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marin [14]
3 years ago
10

A 0.5 kg rock is dropped from a height of 1.0 m above the ground. Approximately how much kinetic energy will be stored in the ro

ck after it has fallen halfway to the ground.
Physics
1 answer:
irina1246 [14]3 years ago
8 0

Answer:

2.45 J

Explanation:

The following data were obtained from the question:

Mass (m) = 0.5 kg

Height (h) = 1 m

Kinetic energy (KE) =?

Next, we shall determine the velocity of the rock after it has fallen half way. This can be obtained as follow:

Initial velocity (u) = 0 m/s

Acceleration due to gravity (g) = 9.8 m/s²

Height (h) = 1/2 = 0.5 m

Final velocity (v) =?

v² = u² + 2gh

v² = 0² + (2 × 9.8 × 0.5)

v² = 9.8

Take the square root of both side

v = √9.8

v = 3.13 m/s

Finally, we shall determine the kinetic energy of the rock after it has fallen half way. This can be obtained as follow:

Mass (m) = 0.5 kg

Velocity (v) = 3.13 m/s

Kinetic energy (KE) =?

KE = ½mv²

KE = ½ × 0.5 × 3.13²

KE = 0.25 × 9.8

KE = 2.45 J

Therefore, the kinetic energy of the rock after it has fallen half way is 2.45 J

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Answer:

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Explanation:

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V = 10,000 cubic feet

also we know that total time is

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Q = \frac{10000}{120}

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A brave but inadequate rugby player is being pushed backward by an opposing player who is exerting a force of 800.0 N on him. Th
ipn [44]

Answer:

a) F_{fric} = 692 N

b) F_{applied} = 932 N

Explanation:

a)

According to newton's second law of motion, acceleration of an object is directly proportional to the net force acting on it. When there is no net force force acting on the body, there is no acceleration. A force is a push or a pull, and the net force ΣF is the total force, or sum of the forces exerted on an object  in all directions.

F_{net}  ∝ a

F_{net} =  ma

F_{applied} - F_{fric} = ma

Given data:

F_{applied} = 800 N

Mass = m = 90 kg

acceleration = a = 1.2 m/s²

F_{fric} = ?

800 - F_{fric} = (90)(1.2)

F_{fric} = 692 N

b)

According to newton's second law of motion,

F_{net}  ∝ a

F_{net} =  ma

F_{applied} - F_{fric} = ma

Given data:

If we assume the same friction and acceleration between player's feet and ground as calculated in part a

F_{fric} = 692 N

acceleration = a = 1.2 m/s²

We take the equal mass to the total mass of both the players because when the winning player push losing player backward, he exert force on the ground not only due to his mass but also due to the mass of losing player.

Mass = M = m₁ + m₂ = 110 kg + 90 kg

= 200 kg

F_{applied} = ?

F_{applied} - 692 N = (200)(1.2)

F_{applied} = 692 + 240

F_{applied} = 932 N

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In the case of Venus, he observed that it presented phases (such as those of the moon) together with a variation in size; observations that are only compatible with the fact that Venus rotates around the Sun and not around Earth.

This is because Venus presented its smaller size when it is in full phase and the largest size when it is in the new one, when it is between the Sun and the Earth.

These images along with other discoveries were presented to the Catholic Church (which supported the <u>geocentric theory</u> for that time) as a proof that completely refutes Ptolemy's geocentric system and affirms <u>Copernicus' heliocentric theory.</u>

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Liula [17]

Answer:

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