A weather balloon is inflated with 0.80 m3 of helium (He) at ground level (pressure is 1.0 atm). The balloon is released and ris
1 answer:
Answer: 0.24 atm
Explanation:
To calculate the new pressure, we use the equation given by Boyle's law. This law states that pressure is directly proportional to the volume of the gas at constant temperature.
The equation given by this law is:
where,
= initial pressure of gas = 1.0 atm
= initial volume of gas =
= final pressure of gas= ?
= final volume of gas =
Putting values in above equation, we get:
Thus the atmospheric pressure of the gas is 0.24 atm
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<h2>a) Average velocity in first 4 seconds is 64 ft/s upward</h2><h2>b) Average velocity in second 4 seconds is 63.5 ft/s downward</h2>
Explanation:
a) Given S(t) = 76 + 128t − 16t²
s(0) = 76 + 128 x 0 − 16 x 0² = 76 ft
s(4) = 76 + 128 x 4 − 16 x 4² = 332 ft
Displacement in 4 seconds = 332 - 76 = 256 ft
Time = 4 - 0 = 4 s
Average velocity in first 4 seconds is 64 ft/s upward
a) Given S(t) = 76 + 128t − 16t²
s(4) = 76 + 128 x 4 − 16 x 4² = 332 ft
s(8) = 76 + 128 x 8 − 16 x 8² = 78 ft
Displacement in 4 seconds = 78 - 332 = -254 ft
Time = 4 - 0 = 4 s
Average velocity in second 4 seconds is 63.5 ft/s downward
Answer:
f = 3.09 Hz
Explanation:
This is a simple harmonic motion exercise where the angular velocity is
w² =
to find the constant (k) of the spring, we use Hooke's law with the initial data
F = - kx
where the force is the weight of the body that is hanging
F = W = m g
we substitute
m g = - k x
k =
we calculate
k =
k = 3.769 10² m
we substitute in the first equation
w² =
w = 19.415 rad / s
angular velocity and frequency are related
w = 2πf
f =
f = 19.415 / 2pi
f = 3.09 Hz