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melomori [17]
3 years ago
7

An object is dropped near the earth's surface. At the end of 4.0 s, how fast is it falling in m/s? (Ignore air resistance.)

Physics
2 answers:
Sati [7]3 years ago
7 0
For free falling bodies, the final velocity may be calculated through the equation,
                                   Vf = gt
Where g is the acceleration due to gravity (9.8 m/s²) and t is the time elapsed. Substituting the known values,
                                   Vf = (9.8 m/s²) x (4 s) = 39.2 m/s
Therefore, the object's velocity is approximately 39.2 m/s. 
DochEvi [55]3 years ago
4 0

The answer would be 39.2 m/s

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You have a 3.00-liter container filled with N₂ at 25°C and 4.45 atm pressure connected to a 2.00-liter container filled with Ar
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Answer : The final pressure in the two containers is, 2.62 atm

Explanation :

Boyle's Law : It is defined as the pressure of the gas is inversely proportional to the volume of the gas at constant temperature and number of moles.

P\propto \frac{1}{V}

Thus, the expression for final pressure in the two containers will be:

PV=P_1V_1+P_2V_2

P=\frac{P_1V_1+P_2V_2}{V}

where,

P_1 = pressure of N₂ gas = 4.45 atm

P_2 = pressure of Ar gas = 2.75 atm

V_1 = volume of N₂ gas = 3.00 L

V_2 = volume of Ar gas = 2.00 L

P = final pressure of gas = ?

V = final volume of gas = (4.45 + 2.75) L = 7.2 L

Now put all the given values in the above equation, we get:

P=\frac{(4.45atm)\times (3.00L)+(2.75atm)\times (2.00L)}{7.2L}

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8 0
3 years ago
If viewed at the correct time and in the right conditions, asteroids, comets, and meteors are all visible to the nked eye.
UkoKoshka [18]

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True

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2 years ago
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Answer:

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Explanation:

Answer:

Explanation:

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6 0
4 years ago
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