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Alexus [3.1K]
2 years ago
5

Rock X is released from rest at the top of a cliff that is on Earth. A short time later, Rock Y is released from rest from the s

ame location as Rock X. Both rocks fall for several seconds before landing on the ground directly below the cliff. Frictional forces are considered to be negligible. After Rock Y is released from rest several seconds after Rock X is released from rest, what happens to the separation distance S between the rocks as they fall but before they reach the ground, and why? Take the positive direction to be downward.
Physics
1 answer:
babunello [35]2 years ago
7 0

Answer:

Separation increases at all times that rock X falls because it falls with a greater speed

Explanation:

For both rocks, let initial velocity ∪=0

To find the displacement at any given time interval of Δt then

S= ∪Δt +0.5gΔt²

Since rock X is first released followed by Y, then X has a greater speed than Y therefore the distance covered by X is longer. This is because despite 0.5gΔt² being same for both rocks at any time Δt but rock X having already attained some velocity, its ∪Δt  is more hence the separation S increases. Conclusively, S increases at all times that rock X falls since rock X falls with a greater velocity than rock Y

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A body is projected upward at an angle of 30 degree to the horizontal at an initial speed of 200ms-.In how many seconds will it
Crazy boy [7]

Answer:

20.41 s

3534.80 m

Explanation:

<em><u>In how many seconds will it reach the ground?</u></em>

We are given the initial velocity of the body, which is 200 m/s at a 30° angle.

We know the acceleration in the vertical direction is -9.8 m/s², assuming that the upwards/right direction is positive and the downwards/left direction is negative.

Since we are using acceleration in the y-direction, let's use the vertical component of the initial velocity.

  • 200 · sin(30) m/s

Let's use the fact that at the top of its trajectory, the body will have a final velocity of 0 m/s.

Now we have one missing variable that we are trying to solve for: time t.

Find the constant acceleration equation that contains v₀, v, a, and t.

  • v = v₀ + at

Substitute known values into the equation.

  • 0 = 200 · sin(30) + (-9.8)t
  • -200 · sin(30) = -9.8t
  • t = 10.20408163

Recall that this is only half of the body's trajectory, so we need to double the time value we found to find the total time the body is in the air.

  • 2t = 20.40816327

The body will reach the ground in 20.41 seconds.

<em><u>How far from the point of projection would it strike? </u></em>

We want to find the displacement in the x-direction for the body.

Let's find the constant acceleration equation that contains time t, that we just found, and displacement (Δx).

  • Δx = v₀t + 1/2at²

Substitute known values into the equation. Remember that we want to use the horizontal component of the initial velocity and that the acceleration in the x-direction is 0 m/s².

  • Δx = (200 · cos(30) · 20.40816327) + 1/2(0)(20.40816327)²
  • Δx = 3534.797567

The body will strike 3534.80 m from the point of projection.

4 0
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Thus, the pressure= 6800/21.19
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A 4kg bird has 8 joules of kinetic energy, how fast is it flying?
Igoryamba
I believe the answer is 2m/s
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2 years ago
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The temperature of a heat engine is 500k some of the heat generated by the engine flows to the surroundings which are at a temp
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1-125/500)x100=efficiency

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1 year ago
The coefficient of friction between the 2-lb block and the surface is μ=0.2. The block has an initial speed of Vβ =6 ft/s and is
Taya2010 [7]

Answer:

x = 0.0685 m

Explanation:

In this exercise we can use the relationship between work and energy conservation

            W = ΔEm

Where the work is

             W = F x

The energy can be found in two points

Initial. Just when the block with its spring spring touches the other spring

           Em₀ = K = ½ m v²

Final. When the system is at rest

            Em_{f} = K_{e1}b +K_{e2} = ½ k₁ x² + ½ k₂ x²

We can find strength with Newton's second law

            ∑ F = F - fr

Axis y

           N- W = 0

           N = W

The friction force has the equation

          fr = μ N

          fr = μ W

  The job

         W = (F – μ W) x

We substitute in the equation

            (F - μ W) x = ½ m v² - (½ k₁ x² + ½ k₂ x²)

           ½ x² (k₁ + k₂) + (F - μ W) x - ½ m v² = 0

We substitute values ​​and solve

           ½ x² (20 + 40) + (15 -0.2 2) x - ½ (2/32) 6² = 0

         x² 30 + 14.4 x - 1,125 = 0

        x² + 0.48 x - 0.0375 = 0

           

We solve the second degree equation

        x = [-0.48 ±√(0.48 2 + 4 0.0375)] / 2

        x = [-0.48 ± 0.617] / 2

        x₁ = 0.0685 m

        x₂ = -0.549 m

The first result results from compression of the spring and the second torque elongation.

The result of the problem is x = 0.0685 m

4 0
2 years ago
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