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astraxan [27]
3 years ago
13

The international space station (ISS) orbits the earth with a velocity 7.6 km/s. How much energy is required to lift a payload o

f 2500 kg from the surface of the earth to the ISS
Physics
1 answer:
Dmitry [639]3 years ago
4 0

Answer:

7.22×10¹⁰ J

Explanation:

From the question given above, the following data were obtained:

Velocity (v) = 7.6 Km/s

Mass (m) = 2500 kg

Energy (E) =?

Next, we shall convert 7.6 Km/s to m/s. This can be obtained as follow:

1 km/s = 10³ m/s

Therefore,

7.6 Km/s = 7.6 Km/s × 10³ m/s / 1 km/s

7.6 Km/s = 7.6×10³ m/s

Finally, we shall determine the energy required as follow:

Velocity (v) = 7.6×10³ m/s

Mass (m) = 2500 kg

Energy (E) =?

E = ½mv²

E = ½ × 2500 × (7.6×10³)²

E = 1250 × 57760000

E = 7.22×10¹⁰ J

Thus, 7.22×10¹⁰ J of energy is required.

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Answer:

The total distance is 381.5 [m]

Explanation:

In order to solve this problem we must use the expressions of kinematics. The clue to solve this problem is that the motorcyclist starts from rest, i.e. its initial speed is zero.

v_{f} =v_{o} +(a*t)

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Vf = final velocity [m/s]

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Vf = 0 + (2*7)

Vf = 14 [m/s]

With this velocity, we can calculate the displacement using the following expression.

v_{f} ^{2} =v_{o} ^{2} +2*a*x

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14² = 0 + (2*7*x)

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The other important clue to solve this problem in the second part is that the final velocity is now the initial velocity.

We must calculate the final velocity.

v_{f}= v_{i} +(a*t)

Vf = final velocity [m/s]

Vi = initial velocity = 14 [m/s]

a = desacceleration = 4 [m/s²]

t = time = 8 [s]

Vf = 24 + (4*8)

Vf = 56 [m/s]

With this velocity, we can calculate the displacement using the following expression.

v_{f} ^{2} =v_{o} ^{2} +2*a*x

where

x = distance traveled [m]

56² = 14² + (2*4*x)

x = 2940/(8)

x = 367.5 [m]

Note: The positive sign in the equations is because the car is accelerating, it means its velocity is increasing.

Therefore the total distance is Xt = 14 + 367.5 = 381.5 [m].

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3 years ago
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