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Luba_88 [7]
3 years ago
8

On a sky coaster (human pendulum) that reaches 10 meters from it's equilibrium position, a man of 120 kg is able to reach a maxi

mum speed of
18m/s

14m/s

1200m/s

11m/s
Physics
1 answer:
dimaraw [331]3 years ago
6 0

Answer:

14 m/s

Explanation:

Using the principle of conservation of energy, the potential energy is converted to kinetic energy, assuming any losses.

Kinetic energy is given by ½mv²

Potential energy is given by mgh

Where m is the mass, v is the velocity, g is acceleration due to gravity and h is the height.

Equating kinetic energy to be equal to potential energy then

½mv²=mgh

V

Making v the subject of the formula

v=√(2gh)

Substituting 9.81 m/s² for g and 10 m for h then

v=√(2*9.81*10)=14.0071410359145 m/s

Rounding off, v is approximately 14 m/s

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Pachacha [2.7K]

Answer:

Mass, m = 125 kg

Explanation:

Let us assume that the question says, "What is the mass of an object whose velocity is 400 m/s and the kinetic energy of 10⁷ J.

The kinetic energy of an object is :

K=\dfrac{1}{2}mv^2\\\\m=\dfrac{2K}{v^2}\\\\m=\dfrac{2\times 10^7}{(400)^2}\\\\m=125\ kg

So, the mass of the object is 125 kg.

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4 years ago
A 4000kg truck has a head-on inelastic collision with a 2500kg truck.
iogann1982 [59]

Answer:it could be B

Explanation:

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3 years ago
two forces A and B act on an object in opposite directions. A is bigger than B. What is the net force acting on an object?
Akimi4 [234]
Imagine a block with A pulling one way and B the other.

If A is bigger than B then obviously it will go in the direction of A.

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Read 2 more answers
A rocket, initially at rest on the ground, accelerates straight upward from rest with constant (net) acceleration 39.2 m/s^2. Th
Vanyuwa [196]

Answer:

9800 m

Explanation:

During acceleration, given:

v₀ = 0 m/s

a = 39.2 m/s²

t = 10.0 s

Find: v and Δy

v = at + v₀

v = (39.2 m/s²) (10.0 s) + 0 m/s

v = 392 m/s

Δy = v₀ t + ½ at²

Δy = (0 m/s) (10.0 s) + ½ (39.2 m/s²) (10.0 s)²

Δy = 1960 m

During free fall, given:

v₀ = 392 m/s

v = 0 m/s

a = -9.8 m/s²

Find: Δy

v² = v₀² + 2aΔy

(0 m/s)² = (392 m/s)² + 2 (-9.8 m/s²) Δy

Δy = 7840 m

Therefore, h = 1960 m + 7840 m = 9800 m.

4 0
3 years ago
Derive an equation that relates the initial release height hxhx of block xx and the speed vsvs of the two-block system after the
erik [133]

Answer:

1 point is earned for stating that the conservation of energy should be applied to this situation.

1 point is earned for stating that the conservation of momentum should be applied to this situation.

Example Response:

Students will need to use both conservation of momentum (for the collision) and conservation of energy (for the slide down the ramp) to be able to determine the relationship between the release height of block X and the speed at which the two-block system travels after they collide and stick together.

Explanation:

Please say thank you.

3 0
2 years ago
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