If the airman had a mass of 80 kg, find the magnitude of the air drag acting on him when he reached terminal velocity of 54 m/s
1 answer:
The magnitude of the air drag is 784 N
Explanation:
An object falling down reaches the terminal velocity when the magnitude of the air drag acting on it becomes equal to the weight of the object. Mathematically, this condition can be written as:
where
is the magnitude of the air drag
m is the mass of the object
g is the acceleration of gravity
In this problem, we have
m = 80 kg is the mass of the airman
is the acceleration of gravity
Substituting into the formula, we find:
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Answer:
The final temperature of both objects is 400 K
Explanation:
The quantity of heat transferred per unit mass is given by;
Q = cΔT
where;
c is the specific heat capacity
ΔT is the change in temperature
The heat transferred by the object A per unit mass is given by;
Q(A) = caΔT
where;
ca is the specific heat capacity of object A
The heat transferred by the object B per unit mass is given by;
Q(B) = cbΔT
where;
cb is the specific heat capacity of object B
The heat lost by object B is equal to heat gained by object A
Q(A) = -Q(B)
But heat capacity of object B is twice that of object A
The final temperature of the two objects is given by
But heat capacity of object B is twice that of object A
Therefore, the final temperature of both objects is 400 K.