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Lena [83]
3 years ago
9

You drop a rock down a well that is 11.5 m deep.How long does it take the rock to hit the bottom of the well ?

Physics
1 answer:
ziro4ka [17]3 years ago
4 0
The acceleration of gravity depends on the distance from
the center of the Earth.  We use 9.8 m/s² for any place that's
on the Earth's surface, or reasonably close.  The 11.5 meters
down into the well certainly qualifies.

The formula we want is: 

                          Distance of fall = (1/2) · (acceleration) · (time)²

                                  11.5 m      = (1/2) (9.8 m/s²) (time)²

Divide each side
by  4.9 m/s² :            (11.5m) / (4.9 m/s²)  =  time²

                                         2.347 sec²        =  time²

                                   time = √(2.347 sec²)  =   1.53 seconds  .
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Read 2 more answers
Three moles of an ideal gas with a molar heat capacity at constant volume of 4.9 cal/(mol∙K) and a molar heat capacity at consta
Sergeu [11.5K]

Answer:

The heat flows into the gas during this two-step process is 120 cal.

Explanation:

Given that,

Number of moles = 3

Heat capacity at constant volume = 4.9 cal/mol.K

Heat capacity at constant pressure = 6.9 cal/mol.K

Initial temperature = 300 K

Final temperature = 320 K

We need to calculate the heat flow in to gas at constant pressure

Using formula of heat

\Delta H_{1}=nC_{p}\times\Delta T

Put the value into the formula

\Delta H_{1}=3\times6.9\times(320-300)

\Delta H_{1}=414\ cal

We need to calculate the heat flow in to gas at constant volume

Using formula of heat

\Delta H_{1}=nC_{v}\times\Delta T

Put the value into the formula

\Delta H_{1}=3\times4.9\times(300-320)

\Delta H_{1}=-294\ cal

We need to calculate the heat flows into the gas during two steps

Using formula of total heat

\Delta H_{T}=\Delta H_{1}+\Delta H_{2}

\Delta H_{T}=414-294

\Delta H_{T}=120\ cal

Hence, The heat flows into the gas during this two-step process is 120 cal.

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