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Galina-37 [17]
3 years ago
14

How Negative Points Worksbtw help me with thisHow much does the sky weigh?​

Physics
2 answers:
DiKsa [7]3 years ago
6 0

Answer:

This is a dumb question.

Explanation:

The 'sky' does not weigh anything.

kipiarov [429]3 years ago
4 0

Answer:

it weighs 237469812734t7162341873498273417234321476281736481273648123764812736481723648273648137468127364872364 million pounds :)

Explanation:

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A light, rigid rod is 55.8 cm long. Its top end is pivoted on a frictionless horizontal axle. The rod hangs straight down at res
VMariaS [17]

To solve this problem we will apply the principle of conservation of energy. For this purpose, potential energy is equivalent to kinetic energy, and this clearly depends on the position of the body. In turn, we also note that the height traveled is twice that of the rigid rod, therefore applying these concepts we will have

KE = PE

\frac{1}{2} mv^2 = mgh

v = \sqrt{2gh}

v = \sqrt{2(9.8)(2(55.8*10^{-2}))}

v = 4.67m/s

Therefore the minimum speed at the bottom is required to make the ball go over the top of the circle is 4.67m/s

4 0
3 years ago
A plane accelerates from 25 m/s to a velocity of 80 m/s in a time of 5 seconds. What
Natali [406]

Answer:

-7

Explanation:

-7 because -7 said so

6 0
3 years ago
A rocket passing the earth with high speed will look shorter for the stationary observer measuring the size on earth than travel
neonofarm [45]

Answer:

True

Explanation:

This can be explained by the special theory of relativity for length contraction.

Length contraction is observed in the direction of motion of an object when an object moves with speed closer to the speed of light.

The length of the rocket in this case, appears shorter to the observer on earth in the stationary reference frame which is improper frame whereas the traveler in the rocket is in the same inertial frame which is proper for the rocket's size measurement.

5 0
3 years ago
How much work is required to lift a 2 kg mass to a height of 10 meters?
scoray [572]
Use the equation potential energy =m*g*h
m-mass,
g-gravitational acceleration,
h-height
Potential energy = 2*10*10
=200
kg {m}^{2}   {s}^{ - 2}
This is the unit of the energy
6 0
3 years ago
A ball is dropped from a height of 20m and renounce with the velocity which is 3/4 of d velocity with which it hits the ground.
Mashutka [201]

Answer:

1.73 seconds

Explanation:

The velocity the ball first hits the ground with is:

v² = v₀² + 2aΔx

v² = (0 m/s)² + 2 (-10 m/s²) (-20 m)

v = -20 m/s

The velocity it rebounds with is 3/4 of that in the opposite direction, or 15 m/s.

The time it takes to return to the ground is:

Δx = v₀ t + ½ at²

0 = (15 m/s) t + ½ (-10 m/s²) t²

0 = t (15 − 5t²)

t = √3

t ≈ 1.73 seconds

3 0
3 years ago
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