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alexgriva [62]
3 years ago
15

An archer puts a 0.30-kg arrow to the bowstring. An average force of 201 N is exerted to draw the string back 1.3 m. Assuming th

at all the energy goes into the arrow, with what speed does the arrow leave the bow?

Physics
2 answers:
Thepotemich [5.8K]3 years ago
8 0

Explanation:

Below is an attachment containing the solution.

vovikov84 [41]3 years ago
3 0

Answer:

41.74 m/s

Explanation:

The energy used to draw the bowstring = the kinetic energy of the arrow.

Fd = 1/2mv²................................ Equation 1

Where F = force, d = distance move string, m = mass of the arrow, v = speed of the arrow.

make v the subject of the equation

v = √(2Fd/m)...................... Equation 2

Given: F = 201 N, m = 0.3 kg, d = 1.3 m.

Substitute into equation 2

v = √(2×201×1.3/0.3)

v = √(1742)

v = 41.74 m/s.

Hence the arrow leave the bow with a speed of 41.74 m/s

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storchak [24]

Answer:

a) V1 = 4V - V2/3 and V2 = 4V -  3V1

b) Δe = 4000V - 4000V2 + 9000V1

Explanation:

Let V represent volume of the box containing the two compartments

V1 represents compartment of the left compartment

V2 represents compartment of the right compartment

Momentum of the compartments before impact:

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Momentum of the compartments after impact:

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a) To obtain the volume of each compartment, that is, V1 and V2, we say:

Momentum before impact = Momentum after impact

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∴ V1 = 4000V - 1000V2/3000 = 4V - V2/3

Also, V2 = 4000V - 3000V1/1000 = 4V - 3V1

b) Change in entropy,Δe = 4000V1 - 1000V2

By substituting the V1 and V2, we have:

4000(4V - V2)/3 - 1000(4V - 3V1)

16000V - 4000V2/3 - 4000V + 3000V1

16000V -  4000V2 - 12000V + 9000V1

∴ Δe = 4000V - 4000V2 + 9000V1

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