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Alchen [17]
3 years ago
15

Is this an example of an elastic collision or inelastic collision? A 200-kg bumper car stopped at an intersection is rear-ended

by an 1100-kg truck
Inoving at 13 m/s. If the car was in neutral and the brakes were off, the bumper car would BOUNCE OFF of the truck into the intersection.​
Physics
1 answer:
Alina [70]3 years ago
4 0

Answer:

inelastic collision.

Explanation:

An elastic collision consists in a collision in wich the bodies implied bounce off. Furthermore, the momentum of the bodies is conserved.

In as inelastic collision the momentum is not conserved, and is possible that the bodies implied in the collision stay joint after the collision.

In this case yo have that both bumper car and truck stayed joint after the collision. Hence, this situation is an inelastic collision.

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A thin layer of liquid methylene iodide (n = 1.756) is sandwiched between two flat, parallel plates of glass (n = 1.50). What mu
Harrizon [31]

Answer:

t = 96.1 nm

Explanation:

For strong reflection through liquid layer we know that the path difference between two reflected light rays must be integral multiple of wavelength

now we know that the path difference of two reflected light from thin liquid layer is given as

2\mu t - \frac{\lambda}{2} = N\lambda

here we know that

\mu = 1.756

t = thickness of layer

N = 0 (for minimum thickness of layer)

\lambda = 675 nm

now we have

2(1.756) t = \frac{675 nm}{2}

t = 96.1 nm

5 0
3 years ago
Using the formula 1/2 (m x v2), what is the kinetic energy of a 4 kg rock falling through the air at 5 m/s
aivan3 [116]

Answer:

KE = 50J

Explanation:

KE = \frac{1}{2}mv^{2} \\\\ KE = \frac{1}{2}(4)(5)^{2} \\\\ KE = 2(25) \\\\ KE = 50J

5 0
3 years ago
A box weighs 25N. How much mass does it have?
Rudiy27

Explanation:

If box weight 25N on ground

MA=F

M(10)=25

M=2.5Kg

3 0
3 years ago
A moving car has momentum. if it moves twice as fast its momentum is ____________ as much
Nikolay [14]
It's momentum is twice as much.
3 0
3 years ago
A(n) 55.5 g ball is dropped from a height of 53.6 cm above a spring of negligible mass. The ball compresses the spring to a maxi
Serggg [28]

Answer:

The spring force constant is  k=243\ \frac{N}{m} .

Explanation:

We are told the mass of the ball is m=0.0555\ kg, the height above the spring where the ball is dropped is h=0.536\ m,  the length the ball compresses the spring is d=0.04897\ m and the acceleration of gravity is 9.8\ \frac{m}{s^{2}} .

We will consider the initial moment to be when the ball is dropped and the final moment to be when the ball stops, compressing the spring. We supose that there is no friction so the initial mechanical energy E_{mi} is equal to the final mechanical energy E_{mf} :

                                                    E_{mf}=E_{mi}

Initially there is only gravitational potential energy because the force of the spring isn't present and the speed is zero. In the final moment there is only elastic potential energy because the height is zero and the ball has stopped. So we have that:

                                                   \frac{1}{2}kd^{2}=mgh

If we manipulate the equation we have that:

                                                    k=\frac{2mgh}{d^{2} }

                                         k=\frac{2\ 0.0555\ kg\ 9.8\frac{m}{s^{2}}\ 0.536\ m}{(0.04897)^{2}m^{2}}

                                              k=\frac{0.58306\ \frac{kgm^{2}}{s^{2}}}{2.398x10^{-3}m^{2}}

                                                     k=243\ \frac{N}{m}

                                                   

                             

5 0
4 years ago
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