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antoniya [11.8K]
3 years ago
6

A boy finds an abandoned mine shaft in the woods, and wants to know how deep the hole is. He drops in a stone, and counts 4 seco

nds before he hears the "plunk" of the stone hitting the bottom of the shaft. Approximately how deep is the shaft? (Assume a gravitational acceleration of 9.8 m/s2 and no significant air resistance.)
Physics
1 answer:
oee [108]3 years ago
4 0
S=(0x4)+(0.5x4.81x4x4)
S=0.78.48

The depth is approximately 78 meters.
(My brain hurts now) :P Good Luck!
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Explanation:

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c. is an electrical force

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3 years ago
To practice Problem-Solving Strategy 11.1 Equilibrium of a Rigid Body. A horizontal uniform bar of mass 2.7 kg and length 3.0 m
Zanzabum

Answer:

T_{1} = 14.88 N

Explanation:

Let's begin by listing out the given variables:

M = 2.7 kg, L = 3 m, m = 1.35 kg, d = 0.6 m,

g = 9.8 m/s²

At equilibrium, the sum of all external torque acting on an object equals zero

τ(net) = 0

Taking moment about T_{1} we have:

(M + m) g * 0.5L - T_{2}(L - d) = 0

⇒ T_{2} = [(M + m) g * 0.5L] ÷ (L - d)

T_{2} = [(2.7 + 1.35) * 9.8 * 0.5(3)] ÷ (3 - 0.6)

T_{2}= 59.535 ÷ 2.4

T_{2} = 24.80625 N ≈ 24.81 N

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Weight of monkey(w) = m * g = 1.35 * 9.8 = 13.23 N

Using sum of equilibrium in the vertical direction, we have:

T_{1} + T_{2} = W + w   ------- Eqn 1

Substituting T2, W & w into the Eqn 1

T_{1} + 24.81 = 26.46 + 13.23

T_{1} = <u>14.88</u> N

8 0
3 years ago
On the planet Xenophous a 1.00 m long pendulum on a clock has a period of 1.32 s. What is the free fall acceleration on Xenophou
Vilka [71]
The period of the pendulum is given by the following equation

T = 2π * sqrt (L/g)

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3 0
3 years ago
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