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3 years ago

5

A bullet is shot downward out of a high building with an initial speed of 10 m/s. How fast is the bullet traveling after 5 secon

ds (by m/s, neglect air friction, g = 9.8 m/s)?
a.) -10
b.) 9.8
c.)-59
d.)-49
e.)-9.8

1 answer:

serious [3.7K]3 years ago

6 0

Answer:

-59

Explanation:

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8. The legs of a young man are each 0.650 meters long. What is his maximum walking speed?

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Answer:

2.52 m/s

Explanation:

When the man takes a step, his foot is stationary while his body revolves around it. At the point when his body is directly above his foot, there will be no normal force at his maximum speed.

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We intend to observe two distant equal brightness stars whose angular separation is 50.0 × 10-7 rad. Assuming a mean wavelength

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Answer:

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Explanation:

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$\theta=1.22\frac{\lambda}{b}$

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b = diameter of the objective

By doing b the subject of the formula and replacing the values of the angle and wavelength you obtain:

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3 years ago

Read 2 more answers

When a condenser discharges electricity, the instantaneous rate of change of the voltage is proportional to the voltage in the c

e-lub [12.9K]

Answer:

460.52 s

Explanation:

Since the instantaneous rate of change of the voltage is proportional to the voltage in the condenser, we have that

dV/dt ∝ V

dV/dt = kV

separating the variables, we have

dV/V = kdt

integrating both sides, we have

∫dV/V = ∫kdt

㏑(V/V₀) = kt

V/V₀ = $e^{kt}$

Since the instantaneous rate of change of the voltage is -0.01 of the voltage dV/dt = -0.01V

Since dV/dt = kV

-0.01V = kV

k = -0.01

So, V/V₀ = $e^{-0.01t}$

V = V₀ $e^{-0.01t}$

Given that the voltage decreases by 90 %, we have that the remaining voltage (100 % - 90%)V₀ = 10%V₀ = 0.1V₀

So, V = 0.1V₀

Thus

V = V₀ $e^{-0.01t}$

0.1V₀ = V₀ $e^{-0.01t}$

0.1V₀/V₀ = $e^{-0.01t}$

0.1 = $e^{-0.01t}$

to find the time, t it takes the voltage to decrease by 90%, we taking natural logarithm of both sides, we have

㏑(0.01) = -0.01t

So, t = ㏑(0.01)/-0.01

t = -4.6052/-0.01

t = 460.52 s

3 0

3 years ago