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Anna11 [10]
3 years ago
11

g A ball is thrown against the wall and bounces back with the same velocity. What type of collision is this

Physics
1 answer:
Softa [21]3 years ago
8 0

Answer:

Perfectly elastic collision

Explanation:

In a closed system, an elastic collision is a type of collision between two bodies, where the total momentum and kinetic energy are conserved.

We are told from the problem that the ball bounces back with its original velocity. For the ball to bounce back to the thrower in the first place, this is our first hint that the collision is elastic. If the collision was inelastic, the ball would most likely have stuck to the wall.

In addition to that, the velocity of the ball remains fairly unchanged even after the collision. This confirms that the kinetic energy it had before the collision is the same as the kinetic energy it has after the collision.  

As a result of this, the collision is perfectly elastic

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In a double-slit experiment, light from two monochromatic light sources passes through the same double slit. The light from the
MatroZZZ [7]

Answer:

The wavelength is \lambda_2 =  534 *10^{-9} \ m

Explanation:

From the question we are told that

   The wavelength of the first light is  \lambda _ 1 =  587 \ nm

    The order of the first light that is being considered is  m_1  =  10

     The order of the second light that is being considered is  m_2  =  11

Generally the distance between the fringes for the first light is mathematically represented as

      y_1 =  \frac{ m_1  * \lambda_1 *  D}{d}

 Here  D is the distance from the screen

 and    d  is the distance of separation of the slit.

      For the second light the distance between the fringes is  mathematically represented as

         y_2 =  \frac{ m_2  * \lambda_2 *  D}{d}

Now given that both of the light are passed through the same double slit

       \frac{y_1}{y_2}  =  \frac{\frac{m_1 *  \lambda_1 * D}{d}  }{\frac{m_2 *  \lambda_2 * D}{d}  } = 1

=>    \frac{ m_1 *  \lambda _1  }{ m_2  *  \lambda_2} =  1

=>     \lambda_2 =  \frac{m_1 *  \lambda_1}{m_2}

=>    \lambda_2 =  \frac{10  *   587 *10^{-9}}{11}

=>   \lambda_2 =  534 *10^{-9} \ m

4 0
3 years ago
A car is stopped for a traffic signal. When the light turns green, the car accelerates, increasing its speed from 0 to 5.40 m/s
tekilochka [14]

Answer:

a) 378Ns

b) 477.27N

Explanation:

Impulse is the defined as the product of the applied force and time taken. This is expressed according to the formula

I = Ft = m(v-u)

m is the mass = 70kg

v is the final velocity = 5.4m/s

u is the initial velocity = 0m/s

Get the impulse

I = m(v-u)

I = 70(5.4-0)

I = 70(5.4)

I = 378Ns

b) Average total force is expressed as

F = ma (Newton's second law)

F = m(v-u)/t

F = 378/0.792

F = 477.27N

Hence the average total force experienced by a 70.0-kg passenger in the car during the time the car accelerates is 477.27N

3 0
3 years ago
If the rate $56 per 7 hours is reduced to a unit rate, the result is dollars per hour.
34kurt
The answer is 8 because multiplying 7 and 8 is 56
4 0
3 years ago
Read 2 more answers
A runner went from 6 m/s and two seconds what was his acceleration
Radda [10]

Answer:

is it 3?

Explanation:

Im taking a guess and just dividing 6 and 2

8 0
3 years ago
You have been called in by your neighborhood safety council to advise them about installing a barrier on a dangerous curve that
Alborosie

Answer: The softer barrier is the better option

Explanation:

1) When is a car is moving at a certain speed, it has a certain amount of momentum (p=mv). A collision against a barrier would cause its momentum to decrease to 0. A change in momentum is Impulse

2) The formula for Impulse: J = f * Δt

J is Impulse

f is the force applied during the time Δt

A tough barrier would produce a smaller Δt, which means more force is applied on the car. (J is always constant)

A softer barrier would apply less force on the car, which means Δt is large.

Answer: The softer barrier is the better option

4 0
3 years ago
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