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3 years ago

An object falls from the Transco Tower in Houston and takes 15 seconds to reach the ground.

Physics

1 answer:

Grace [21]3 years ago

8 0

Answer:

Impact velocity = $v_{f} =$ 147m/s

Explanation:

Impact velocity means the velocity of an object when it just falls on the ground. It is the maximum velocity obtain by an object before it stopped by something.

Given data:

Initial velocity = $v_{i} = 0$

Time = t = 15 s

Gravitational acceleration = g = 9.8 m/s²

Final velocity = impact velocity = $v_{f}=?$

According to first equation of motion

$v_{f} = v_{i} + gt$

As inial velocity is zero so

$v_{f} = gt$

$v_{f} = (9.8)(15)$

$v_{f} = 147m/s$

Impact velocity = 147 m/s

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b) The relative percentage change of power output is 1.67%

c) The intensity of the radiation on Mars is $540 W/m^2$

Explanation:

The intensity of electromagnetic radiation is given by

$I=\frac{P}{A}$

where

P is the power output

A is the surface area considered

In this problem, we have

$I=1360 W/m^2$ is the intensity of the solar radiation at the Earth

The area to be considered is area of a sphere of radius

$r=1.5\cdot 10^{11} m$ (distance Earth-Sun)

Therefore

$A=4\pi r^2 = 4 \pi (1.5\cdot 10^{11})^2=2.8\cdot 10^{23}m^2$

And now, using the first equation, we can find the total power output of the Sun:

$P=IA=(1360)(2.8\cdot 10^{23})=3.845\cdot 10^{26} W$

The energy of the solar radiation is directly proportional to its frequency, given the relationship

$E=hf$

where E is the energy, h is the Planck's constant, f is the frequency.

Also, the power output of the Sun is directly proportional to the energy,

$P=\frac{E}{t}$

where t is the time.

This means that the power output is proportional to the frequency:

$P\propto f$

Here the frequency increases by 1 MHz: the original frequency was

$f_0 = 60 MHz$

so the relative percentage change in frequency is

$\frac{\Delta f}{f_0}\cdot 100 = \frac{1}{60}\cdot 100 =1.67\%$

And therefore, the power also increases by 1.67 %.

In this second case, we have to calculate the new power output of the Sun:

$P' = P + \frac{1.67}{100}P =1.167P=1.0167(3.845\cdot 10^{26})=3.910\cdot 10^{26} W$

Now we want to calculate the intensity of the radiation measured on Mars. Mars is 60% farther from the Sun than the Earth, so its distance from the Sun is

$r'=(1+0.60)r=1.60r=1.60(1.5\cdot 10^{11})=2.4\cdot 10^{11}m$