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madam [21]
3 years ago
6

A spotlight on the ground shines on a wall 15 meters away. A man 1.8 meters tall walks away from the spotlight toward the buildi

ng. Write the height of the shadow (s) as a function of the distance (d) from the man to the building.

Physics
1 answer:
Alex17521 [72]3 years ago
6 0

Answer:

s=(15-x)m

Explanation:

Let the distance from spotlight to wall be 15m, and distance from the man to the building be x.

#Therefore the height of the shadow as a function of the above is s=(15-x )m

Hence, height of the shadow is expressed as s=(15-x)m

#See attached photo for illustration

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Things that are moving out of phase create an interference pattern<br> TURE or False??
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A satellite is in a circular orbit about the earth (ME = 5.98 x 10^24 kg). The period of the satellite is 1.26 x 10^4 s. What is
Soloha48 [4]

Answer: V=5839.051m/s  

Explanation:

According to the <u>Third Kepler’s Law</u> of Planetary motion:

T^{2}=\frac{4\pi^{2}}{GM}a^{3}   (1)

Where;:

T=1.26(10)^{4}s is the period of the satellite

G is the Gravitational Constant and its value is 6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}

M=5.98(10)^{24}kg is the mass of the Earth

a  is the semimajor axis of the orbit the satllite describes around the Earth (as we know it is a circular orbit, the semimajor axis is equal to the radius of the orbit).

On the other hand, the orbital velocity V is given by:

V=\sqrt{\frac{GM}{a}}   (2)

Now, from (1) we can find a, in order to substitute this value in (2):

a=\sqrt[3]{\frac{T^{2}GM}{4\pi}^{2}}   (3)

a=\sqrt[3]{\frac{(1.26(10)^{4}s)^{2}(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.98(10)^{24}kg)}{4\pi}^{2}}   (4)

a=11705845.57m   (5)

Substituting (5) in (2):

V=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.98(10)^{24}kg)}{11705845.57m}}   (6)

V=5839.051m/s   (7)  This is the speed at which the satellite travels

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Feathers and a bowling ball are dropped in a vacuum, airless environment. Which one will hit the ground first?
Reil [10]

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