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lukranit [14]
3 years ago
9

A clay ball with a mass of 0.48kg has an initial speed of 4.08m/s it strikes a 3.04kg clay ball at rest and the two balls stick

together and remain stationary. What is the decrease in kinetic energy of the 0.48kg ball
Physics
1 answer:
Rom4ik [11]3 years ago
5 0
Final velocity = 0, thus final kinetic energy is 0
Initial kinetic energy:
0.5mv²
= 0.5 x 0.48 x 4.08²
= 4.0 J
Decrease in kinetic energy = 4 - 0 = 4 Joules
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If you shine a light of frequency 375hz on a double slit setup , and you measure the slit separation to be 950 nm and the screen
amm1812

Answer:

y = 33.93 10⁵ m

Explanation:

This is an interference exercise, for the contributory interference is described by the expression

           d sin θ = m λ

let's use trigonometry for the angle

           tan θ = y / L

how the angles are small

          tan θ = sin θ / cos tea = sin θ

 

we substitute

         sin θ = y / L

        d y / L = m λ

         y = m λ L / d

the light fulfills the relation of the waves

       c = λ f

       λ = c / f

       λ = 3 10⁸ /375

       λ = 8 10⁵ m

first order m = 1

let's calculate

        y = 1  8 10⁵  4030 10-9 / 950 10-9

        y = 33.93 10⁵ m

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Answer:

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Explanation:

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3 years ago
1. A 180 kg motorcycle travels in a straight line on a horizontal road. The relationship between
Sergio039 [100]
Eh not really sure bout this one
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If a topographic map included a 6,000 ft. mountain next to an area of low hills, which would best describe the contour lines on
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8 0
3 years ago
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A skydiver jumps out of a hovering helicopter. A few seconds later, another diver jumps out, so they both fall along the same ve
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Answer:

distance difference would a) increase

speed difference would f) stay the same

Explanation:

Let t be the time the 2nd skydiver takes to travel, since the first skydiver jumped first, his time would be t + Δt where Δt represent the duration between the the first skydiver and the 2nd one. Remember that as t progress (increases), Δt remain constant.

Their equations of motion for distance and velocities are

s_2 = gt^2/2

s_1 = g(t + \Delta t)^2/2

v_2 = gt

v_1 = g(t + \Delta t)

Their difference in distance are therefore:

\Delta s = s_1 - s_2 = g(t + \Delta t)^2/2 - gt^2/2

\Delta s = g/2((t + \Delta t)^2 - t^2)

\Delta s = g/2(t + \Delta t - t)(t + \Delta t + t) (AsA^2 - B^2 = (A-B)(A+B)

\Delta s = g\Delta t/2(2t + \Delta t)

So as time progress t increases, Δs would also increases, their distance becomes wider with time.

Similarly for their velocity difference

\Delta v = v_1 - v_2 = g(t + \Delta t) - gt

\Delta v = gt + g\Delta t - gt = g\Delta t

Since g and Δt both are constant, Δv would also remain constant, their difference in velocity remain the same.

This of this in this way: only the DIFFERENCE in speed stay the same, their own individual speed increases at same rate (due to same acceleration g). But the first skydiver is already at a faster speed (because he jumped first) when the 2nd one jumps. The 1st one would travel more distance compare to the 2nd one in a unit of time.

8 0
3 years ago
Read 2 more answers
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