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Leokris [45]
3 years ago
6

A soccer ball is released from rest at the top of a grassy incline. After 4.1 seconds, the ball travels 43 meters and 1.0 s afte

r this, the ball reaches the bottom of the incline. What was the magnitude of the ball's acceleration, assume it to be constant? Express your answer using two significant figures.
Physics
1 answer:
gavmur [86]3 years ago
5 0

Answer:

The acceleration of the ball as it moves down the grassy incline, if constant, is 5.1 m/s²

Explanation:

Initial velocity, u = 0m/s

Acceleration = ?

Vertical distance covered as at t=4.1s, H = 43m

Using the equations of motion,

H = ut + 0.5at²

43 = 0 + 0.5a (4.1)²

a = 43/(0.5×4.1²) = 5.1 m/s²

Hope this Helps!!

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Answer:

1.62 L

Explanation:

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\frac{V_1}{T_1}=\frac{V_2}{T_2}

where in this problem we have

V_1 = 1.75 L is the initial volume of the gas

T_1 = 26.0^{\circ}+273=299 K is the initial temperature of the gas

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Solving the equation for V2, we find

V_2 = \frac{V_1}{T_1} T_2 = \frac{1.75 L}{299 K}(276 K)=1.62 L

8 0
3 years ago
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Answer:

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Explanation:

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An ideal gas, initially at a pressure of 11.2 atm and a temperature of 299 K, is allowed to expand adiabatically until its volum
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Answer:

The pressure is  P_2  =  4.25 \ a.t.m

Explanation:

From the question we are told that

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Let the first volume be  V_1 Then the final volume will be  2 V_1

 Generally for a diatomic  gas

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Here r is the radius of the molecules which is  mathematically represented as

    r =  \frac{C_p}{C_v}

Where C_p \  and\   C_v are the molar specific heat of a gas at constant pressure and  the molar specific heat of a gas at constant volume with values

     C_p=7 \  and\   C_v=5

=>   r =  \frac{7}{5}

=>  11.2*( V_1 ^{\frac{7}{5} } ) =  P_2  *  (2 V_1 ^{\frac{7}{5} } )

=>   P_2  =  [\frac{1}{2} ]^{\frac{7}{5} } * 11.2

=>  P_2  =  4.25 \ a.t.m

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