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 pressure at sea level (the atmospheric pressure)
 is the initial volume
 is the initial volume
 is the final pressure
 is the final pressure
 is the final volume
 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)
  (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)
  (2)
 
  
 
  
Hence, he requires 1 gram of mass.