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.
Answer:
A. Mass
Explanation:
Inertia of an object is the resistance of the object to any change in its state of motion: it means that if an object is at rest, it tends to stay at rest for inertia (unless a net force acts on it), and if it is moving, it tends to continue moving with the same velocity, for inertia.
The inertia also describes how difficult it is to stop/accelerate an object, and it is directly proportional to the mass of the object: in fact, the larger the mass of an object, the more difficult it is to change its state of motion, and this means it has greater inertia.
Gravitational acceleration (Ga) is inversely proportional to k / Distance^2
so Ga * Distance^2 = K
On the surface of Earth acceleration due to gravity is about 9.8m/s^2 with an average distance to the earths core of about 6371 km (Wolfram alpha).
So k = 9.8 * 6371^2
I'm presuming that your distance of 116 is km
As
Ga = k / distance^2
Ga = ((9.8 * 6371^2) / (6371 + 116)^2 ) = 397778481.8 / 42081169
= 9.45 m/s^2 to 2sf
Answer:
The power transmitted to the pulley is 0.0455 hp.
Explanation:
Given;
mass attached to the rope, m = 5 lb
radius of the pulley, r = 3 in
constant rate of fall of the mass, v = 5 ft/s
acceleration due to gravity, g = 32.2 ft/s²
1 lbf = 32.2 lb.ft/s²
The power transmitted to the pulley is calculated as;
P = Fv
P = (mg)v
in horse power, the power transmitted is calculated as;
Therefore, the power transmitted to the pulley is 0.0455 hp.
