A bullet starting from rest accelerates uniformly at a rate of 1,250 meters per square second. What is the bullet's speed after
1 answer:
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Answer:
a) I = 2279.5 N s , b) F = 3.80 10⁵ N, c) I = 3125.5 N s and d) F = 5.21 10⁵ N
Explanation:
The impulse is equal to the variation in the amount of movement.
I =∫ F dt = Δp
I = m - m v₀
Let's calculate the final speed using kinematics, as the cable breaks the initial speed is zero
² = V₀² - 2g y
² = 0 - 2 9.8 30.0
= √588
= 24.25 m/s
a) We calculate the impulse
I = 94 24.25 - 0
I = 2279.5 N s
b) Let's join the other expression of the impulse to calculate the average force
I = F t
F = I / t
F = 2279.5 / 6 10⁻³
F = 3.80 10⁵ N
just before the crash the passenger jumps up with v = 8 m / s, let's take the moments of interest just when the elevator arrives with a speed of 24.25m/s down and as an end point the jump up to vf = 8 m / n
c) I = m - m v₀
I = 94 8 - 94 (-24.25)
I = 3125.5 N s
d) F = I / t
F = 3125.5 / 6 10⁻³
F = 5.21 10⁵ N
There are some missing data in the text of the exercise. Here the complete text:
"<span>A sample of 20.0 moles of a monatomic ideal gas (γ = 1.67) undergoes an adiabatic process. The initial pressure is 400kPa and the initial temperature is 450K. The final temperature of the gas is 320K. What is the final volume of the gas? Let the ideal-gas constant R = 8.314 J/(mol • K). "
Solution:
First, we can find the initial volume of the gas, by using the ideal gas law:
</span>
<span>where
p is the pressure
V the volume
n the number of moles
R the gas constant
T the absolute temperature
Using the initial data of the gas, we can find its initial volume:
</span>
<span>
Then the gas undergoes an adiabatic process. For an adiabatic transformation, the following relationship between volume and temperature can be used:
</span>
<span>where </span>
for a monoatomic gas as in this exercise. The previous relationship can be also written as
where i labels the initial conditions and f the final conditions. Re-arranging the equation and using the data of the problem, we can find the final volume of the gas:
![V_f = V_i \sqrt[\gamma-1]{ \frac{T_i}{T_f} }=(0.187 m^3) \sqrt[0.67]{ \frac{450 K}{320 K} }=0.310 m^3 = 310 L](https://tex.z-dn.net/?f=V_f%20%3D%20V_i%20%20%5Csqrt%5B%5Cgamma-1%5D%7B%20%5Cfrac%7BT_i%7D%7BT_f%7D%20%7D%3D%280.187%20m%5E3%29%20%5Csqrt%5B0.67%5D%7B%20%5Cfrac%7B450%20K%7D%7B320%20K%7D%20%7D%3D0.310%20m%5E3%20%3D%20310%20L%20%20)
So, the final volume of the gas is 310 L.
"<span>A sample of 20.0 moles of a monatomic ideal gas (γ = 1.67) undergoes an adiabatic process. The initial pressure is 400kPa and the initial temperature is 450K. The final temperature of the gas is 320K. What is the final volume of the gas? Let the ideal-gas constant R = 8.314 J/(mol • K). "
Solution:
First, we can find the initial volume of the gas, by using the ideal gas law:
</span>
<span>where
p is the pressure
V the volume
n the number of moles
R the gas constant
T the absolute temperature
Using the initial data of the gas, we can find its initial volume:
</span>
<span>
Then the gas undergoes an adiabatic process. For an adiabatic transformation, the following relationship between volume and temperature can be used:
</span>
<span>where </span>
where i labels the initial conditions and f the final conditions. Re-arranging the equation and using the data of the problem, we can find the final volume of the gas:
So, the final volume of the gas is 310 L.