction helps to maintain the slipping of the vehicle on the road hence lays a very important role.

Maximum velocity of a road with friction is given by the formula,

v = μRg

where, v is the maximum velocity

μ is the coefficient of static friction

R is the radius of the circle road

g is the acceleration due to gravity

Given,

μ = 0.20

R = 75m

g = 9.8m/s²

On substituting the given values in the above formula,

v = 0.20× 75 ×9.8

v = 147m/s

So, the Maximum velocity of the wet road is 147m/s.

Learn more about Velocity here, brainly.com/question/18084516

#SPJ4

3 0

2 years ago