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kvv77 [185]
3 years ago
14

An iron cube has edge 15cm long at 20c .What will be? The new surface area of a face when the temperature rises to 80c

Physics
1 answer:
grigory [225]3 years ago
3 0

Answer:

271cm^2

Explanation:

volume 1= 15^3 =3375

temp. 1. = 20+273 = 293

temp. 2. = 50+273 = 353

volume 2 =?

According to Charles law

volume is proportional to temperature

v2 = v1 * t2 / t1

v2 = 3375 * 353 / 293

v2 = 4066cm^3

v = area * length

4066 = area * 15

area = 4066/15 = 271cm^2

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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
Sergio039 [100]

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.

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3 years ago
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Answer:

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y_{aceribo} = 1.1*10^6 \ \ m

Explanation:

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y_{hubble} = 1.22*400*10^{-9}*3.8*10^8/2.4

y_{hubble} = 77\ \ m

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y = 1.22 λs/D

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λ = 75 cm = 0.75 m

D = 305 m

y_{acerbo} = 1.22*0.75 *3.8*10^8/305

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The answer would be a radio wave
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