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elena55 [62]
4 years ago
15

1. Earth is approximately a sphere of radius 6.37х106 m. what are (a) its circumference in kilometers, (b) its surface are in sq

uare kilometers, and (c) its volume in cubic kilometers?
Physics
2 answers:
kvasek [131]4 years ago
7 0

Explanation:

Given that,

Radius of Earth, r=6.37\times 10^6\ m

(a) It is not possible to find the circumference of the sphere. But if we consider an image of the Earth, its circumference is given by :

C=2\pi r

C=2\pi \times 6.37\times 10^6

C=40.01\times 10^6\ m

C=40.01\times 10^3\ km

(b) Surface area of a sphere is given by :

A=4\pi r^2

A=4\pi (6.37\times 10^6)^2

A=5.09\times 10^{14}\ m^2

A=5.09\times 10^{11}\ km^2

(c) Volume of the sphere is given by :

V=\dfrac{4}{3}\pi r^3

V=\dfrac{4}{3}\pi (6.37\times 10^6)^3

V=1.082\times 10^{21}\ m^3

V=1.082\times 10^{18}\ km^3

Hence, this is the required solution.

Minchanka [31]4 years ago
5 0

As we know that circumference of the sphere is given as

C = 4\pi R

here we know that

R = 6.37 \times 10^6 m

now we have

C = 4\pi (6.37 \times 10^6)

C = 8\times 10^7 m

C = 8 \times 10^4 km

PART B)

surface area of the sphere is given as

A = 4\pi R^2

R = 6.37 \times 10^3 km

A = 4\pi (6.37\times 10^3)^2

A = 5.1 \times 10^8 km^2

PART C)

Volume of the sphere is given as

V = \frac{4}{3}\pi R^3

here we have

V = \frac{4}{3}\pi(6.37 \times 10^3)^3

V = 1.1 \times 10^{12} km^3

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In each case the momentum before the collision is: (2.00 kg) (2.00 m/s) = 4.00 kg * m/s
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Answer:

Check Explanation.

Explanation:

Momentum before collision = (2)(2) + (2)(0) = 4 kgm/s

a) Scenario A

After collision, Mass A sticks to Mass B and they move off with a velocity of 1 m/s

Momentum after collision = (sum of the masses) × (common velocity) = (2+2) × (1) = 4 kgm/s

Which is equal to the momentum before collision, hence, momentum is conserved.

Scenario B

They bounce off of each other and move off in the same direction, mass A moves with a speed of 0.5 m/s and mass B moves with a speed of 1.5 m/s

Momentum after collision = (2)(0.5) + (2)(1.5) = 1 + 3 = 4.0 kgm/s

This is equal to the momentum before collision too, hence, momentum is conserved.

Scenario C

Mass A comes to rest after collision and mass B moves off with a speed of 2 m/s

Momentum after collision = (2)(0) + (2)(2) = 0 + 4 = 4.0 kgm/s

This is equal to the momentum before collision, hence, momentum is conserved.

b) Kinetic energy is normally conserved in a perfectly elastic collision, if the two bodies do not stick together after collision and kinetic energy isn't still conserved, then the collision is termed partially inelastic.

Kinetic energy before collision = (1/2)(2.00)(2.00²) + (1/2)(2)(0²) = 4.00 J.

Scenario A

After collision, Mass A sticks to Mass B and they move off with a velocity of 1 m/s

Kinetic energy after collision = (1/2)(2+2)(1²) = 2.0 J

Kinetic energy lost = (kinetic energy before collision) - (kinetic energy after collision) = 4 - 2 = 2.00 J

Kinetic energy after collision isn't equal to kinetic energy before collision. This collision is evidently totally inelastic.

Scenario B

They bounce off of each other and move off in the same direction, mass A moves with a speed of 0.5 m/s and mass B moves with a speed of 1.5 m/s

Kinetic energy after collision = (1/2)(2)(0.5²) + (1/2)(2)(1.5²) = 0.25 + 3.75 = 4.0 J

Kinetic energy lost = 4 - 4 = 0 J

Kinetic energy after collision is equal to kinetic energy before collision. Hence, this collision is evidently elastic.

Scenario C

Mass A comes to rest after collision and mass B moves off with a speed of 2 m/s

Kinetic energy after collision = (1/2)(2)(0²) + (1/2)(2)(2²) = 4.0 J

Kinetic energy lost = 4 - 4 = 0 J

Kinetic energy after collision is equal to kinetic energy before collision. Hence, this collision is evidently elastic.

c) An impossible outcome of such a collision is that A stocks to B and they both move off together at 1.414 m/s.

In this scenario,

Kinetic energy after collision = (1/2)(2+2)(1.414²) = 4.0 J

This kinetic energy after collision is equal to the kinetic energy before collision and this satisfies the conservation of kinetic energy.

But the collision isn't possible because, the momentum after collision isn't equal to the momentum before collision.

Momentum after collision = (2+2)(1.414) = 5.656 kgm/s

which is not equal to the 4.0 kgm/s obtained before collision.

This is an impossible result because in all types of collision or explosion, the second law explains that first of all, the momentum is always conserved. And this evidently violates the rule. Hence, it is not possible.

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