Answer: 0.687 m/s
Explanation:
Given
Mass of object, m = 76.7 g = 0.767 kg
Force constant of the spring, k = 3.34 N/m
Amplitude of the spring, x = 38 cm = 0.38 m
EPE(i) = KE(f) + EPE(f)
1/2kx(i)² = 1/2mv² + 1/2kx(f)²
1/2 * 3.34 * 0.38² = (1/2 * 0.767 * v²) + [1/2 * 3.34 * (0.38/2)²]
1/2 * 3.34 * 0.1444 = (1/2 * 0.767v²) + 1/2 * 3.34 * 0.0361
1/2 * 0.482 = 1/2 * 0.767v² + 1/2 * 0.121
0.241 = 1/2 * 0.767v² + 0.06
1/2 * 0.767v² = 0.181
0.767v² = 0.181 * 2
0.767v² = 0.362
v² = 0.362 / 0.767
v² = 0.472
v = √0.472
v = 0.687 m/s
Explanation:
It is given that initially pressure of ideal gas is 4.00 atm and its temperature is 350 K. Let us assume that the final pressure is and final temperature is .
(a) We know that for a monoatomic gas, value of is \frac{5}{3}[/tex].
And, in case of adiabatic process,
= constant
also, PV = nRT
So, here = 350 K, , and
Hence,
= 267 K
Also, = 4.0 atm, , and
= 2.04 atm
Hence, for monoatomic gas final pressure is 2.04 atm and final temperature is 267 K.
(b) For diatomic gas, value of is \frac{7}{5}[/tex].
As, = constant
also, PV = nRT
= 350 K, , and
= 289 K
And, = 4.0 atm, , and
= 2.27 atm
Hence, for diatomic gas final pressure is 2.27 atm and final temperature is 289 K.
Answer:
k = 52.2 N / m
Explanation:
For this exercise we are going to use the conservation of mechanical energy.
Starting point. When it is 30 m high
Em₀ = K + U = ½ m v² + m g h
Final point. Right when you hit the water
= K_{e} = ½ k x²
in this case the distance the bungee is stretched is 30 m
x = h
as they indicate that there are no losses, energy is conserved
Em₀ = Em_{f}
½ m v² + m g h = ½ k h²
k =
let's calculate
k =
k = 52.2 N / m
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as we know that the velocity vectors are at right angles
magnitude = ?
hypotenuse of a right
triangle.
v^2 = 90^2 + 4^2
v^2 = 8116
Taking the square root of both sides here we get,
v = 90.1 m/s
hope it helps
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