The total power emitted by an object via radiation is:
where:
A is the surface of the object (in our problem,
is the emissivity of the object (in our problem,
)
is the Stefan-Boltzmann constant
T is the absolute temperature of the object, which in our case is
Substituting these values, we find the power emitted by radiation:
So, the correct answer is D.
Answer:
Rs. 432*10^3 (In kilowatts per hour)
I hope it will be useful.
Answer:It is actually the South Magnetic pole
Explanation:The magnetic pole near earth's geographic north pole is actually the south magnetic pole. When it comes to magnets, opposites attract. This fact means that the north end of a magnet in a compass is attracted to the south magnetic pole, which lies close to the geographic north pole.
Answer:
The force is 
Explanation:
From the question we are told that
The mass of the block is 
The coefficient of static friction is 
The coefficient of kinetic friction is 
The normal force acting on the block is
substituting values
Given that the force we are to find is the force required to get the block to start moving then the force acting against this force is the static frictional force which is mathematically evaluated as
substituting values
Now for this block to move the force require is equal to 
i.e
=>
Answer:
x ≤ 3.6913 m
Explanation:
Given
Mrod = 44.0 kg
L = 4.90 m
Tmax = 1450 N
Mman = 69 kg
A: left end of the rod
B: right end of the rod
x = distance from the left end to the man
If we take torques around the left end as follows
∑τ = 0 ⇒ - Wrod*(L/2) - Wman*x + T*Sin 30º*L = 0
⇒ - (Mrod*g)*(L/2) - (Mman*g)*x + Tmax*Sin 30º*L = 0
⇒ - (44*9.8)*(4.9/2) - (69*9.8)*x + (1450)*(0.5)*(4.9) = 0
⇒ x ≤ 3.6913 m
