The right answer for the question that is being asked and shown above is that: "The image produced is virtual and of the same size as the object." the image if the object is shifted closer to the lens to a point one focal length away from it is that The image produced is virtual and of the same size as the object.<span>
</span>
Answer:
a. 1.027 x 10^7 m/s b. 3600 V c. 0 V and d. 1.08 MeV
Explanation:
a. KE =1/2 (MV^2) where the M is mass of electron
b. E = V/d
c. V= 0 V (momentarily the pd changes to zero)
d KE= 300*3600 v = 1.08 MeV
Answer:
a.) 1567.2 m/s
b.) 149.4 m/s
Explanation:
Given that a 26 kg body is moving through space in the positive direction of an x axis with a speed of 350 m/s when, due to an internal explosion, it breaks into three parts. One part, with a mass of 7.8 kg, moves away from the point of explosion with a speed of 180 m/s in the positive y direction. A second part, with a mass of 8.8 kg, moves in the negative x direction with a speed of 640 m/s.
The x-component of the third part can be calculated by assuming that it moves in a positive x axis.
The third mass = 26 - ( 7.8 + 8.8)
The third mass = 26 - 16.6
The third mass = 9.4kg
since momentum is conserved, the momentum before explosion will be equal to sum of the momentum after explosion
26 x 350 = -8.8 x 640 + 9.4V
9100 = -5632 + 9.4V
9.4V = 9100 + 5632
9.4V = 14732
V = 14732/9.4
V = 1567.2 m/s
(b) y-component of the velocity of the third part will be
7.8 x 180 = 9.4 V
1404 = 9.4V
V = 1404/9.4
V = 149.4 m/s
A) use v=u+at for both
First section, v=27, u=0, a=2.4. You should get 11seconds.
Second section, v=0, u=27, a=-1.3. You should get 21seconds.
This means that the total time is 22seconds.
b) You can either use s=ut+0.5at^2 or v^2=u^2+2as. Personally, I would use the second one as you are not relying on your previous answer.
First section, v=27, u=0, a=2.4. You should get 152m.
Second section, v=0, u=27, a=-1.3. You should get 280m.
This makes your overall displacement 432m.
