Answer:

Explanation:
Mass of block=10 kg
Applied horizontal force =F=20 N
Friction force=f=10 N
We have to find the acceleration of block.
Net force=Applied horizontal force-friction force

Where F= Horizontal force
f=Friction force
m=Mass of object
a=Acceleration of object


Hence, the acceleration of the block=
Answer:
I think its A plz tell me if im right
To solve this problem we will apply the concepts related to the thermal efficiency given in an engine of the Carnot cycle. Here we know that efficiency is given under the equation

Where,
Temperature of Cold Body
Temperature of Hot Body
= Efficiency
According to the statement our values are:


Replacing we have that




Therefore the temperature of the heat source is 300K
Explanation:
Below is an attachment containing the solution.
The answer to the statement is true because the day is of the logical proportion it has to be time.