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ivann1987 [24]
3 years ago
15

A sky diver jumps from a reasonable height above the ground. The air resistance she experiences is proportional to her velocity,

and the constant of proportionality is k = 0.19. It can be shown that the downward velocity of the sky diver at time t is given by v(t) = A(1 − e^-kt) where t is measured in seconds and v(t) is measured in feet per second (ft/s). Suppose A = 64.
(a) Find the initial velocity of the sky diver.
(b) Find the velocity after 5 s and after 15 s. (Round your answers to one decimal place.) ...?
Physics
1 answer:
ch4aika [34]3 years ago
7 0

The initial velocity of the sky diver would be t=0. So put that into the equation:

<span>v(0)=64(1−<span>e<span>(−0.19)(0)</span></span>)

and for b it will be
</span><span>(5)=64(1−<span>e<span>(−0.19)(5)</span></span>)</span><span>, and likewise for t=15. You just need to throw it in a calculator.</span>

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A 95kg clock initially at rest on a horizontal floor requires a 560N horizontal force to set it in motion. After the clock is in
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Complete Question

A 95 kg clock initially at rest on a horizontal floor requires a 650 N horizontal force to set it in motion. After the clock is in motion, a horizontal force of 560 N keeps it moving with a constant velocity. Find the coefficient of static friction and the coefficient of kinetic friction.

Answer:

The value for static friction is \mu_s =  0.60

The value for static friction is \mu_k =  0.70

Explanation:

From the question we are told that

The mass of the clock is m  =  95 \  kg

The first horizontal force is F _s  =  560 \  N

    The second horizontal force is    F _k  =  650  \  N

Generally the static frictional force is equal to the first  horizontal force

So

     F _s  =  m  *  g  *  \mu_s

=>   560  =  95  *  9.8  *  \mu_s

=>    \mu_s =  0.60

Generally the kinetic frictional force is equal to the second horizontal force

So

      F _k  =  m  *  g  *  \mu_k

      650 =  95  *  9.8  *  \mu_k

     \mu_k =  0.70

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80 grams of iron at 100°C is dropped into 200 of water at 20°C contained in an iron vessel of mass 50 gram find the resulting te
vodka [1.7K]

Answer:

the resulting temperature is 23.37 ⁰C

Explanation:

Given;

mass of the iron, m₁ = 80 g = 0.08 kg

mass of the water, m₂ = 200 g = 0.2 kg

mass of the iron vessel, m₃ = 50 g = 0.05 kg

initial temperature of the iron, t₁ = 100 ⁰C

initial temperature of the water, t₂ = 20 ⁰C

specific heat capacity of iron, c₁ = 462 J/kg⁰C

specific heat capacity of water, c₂ = 4,200 J/kg⁰C

let the temperature of the resulting mixture = T

Apply the principle of conservation of energy;

heat lost by the hot iron = heat gained by the water

m_1c_1 \Delta t_1 = m_2c_2\Delta t_2\\\\m_1c_1 (100 - T) = m_2c_2 (T- 20)\\\\0.08 \times 462 (100-T) = 0.2 \times 4,200 (T-20)\\\\36.96 (100-T) = 840 (T-20) \\\\100 - T = 22.72 (T-20)\\\\100-T = 22.72 T - 454.4 \\\\554.4 = 23.72T\\\\T = \frac{554.4}{23.72} \\\\T = 23.37 \ ^0C

Therefore, the resulting temperature is 23.37 ⁰C

6 0
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