Answer:
True. The two laws of thermal radiation state; 1) "Each square meter of a hotter object
Explanation:
D. physical property
the bonds between molecules of mercury are breaking so it's physical and it's not changing the chemical composition of the substance
Answer:

Explanation:
By energy conservation we know that spring energy is converted into kinetic energy of the block
so we will have

so we will have

now we will have same thing for another mass 4m which moves out with speed 5v
so we have

now from above two equations we have

so we have

Answer:
Electric field, E = 40608.75 N/C
Explanation:
It is given that,
Mass of electrons, 
Initial speed of electron, u = 0
Final speed of electrons, 
Distance traveled, s = 6.3 cm = 0.063 m
Firstly, we will find the acceleration of the electron using third equation of motion as :



Now we will find the electric field required in the tube as :



E = 40608.75 N/C
So, the electric field required in the tube is 40608.75 N/C. Hence, this is the required solution.