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den301095 [7]
3 years ago
15

How much power does a 2000 kg car need to accelerate from 20 m/s to 35 m/s in 7 seconds?

Physics
1 answer:
Alexus [3.1K]3 years ago
3 0

firstly you get your acceleration with the formula, a=v-u/t. Then you use the formula for kinetic energy 1/2mv^2

then you can finally get the answer for power by dividing your previous answer by the time

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If you increase the length ? of a pendulum by a factor of 9, how will the period t increase?
garik1379 [7]
T = 2 * pie √(L/g)

so, if length is increased by 9

then time period is increased by √9 = 3

hope it helped :)
3 0
3 years ago
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When using the right-hand rule to determine the direction of the magnetic field around a current-carrying wire, which part of th
Margarita [4]
Your answer will be fingers

5 0
3 years ago
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A spacecraft is in a circular Earth orbit at an altitude of 6000 km . By how much will its altitude decrease if it moves to a ne
blagie [28]

Answer:

a) 2148 km = 2150 km

b) 840 km

Explanation:

The force keeping the satellite in circular motion is the force given by Netwon's gravitational law

Centripetal force = (mv²/r)

Force due to Newton's law of gravitation = (GMm/r²)

where m = mass of satellite

M = mass of the earth

G = Gravitational constant

v = velocity of the satellite

r = radius of circular orbit

(mv²/r) = (GMm/r²)

v² = (GM/r)

Meaning that the square of the velocity of orbit is inversely proportional to the radius of circular orbit. (Since G and M are constants)

v² = (k/r)

when v = v₀, r = 6000 + 6378 = 12,378 km (the radius of orbit = 6000 km + radius of the earth)

v₀² = (k/12,378)

K = 12378v₀²

When the velocity increases by 10%, v₁ = 1.1v₀, the square of the new velocity = (1.1v₀)² = 1.21v₀² and the new radius of orbit = r₁

1.21v₀² = (k/r₁)

r₁ = (k/1.21v₀²)

Recall, k = 12378v₀²

r₁ = 12378v₀² ÷ 1.21v₀² = 10,229.75 km

10,229.75 km = (10,229.75 - 6378) km altitude above the Earth's surface

New altitude of orbit = 3851.75 km

Decrease in altitude = 6000 - 3851.75 = 2148 km

b) The period of orbit is related to the radius of orbit through Kepler's Law

T² ∝ R³

T² = kR³

When the period of orbit is T₀, Radius of orbit = R₀ = (6000 + 6378) = 12378 km (Earth's radius = 6378 km)

T₀² = kR₀³

T₀² = k(12378)³

k = (T₀²) ÷ (12378)³

When the period reduces by 10%, T₁ = 0.90T₀ and the new radius of orbit = R₁

T₁² = kR₁³

(0.90T₀)² = kR₁³

0.81T₀² = kR₁³

R₁³ = (0.81T₀²) ÷ k

Recall, k = (T₀²)/(12378)³

R₁³ = (0.81T₀²) ÷ [(T₀²)/(12378)³]

R₁³ = 1,536,160,005,663.1

R₁ = ∛(1,536,160,005,663.1) = 11,538.4 km

New Altitude = R₁ - (Radius of the Earth)

= 11,538.4 - 6378 = 5160.4 km

Decrease in altitude = 6000 - 5160.4 = 839.6 km = 840 km

Hope this Helps!!!

3 0
3 years ago
Hello please help i’ll give brainliest
Zarrin [17]

Answer:

C because it is in earth's mantle

3 0
3 years ago
What real-world examples show no work begin done? Can you think of examples other than resisting the force of gravity?
irga5000 [103]
Oh my gosh !  Resisting the force of gravity always DOES involve doing work.
If no work is being done, then you're NOT resisting the force of gravity.

Example:

-- ball rolling on the floor . . . no work
-- ball rolling up a ramp . . . work being done
-- ball rolling down a ramp . . . work being done, BY gravity

6 0
3 years ago
Read 2 more answers
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