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Rudiy27
2 years ago
11

a train has an initial velocity of 30 m/s. If the train accelerates uniformly at a rate of 6.3 m/s ^ for 2.8 seconds what is the

trains final velocity?
Physics
1 answer:
Annette [7]2 years ago
3 0

T

Answer:

the velocity is a second final to initial velocity of 39

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What provides most of the force to allow the runners to move?
kobusy [5.1K]

Answer:

When you are running the most important force that you should understand is friction. Friction is a force that opposes movement between two objects, but for runners friction makes you faster. Friction gives you a better and more efficient way to use your energy into speed.

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After the collapse of a nebular cloud, atoms begin gravitating together to form a condensed center. What happens next in the sta
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How can you tell that this paragraph uses a quotation?<br> Check all that apply?
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8 0
3 years ago
The flywheel is rotating with an angular velocity ω0 = 2.37 rad/s at time t = 0 when a torque is applied to increase its angular
nika2105 [10]

Answer:

ω = 12.023 rad/s

α = 222.61 rad/s²

Explanation:

We are given;

ω0 = 2.37 rad/s, t = 0 sec

ω =?, t = 0.22 sec

α =?

θ = 57°

From formulas,

Tangential acceleration; a_t = rα

Normal acceleration; a_n = rω²

tan θ = a_t/a_n

Thus; tan θ = rα/rω² = α/ω²

tan θ = α/ω²

α = ω²tan θ

Now, α = dω/dt

So; dω/dt = ω²tan θ

Rearranging, we have;

dω/ω² = dt × tan θ

Integrating both sides, we have;

(ω, ω0)∫dω/ω² = (t, 0)∫dt × tan θ

This gives;

-1[(1/ω_o) - (1/ω)] = t(tan θ)

Thus;

ω = ω_o/(1 - (ω_o × t × tan θ))

While;

α = dω/dt = ((ω_o)²×tan θ)/(1 - (ω_o × t × tan θ))²

Thus, plugging in the relevant values;

ω = 2.37/(1 - (2.37 × 0.22 × tan 57))

ω = 12.023 rad/s

Also;

α = (2.37² × tan 57)/(1 - (2.37 × 0.22 × tan 57))²

α = 8.64926751525/0.03885408979 = 222.61 rad/s²

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3 years ago
If the second harmonic of a certain string is 42 Hz, what is the fundamental frequency of the string?
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