Answer:
1) Weight on Mercury

Explanation:
do the same to the rest and use your calculator to find the weight in N.
Answer:
A photon of wavelength 103 nm is released
Explanation:
When an electron in an atom jumps from a higher energy level to a lower energy level, it releases a photon whose energy is equal to the difference in energy between the two levels.
For example, if we are talking about a hydrogen atom, the energy of the levels are:

So, the energy of the photon released when the electron jumps from the level n=3 to n=1 is

In Joules,

We can also find the wavelength of this photon, using the equation:

The air particle inside the balloon will collide more with each other and the temperature inside the balloon will increase.
As a person squeezed and applies the pressure to the outside of a balloon, the air particle inside the balloon gains energy and collide with each other, the particle of the air also try leave the balloon surface will implies equal pressure on the wall of the balloon, as the pressure outside the balloon increase, the inside pressure will also increase.
Answer:
Less than.
Explanation:
We have the positive charged metal sphere and we have to determine the electric field at a point near to it. In order to find that if we bring the positive test charge at that point then as we know that "like charges repel" so their electric field lines will repel each other resulting in a weaker electric field.
However if we bring the negative test charge at that point then of course there will be attraction and also the the electric field lines will direct from the positive to negative resulting in a stronger electric field between them. So there will be larger electric field then before.
"In this case, It can be concluded that electric field will be less than it was at this point before the test charge was present."
Answer:
541.14 m/s
Explanation:
We are given that
Mass of cannon=
Mass of shell,
Initial velocity of shell,v=547 m/s
We have to find the velocity of shell fired from this loose cannon.
According to law of conservation of momentum

Initial momentum of system=0


When the cannon is bolted to the ground then only shell moves and kinetic energy of system equals to kinetic energy of shell
Kinetic energy of shell,
K.E of shell=
K.E of shell=
K.E of shell=
2K.E of shell=
Velocity of shell fired from this loose cannon,v_2=


Hence, the velocity of shell fired from this loose cannon would be 541.14 m/s