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lana [24]
3 years ago
11

A 0.150-kg ball traveling horizontally on a frictionless surface approaches a very massive stone at 20.0 m/s perpendicular to wa

ll and rebounds with 70.0% of its initial kinetic energy. What is the magnitude of the change in momentum of the stone?
Physics
1 answer:
Hitman42 [59]3 years ago
3 0

Answer:

The magnitude of the change in momentum of the stone is 5.51kg*m/s.

Explanation:

the final kinetic energy = 1/2(0.15)v^2

                1/2(0.15)v^2  = 70%*1/2(0.15)(20)^2

                              v^2 = 21/0.075

                              v^2 = 280

                                 v = 16.73 m.s

if u is the initial speed and v is the final speed, then:

u = 20 m/s and v = - 16.73m/s

change in momentum = m(v-u)

                                     = 0.15(- 16.73-20)

                                    = -5.51 kg*m.s

Therefore, The magnitude of the change in momentum of the stone is 5.51kg*m/s.

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Which of the following is true of education in 1950
Maksim231197 [3]
Information I learned from history class Education in the 1950's expanded from previous decades. They no longer focused purely on reading, writing and arithmetic. History and science became a main part of the cirriculum. Also, enrollment skyrocketed as the baby-boomers began enrolling in elementary school. One interesting thing that categorized this generation was the presence of fallout tests. Schools would require the students to go through a fake atomic bomb attack in which they would hide under their desks (which was completely pointless in protecting them from radiation, it was more of an emotional security for the parents and teachers, but scared the hell out of the students). Socially, children were taught to conform and to be normal. Standing out or questioning authority was bad. Sex was taught, though minimally. They explained the penis and vagina. Sexually transmitted diseases were focused on greatly so as to "scare" the students out of premarital sex.
8 0
3 years ago
At what speed, as a fraction of c, will a moving rod have a length 65% that of an identical rod at rest
dezoksy [38]

Answer:

v/c = 0.76

Explanation:

Formula for Length contraction is given by;

L = L_o(√(1 - (v²/c²))

Where;

L is the length of the object at a moving speed v

L_o is the length of the object at rest

v is the speed of the object

c is speed of light

Now, we are given; L = 65%L_o = 0.65L_o, since L_o is the length at rest.

Thus;

0.65L_o = L_o[√(1 - (v²/c²))]

Dividing both sides by L_o gives;

0.65 = √(1 - (v²/c²))

Squaring both sides, we have;

0.65² = (1 - (v²/c²))

v²/c² = 1 - 0.65²

v²/c² = 0.5775

Taking square root of both sides gives;

v/c = 0.76

6 0
3 years ago
Match each characteristic with the type of wave it corresponds to. Longitudinal waves Transverse waves The direction the wave tr
zubka84 [21]

Answer:The direction the wave travels is parallel to the displacement of the medium.LW.

Electromagnetic waves are an example of this type of wave. TW

Sound waves are an example of this type of wave.LW

Ir the wave propagates in the x‑direction, the medium is also disturbed in the x‑direction. LW

If the wave propagates in the x‑direction, the medium is disturbed in the y‑ and/or z‑direction. TW

The direction the wave travels is perpendicular to the displacement of the medium.TW

Explanación:

the main difference of Transversal AND Longitudinal waves (TW AND LW) is that the disturbation direction in the media with respect to the travel direction is well different for both cases. For TW the perturbation oscilatiew perpendicular to propagation direction of the wave.

In constrant for LW the disturbation oscilates in the same direction as the wave is propagating.

3 0
3 years ago
Read 2 more answers
que tiempo tarda juan en su carro al llegar al trabajo si lleva una velocidad de 25 m/s a una distancia de 5m
kramer

Answer:

0.2s

Explanation:

Given parameters:

Speed  = 25m/s

Distance  = 5m

Unknown:

Time taken for Juan to get to his work  = ?

Solution:

To solve this problem we use the relationship below:

 

   Speed  = \frac{distance}{time}  

  Time  =  \frac{distance }{speed}  

So;

   Time  = \frac{5}{25}   = 0.2s

4 0
3 years ago
Let’s calculate Ms Jackson’s acceleration as she comes to a stop. Remember that the formula for acceleration is the change in ve
Natali [406]

Answer:

-1.5 m/s^2

Explanation:

The acceleration of an object is equal to the change in velocity of the object divided by the time taken for the object to change the velocity:

a=\frac{v-u}{t}

where:

u is the initial velocity

v is the final velocity

t is the time taken for the velocity to change from u to v

For Ms Jackson in this problem, we have:

u = 6 m/s (initial velocity)

v = 0 m/s (final velocity)

t = 4 s (time elapsed)

Substituting into the equation, we find the acceleration:

a=\frac{0-6}{4}=-1.5 m/s^2

And the negative sign indicates that the direction of the acceleration is opposite to the direction of motion.

4 0
3 years ago
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