Answer:
The lose of thermal energy is, Q = 22500 J
Explanation:
Given data,
The mass of aluminium block, m = 1.0 kg
The initial temperature of block, T = 50° C
The final temperature of the block, T' = 25° C
The change in temperature, ΔT = 50° C - 25° C
= 25° C
The specific heat capacity of aluminium, c = 900 J/kg°C
The formula for thermal energy,
<em>Q = mcΔT</em>
= 1.0 x 900 x 25
= 22500 J
Hence, the lose of thermal energy is, Q = 22500 J
Get to school and learn boi
<span>The Gravitational Force of an object is a measure of the amount of matter it contains. on the other hand __Matter__ is a measure of the gravitational force on an object. I hope it helps :)</span>
Explanation:
Assuming the wall is frictionless, there are four forces acting on the ladder.
Weight pulling down at the center of the ladder (mg).
Reaction force pushing to the left at the wall (Rw).
Reaction force pushing up at the foot of the ladder (Rf).
Friction force pushing to the right at the foot of the ladder (Ff).
(a) Calculate the reaction force at the wall.
Take the sum of the moments about the foot of the ladder.
∑τ = Iα
Rw (3.0 sin 60°) − mg (1.5 cos 60°) = 0
Rw (3.0 sin 60°) = mg (1.5 cos 60°)
Rw = mg / (2 tan 60°)
Rw = (10 kg) (9.8 m/s²) / (2√3)
Rw = 28 N
(b) State the friction at the foot of the ladder.
Take the sum of the forces in the x direction.
∑F = ma
Ff − Rw = 0
Ff = Rw
Ff = 28 N
(c) State the reaction at the foot of the ladder.
Take the sum of the forces in the y direction.
∑F = ma
Rf − mg = 0
Rf = mg
Rf = 98 N
Answer:
The increase in the internal energy of the system is 360 Joules.
Explanation:
Given that,
Heat supplied to a system, Q = 292 J
Work done on the system by its surroundings, W = 68 J
We need to find the increase in the internal energy of the system. It can be given by first law of thermodynamics. It is given by :

So, the increase in the internal energy of the system is 360 Joules. Hence, this is the required solution.