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Brums [2.3K]
3 years ago
11

Suppose that the resistive force of the air on a skydiver can be approximated by f = −bv2. If the terminal velocity of an 82.0 k

g skydiver is 33.4 m/s, what is the value of b (in kg/m)?
Physics
1 answer:
antiseptic1488 [7]3 years ago
6 0

Answer:

The value of b 0.7351 kg/m

Explanation:

Given that;

Mass of sky diver = 82 kg

Velocity = 33.4 m/s

f = −bv²

It is a resistance force, therefore the negative sign is ignored.

since; f = mg

∴ mg = bv²

b = mg / v ²  ........ (1)

At terminal velocity a = 0

Put parameters in (1)

b = (82 × 10) / (33.4)²

b = 820 / 1,115.56

b = 0.7351

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Answer:

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Explanation:

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After 22 kg are added the period becomes 33 sec

Time period of spring mass system is

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From the relation we can see that

\frac{T_1}{T_2}=\sqrt{\frac{m}{m+22}}

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A 7.80-g bullet moving at 540 m/s strikes the hand of a superhero, causing the hand to move 5.10 cm in the direction of the bull
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Answer:

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The average force exerted on the bullet to stop it is:

F = \frac{m\cdot [v_{A}^{2}-v_{B}^{2}]}{2\cdot \Delta s}

F = \frac{(7.8\times 10^{-3}\,kg)\cdot [(540\,\frac{m}{s} )^{2}-(0\,\frac{m}{s} )^{2}]}{2\cdot (0.051\,m)}

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