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Mice21 [21]
3 years ago
12

A coin is thrown with a velocity of 0 m/s down a dry well and hits bottom in 1.2s, what’s the depth of the well?

Physics
1 answer:
pentagon [3]3 years ago
5 0

Answer:

The well is 7.1 meters deep.

Explanation:

The formula to use here is the distance in a uniformly accelerated motion:

d = \frac{1}{2}at^2+v_0t+d_0

where d stands for distance, t for time, a for acceleration, v0 and d0 for initial velocity and distance, respectively. Since the initial distance and velocity are both zero, we are left with the first term. The coin is in free fall and so it is accelerated by gravity:

d = \frac{1}{2}at^2= \frac{1}{2}gt^2=\frac{1}{2}9.8\frac{m}{s^2}1.2^2s^2=7.1m

The well is 7.1 meters deep.

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3 years ago
A batter hits a pop fly, and the baseball (with a mass of 148 g) reaches an altitude of 265 ft. If we assume that the ball was 3
den301095 [7]

Answer:

The increase in potential energy of the ball is 115.82 J

Explanation:

Conceptual analysis

Potential Energy (U) is the energy of a body located at a certain height (h) above the ground and is calculated as follows:

U = m × g × h

U: Potential Energy in Joules (J)

m: mass in kg

g: acceleration due to gravity in m/s²

h: height in m

Equivalences

1 kg = 1000 g

1 ft = 0.3048 m

1 N = 1 (kg×m)/s²

1 J = N × m

Known data

h_2 = 265ft * \frac{0.3048m}{ft} = 80.77m

h_1 = 3ft * \frac{0.3048m}{ft} = 0.914m

m = 148g*\frac{1kg}{1000g} = 0.148kg

g = 9.8 \frac{m}{s^2}

Problem development

ΔU: Potential energy change

ΔU = U₂ - U₁

U₂ - U₁ = mₓgₓh₂ - mₓgₓh₁

U₂ - U₁ = mₓg(h₂ - h₁)

U_2 - U_1 = 0.148kg * 9.8 \frac{m}{s^2}*(80.77m - 0.914m) = 115.82 N * m = 115.82J

The increase in potential energy of the ball is 115.82 J

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Answer:

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