B: Energy lose
i say this because in order to change they lose energy.
<span>E=hc/wav. len
E = (6.62 x 10^-34 x 3 x 10^8)/0.0275 x 10^-9
E = 7.22182 x 10^-15 J
To convert to eV divide by 1.6 x 10^-19
E = 7.22182 x 10^-15/1.6 x 10^-19 eV
E =45.36 x 10^3 eV
Th energy, E, of a single x-ray photon in eV is = 45.36keV.
Number of photons, n = total energy/ energy of photon
n = 3.85 x 10^-6/7.22182 x 10^-15
n = 5.33 x 10^8 photons </span>
Answer:
The canon B hits the ground fast.
Explanation:
Given that,
Speed of cannon A = 85 m/s
Speed of cannon B= 100 m/s
Speed of cannon C = 75 m/s
We need to calculate the cannonballs will hit the ground with the greatest speed
Using conservation of energy
The final kinetic energy of canon depends on initial kinetic energy and potential energy.
The final velocity depends upon initial velocity and initial height.
So, the initial velocity of canon B is high.
Hence, The canon B hits the ground fast.
Consider a long train moving at speed v. Now consider a passenger throwing a ball inside this train, towards the back of the train, with same velocity v (but in the opposite direction of the train movement).
- A passenger inside the train will see the ball moving with speed v
- For an observer outside the train, however, the ball will appear as still. In fact, for him the ball will have a speed v (given by the movement of the train) -v (velocity of the ball but moving in the opposite direction), so the net velocity will be v+(-v)=0.
Answer:
For the complete question provided in explanation, if the elevator moves upward, then the apparent weight will be 1035 N. While for downward motion the apparent weight will be 435 N.
Explanation:
The question is incomplete. The complete question contains a velocity graph provided in the attachment. This is the velocity graph for an elevator having a passenger of 75 kg.
From the slope of graph it is clear that acceleration at t = 1 sec is given as:
Acceleration = a = (4-0)m/s / (1-0)s = 4 m/s^2
Now, there are two cases:
1- Elevator moving up
2- Elevator moving down
For upward motion:
Apparent Weight = m(g + a)
Apparent Weight = (75 kg)(9.8 + 4)m/s^2
<u>Apparent Weight = 1035 N</u>
For downward motion:
Apparent Weight = m(g - a)
Apparent Weight = (75 kg)(9.8 - 4)m/s^2
<u>Apparent Weight = 435 N</u>