Answer:
The temperature is 
Explanation:
From the question ewe are told that
The rate of heat transferred is 
The surface area is 
The emissivity of its surface is 
Generally, the rate of heat transfer is mathematically represented as

=> ![T = \sqrt[4]{\frac{P}{e* \sigma } }](https://tex.z-dn.net/?f=T%20%20%3D%20%20%5Csqrt%5B4%5D%7B%5Cfrac%7BP%7D%7Be%2A%20%5Csigma%20%7D%20%7D)
where
is the Boltzmann constant with value 
substituting value
![T = \sqrt[4]{\frac{13.1}{ 0.287* 5.67 *10^{-8} } }](https://tex.z-dn.net/?f=T%20%20%3D%20%20%5Csqrt%5B4%5D%7B%5Cfrac%7B13.1%7D%7B%200.287%2A%205.67%20%2A10%5E%7B-8%7D%20%7D%20%7D)

Answer:
The direction of the contact forces acting on a body is not necessarily perpendicular to the contact surface. The resolution of contact forces in two components i.e. perpendicular to contact surface and along surface. Perpendicular component is normal force and parallel component is friction.
Explanation:
Answer:
A. 16.9 m
Explanation:
I think this is the answer i am not sure
but hope it helps
Answer:
The right solution is "24.39 per sec".
Explanation:
According to the question,
⇒ 

The time will be:
⇒ 



hence,
⇒ 
