Answer:
a) 147.95 Mpc
Explanation:
Using Hubble's formula
where;
v = radical velocity
= Hubble's constant
d = distance
Given that:
The average radial velocity of galaxies in the Hercules cluster v = 10,800 km/s
Also using = 73 km/s/ Mpc ; we make distance d the subject of the formula:
Then distance d can be written as:
d = 147.95 Mpc
b)
Now, if the Hubble constant had a smaller value, then for a given velocity the distance to the galaxy will increase because distance d is inversely proportional to
i.e
d ∝ 
That it's more puashe on the back of the canoe and that effects the back of the canoe to fall back
Answer:
The final kinetic energy is 
Explanation:
From the question we are told that
The electric field is 
The charge on the object is 
The mass of the object is 
The distance moved by the object is 
The workdone on the object by the fields is mathematically represented as
![W = [qE + mg]d](https://tex.z-dn.net/?f=W%20%3D%20%20%5BqE%20%2B%20mg%5Dd)
Now this workdone is equivalent to the final kinetic energy so
![K = W = [qE + mg]d](https://tex.z-dn.net/?f=K%20%3D%20W%20%3D%20%20%5BqE%20%2B%20mg%5Dd)
substituting values
![K = W = [4.5*10^{-3 } *100 + 0.68 * 9.8]* 1](https://tex.z-dn.net/?f=K%20%3D%20W%20%3D%20%20%5B4.5%2A10%5E%7B-3%20%20%7D%20%2A100%20%20%2B%200.68%20%2A%209.8%5D%2A%201)
It’s A because it stays in motion whenever you drop it
Answer:
When an electron is hit by a photon of light, it absorbs the quanta of energy the photon was carrying and moves to a higher energy state. One way of thinking about this higher energy state is to imagine that the electron is now moving faster, (it has just been "hit" by a rapidly moving photon)
A photon is a quantum of EM radiation. Its energy is given by E = hf and is related to the frequency f and wavelength λ of the radiation by. E=hf=hcλ(energy of a photon) E = h f = h c λ (energy of a photon) , where E is the energy of a single photon and c is the speed of light.
