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pentagon [3]
3 years ago
10

You drop a rock down a well that is 5.4 m deep. How long does it take the rock to hit the bottom of the well?

Physics
2 answers:
denis23 [38]3 years ago
5 0

Answer:

it is 1.05 as the other person answered  :)

Explanation:

Natali5045456 [20]3 years ago
4 0
Equation of motion:

y_{f}=y_{o}+v_{o}t+ \frac{1}{2} at^{2}

Since initial velocity is zero, the second term goes away:

y_{f}=y_{o}+0+ \frac{1}{2} at^{2}

y_{f}=y_{o}+\frac{1}{2} at^{2}

y_{f}-y_{o}= \frac{1}{2} at^{2}

y_{f}-y_{o}=5.4m

5.4m= \frac{1}{2} at^{2}

\frac{2(5.4m) }{a} = t^{2}

a = g = 9.81  \frac{m}{ s^{2}}

\frac{2(5.4m) }{9.81 \frac{m}{ s^{2} } } = t^{2}

1.1 s^{2} = t^{2}\sqrt{1.1 s^{2}} =  \sqrt{t^{2}}

<u><em>t = 1.05s</em></u>
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ow long must a simple pendulum be if it is to make exactly ten swings per second? (That is, one complete vibration takes exactly
Igoryamba
The period T of a pendulum is given by:
T=2 \pi  \sqrt{ \frac{L}{g} }
where L is the length of the pendulum while g=9.81 m/s^2 is the gravitational acceleration.

In the pendulum of the problem, one complete vibration takes exactly 0.200 s, this means its period is T=0.200 s. Using this data, we can solve the previous formula to find L:
L=g ( \frac{T}{2\pi} )^2=(9.81 m/s^2)( \frac{0.2 s}{2 \pi} )^2=1 \cdot 10^{-3} m=1 mm
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What os the term for the smallest bit into which chemical substance can be divided and still have the properties of that substan
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Here's an example:

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3 years ago
Which was an idea use to distribute the theory of plate tectonics
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8 0
4 years ago
According to Hooke's law
SOVA2 [1]

Answer:

According to Hook's law, we know,

strain/stress =Constant

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7 0
3 years ago
A sealed tank containing seawater to a height of 12.8 m also contains air above the water at a gauge pressure of 2.90 atm . Wate
exis [7]

Answer:

The velocity of water at the bottom, v_{b} = 28.63 m/s

Given:

Height of water in the tank, h = 12.8 m

Gauge pressure of water, P_{gauge} = 2.90 atm

Solution:

Now,

Atmospheric pressue, P_{atm} = 1 atm = 1.01\tiems 10^{5} Pa

At the top, the absolute pressure, P_{t} = P_{gauge} + P_{atm} = 2.90 + 1 = 3.90 atm = 3.94\times 10^{5} Pa

Now, the pressure at the bottom will be equal to the atmopheric pressure, P_{b} = 1 atm = 1.01\times 10^{5} Pa

The velocity at the top, v_{top} = 0 m/s, l;et the bottom velocity, be v_{b}.

Now, by Bernoulli's eqn:

P_{t} + \frac{1}{2}\rho v_{t}^{2} + \rho g h_{t} = P_{b} + \frac{1}{2}\rho v_{b}^{2} + \rho g h_{b}

where

h_{t} -  h_{b} = 12.8 m

Density of sea water, \rho = 1030 kg/m^{3}

\sqrt{\frac{2(P_{t} - P_{b} + \rho g(h_{t} - h_{b}))}{\rho}} =  v_{b}

\sqrt{\frac{2(3.94\times 10^{5} - 1.01\times 10^{5} + 1030\times 9.8\times 12.8}{1030}} =  v_{b}

v_{b} = 28.63 m/s

5 0
4 years ago
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