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matrenka [14]
3 years ago
7

Astronomersestimatethata2.0-km-diameterasteroidcollides with the Earth once every million years. The collision could pose a thre

at to life on Earth. (a) Assume a spherical asteroid has a mass of 3200 kg for each cubic meter of volume and moves toward the Earth at 15 km????s. How much destructive energy could be released when it embeds itself in the Earth? (b) For comparison, a nuclear bomb could release about 4.0 * 1016 J. How many such bombs would have to explode

Physics
2 answers:
tangare [24]3 years ago
7 0

Answer:

a) 1.6*10^21 Joules.

b) 40,000

Explanation:

part a )

maximum destructive energy that can be released is the case when all the kinetic energy of the asteroid is consumed.

therefore E = 1/2 m v^2

m= density * volume

= 3100* (4/3* pi * 1000^3 ) = 12978666666666.67 kg

given v = 16000m/s

therefore

E= 1/2 * 12978666666666.67 * 16000 * 16000

= 1.6 x 10^21 Joules!

part B)

each bomb is capable of 4 x 10^16 joules

therefore no of bombs that are needed to produce the required energy are

1.6 x 10^21 / 4 x 10^16 = 40,000

that is 40,000 such nuclear bombs are required!

hichkok12 [17]3 years ago
6 0

Explanation:

Below is an attachment containing the solution.

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Answer: (a)F=7(10)^{-7}N

              (b)F=1.344(10)^{-6}N  

              (c) The force of Jupiter on the baby is slightly greater than the the force of the father on the baby.

Explanation:

According to the law of universal gravitation, which is a classical physical law that describes the gravitational interaction between different bodies with mass:

F=G\frac{m_{1}m_{2}}{r^2}   (1)

Where:

F is the module of the force exerted between both bodies

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

m_{1} and m_{2} are the masses of both bodies.

r is the distance between both bodies

Knowing this, let's begin with the answers:

<h2 /><h2>(a) Gravitational force Father exertes on baby</h2>

Using equation (1) and taking into account the mass of the father m_{1}=100kg, the mass of the baby m_{2}=4.20kg and the distance between them r=0.2m, the force F_{F}  exerted by the father is:

F_{F}=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}\frac{(100kg)(4.20kg)}{(0.2m)^2}   (2)

F_{F}=0.0000007N=7(10)^{-7}N   (3)

<h2>(b) Gravitational force Jupiter exertes on baby</h2>

Using again equation (1) but this time taking into account the mass of Jupiter m_{J}=1.898(10)^{27}kg, the mass of the baby m_{2}=4.20kg and the distance between Jupiter and Earth (where the baby is) r_{E}=6.29(10)^{11}m, the force F_{J}  exerted by the Jupiter is:

F_{J}=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}\frac{(1.898(10)^{27}kg)(4.20kg)}{(6.29(10)^{11}m)^2}   (4)

F_{J}=0.000001344N=1.344(10)^{-6}N   (5)

<h2>(c) Comparison</h2>

Now, comparing both forces:

F_{J}=0.000001344N=1.344(10)^{-6}N   and F_{F}=0.0000007N=7(10)^{-7}N  we can see F_{J} is greater than F_{F}. However, the difference is quite small as well as the force exerted on the baby.

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