Physics, 25.12.2020 02:00 mia3128

A tourist stands at the top of the Grand Canyon, holding a rock, overlooking the valley below. Find the final velocity, and displacement of the rock after 4.0 seconds give the three initial conditions listed below. a. The rock is dropped vertically from rest
b. The rock is tossed vertically upward with 8.0 m/s of speed
c. The rock is tossed vertically downward with 8.0 m/s of speed

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Answer from: ErrorNameTaken505

a. -39.2 m/s; -78.4 m

b. -31.2 m/s; -46.4 m

c. -47.2 m/s; -110.4 m

Explanation:

Part (a)

We are given/can infer these variables:

t = 4.0 sa = -9.8 m/s² v_0 = 0 m/s

We want to find the displacement and the final velocity of the rock.

Δx = ?v = ?

We can use this equation to find the final velocity:

v = v_0 + at

Plug in the known variables into this equation.

v = 0 + (-9.8)(4.0) v = -9.8 * 4.0 v = -39.2 m/s

The final velocity of the rock is -39.2 m/s.

Now we can use this equation to find the displacement of the rock:

Δx = v_0 t + 1/2at²

Plug in the known variables into this equation.

Δx = 0 * 4.0 + 1/2(-9.8)(4.0)² Δx = 1/2(-9.8)(4.0)² Δx = -4.9 * 16 Δx = -78.4 m

The displacement of the rock is -78.4 m.

Part (b)

We are given/can infer these variables:

v_0 = 8.0 m/sa = -9.8 m/s²t = 4.0 s

We can use this equation to find the final velocity:

v = v_0 + at

Plug in the known variables into this equation.

v = 8.0 + (-9.8)(4.0) v = 8.0 + -39.2 v = -31.2 m/s

The final velocity of the rock is -31.2 m/s.

We can use this equation to find the displacement:

Δx = v_0 t + 1/2at²

Plug in known variables:

Δx = 8.0(4.0) + 1/2(-9.8)(4.0)² Δx = 32 - 4.9(16)Δx = -46.4 m

The displacement of the rock is -46.4 m.

Part (c)

We are given/can infer these variables:

v_0 = -8.0 m/sa = -9.8 m/s²t = 4.0 s

We can use this equation to find the final velocity:

v = v_0 + at

Plug in the known variables into this equation.

v = -8.0 + (-9.8)(4.0) v = -8.0 - 39.2 v = -47.2 m/s

The final velocity of the rock is -47.2 m/s.

We can use this equation to find the displacement:

Δx = v_0 t + 1/2at²

Plug in known variables:

Δx = -8.0(4.0) + 1/2(-9.8)(4.0)²Δx = -32 - 4.9(16)Δx = -110.4 m

The displacement of the rock is -110.4 m.

Answer from: Quest

a small object of mass 3.66 g and charge -17.6 µc is suspended motionless above the ground when immersed in a uniform electric field perpendicular to the ground. what are the magnitude and direction of the electric field?

Answer from: Quest

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