This would be typical of an elastic collision.
Weight = mg, g ≈ 9.8 m/s²
Weight = 2.2 * 9.8 ≈ 21.56 N
The cliff is 2042 ft away.
We know that the speed of sound in air is directly proportional to the absolute temperature.
First convert the Fahrenheit temperature to Celsius;
°C = 5/9(44.5 - 32)
°C = 6.9 °C
Applying the formula;
V1/V2 = √T1/T2
Where; V1 = velocity of sound in air at 0°C
V2 = Velocity of sound in air at 6.9 °C
1087/V2 = √273/279.9
V2= 1101 ft/s
Given that; V = 2s/t
Where s is the distance of the cliff
t is the time taken
1101 ft/s = 2s/3.71 s
s = 1101 ft/s × 3.71 s/2
s = 2042 ft
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A sample of nitrogen gas has a volume of 5.0 ml at a pressure of 1.50 atm. what is the pressure exerted by the gas if the volume increases to 30.0 ml, at constant temperature is 0.25atm.
On constant temperature, the pressure and volume relation become constant before and after the change in quantitities have occurred.
According to Boyle's Law,
P₁V₁ = P₂V₂
where, P₁ is pressure exerted by the gas initially
V₁ is the volume of gas initially
P₂ is pressure exerted by the gas finally
V₂ is the volume of gas finally
Given,
P₁ = 1.5 atm
V₁ = 5 ml
V₂ = 30 ml
P₂ =?
On substituting the given values in the above equation:
P₁V₁ = P₂V₂
1.5 atm × 5 ml = P₂ × 30 ml
P₂ = 0.25 atm
Hence, pressure exerted by the gas is 0.25atm.
Learn more about Boyle's Law here, brainly.com/question/1437490
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Answer: 1.91*10^8 N/m²
Explanation:
Given
Radius of the steel, R = 10 mm = 0.01 m
Length of the steel, L = 80 cm = 0.8 m
Force applied on the steel, F = 60 kN
Stress on the rod, = ?
Area of the rod, A = πr²
A = 3.142 * 0.01²
A = 0.0003142
Stress = Force applied on the steel/Area of the steel
Stress = F/A
Stress = 60*10^3 / 0.0003142
Stress = 1.91*10^8 N/m²
From the calculations above, we can therefore say, the stress on the rod is 1.91*10^8 N/m²
