The volume of the balloon will halve
Explanation:
Boyle's law states that for an ideal gas kept at constant temperature, the pressure of the gas is proportional to its volume. Mathematically,
where
p is the gas pressure
V is the volume
The equation can also be rewritten as
And if we apply it to the gas inside the balloon in this problem (assuming its temperature is constant), we have:
is the initial pressure at sea level (the atmospheric pressure)
is the initial volume
is the final pressure
is the final volume
Substituting into the equation, we find:
Which means that the volume of the balloon will halve.
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The SI unit of length or distance is the meter.
The mass of the hoop is the only force which is computed by:F net = 2.8kg*9.81m/s^2 = 27.468 N
the slow masses that must be quicker are the pulley, ring, and the rolling sphere.
The mass correspondent of M the pulley is computed by torque τ = F*R = I*α = I*a/R F = M*a = I*a/R^2 --> M = I/R^2 = 21/2*m*R^2/R^2 = 1/2*m
The mass equal of the rolling sphere is computed by: the sphere revolves around the contact point with the table. So using the proposition of parallel axes, the moment of inertia of the sphere is I = 2/5*mR^2 for spin about the midpoint of mass + mR^2 for the distance of the axis of rotation from the center of mass of the sphere. I = 7/5*mR^2 M = 7/5*m
the acceleration is then a = F/m = 27.468/(2.8 + 1/2*2 + 7/5*4) = 27.468/9.4 = 2.922 m/s^2
Answer:
He requires 1 gram of mass.
Explanation:
The density is defined as:
(1)
Where m is the mass and V is the volume.
Then, m can be isolated from equation 1 in order to determine the mass.
(2)
Hence, he requires 1 gram of mass.
