Answer:
-5 V
Explanation:
The charged particle (which is positively charged) moves from point A to B, and its kinetic energy increases: it means that the particle is following the direction of the field, so its potential energy is decreasing (because it's been converted into potential energy), therefore it is moving from a point at higher potential (A) to a point at lower potential (B). This means that the value
vb−va
is negative.
We can calculate the potential difference between the two points by using the law of conservation of energy:

where:
is the change in kinetic energy of the particle
is the charge of the particle
is the potential difference
Re-arranging the equation, we can find the value of the potential difference:

Answer:
a) KE = 888.26J
b) N = 294.5 turns
Explanation:
For the kinetic energy:

The inertia is:

So, the kinetic energy will be:

Now, friction force is:
Ff = μ*N = 0.80*5N = 4N
The energy balance would be:
Kf - Ko = Wf where Kf=0; Ko = 888.26J; and Wf is the work done by friction force.
Wf = -Ff*d = -Ff*N*2*π*R where N is the amount of turns it gives.
Replacing these values into the energy balance:
0-888.26=-4*N*2*π*0.12
-888.26=-0.96*π*N
N=294.5 turns
Answer:
26.5 minutes
Explanation:
When the airplane is flying due West from Denver to Reno, the due-East wind with speed of 80km/h would reduce the ground speed by 80 km/h.
Its Denver to Reno ground speed is 900 - 80 = 720 km/h
The time it takes to cover 1200km at this speed is 1200 / 720 = 1.67 hours
On the other hand, when it returns from Reno to Denver in the due-East direction, the due-East wind with speed of 80km/h would add to the ground speed by 80 km/h
Its Reno to Denver ground speed is 900 + 80 = 980 km/h
The time it takes to cover 1200 km at this speed is 1200 / 980 = 1.22 hours
The difference it flight time would be 1.67 - 1.22 = 0.44 hours or 26.5 minutes
<u>Given </u><u>:</u><u>-</u>
- An elevator is moving vertically up with an acceleration a.
<u>To </u><u>Find</u><u> </u><u>:</u><u>-</u>
- The force exerted on the floor by a passenger of mass m .
<u>Solution</u><u> </u><u>:</u><u>-</u>
As the man is in a accelerated frame that is <u>non </u><u>inertial</u><u> frame</u><u> </u>, we would have to think of a pseudo force .
- The direction of this force is opposite to the direction of acceleration the frame and its magnitude is equal to the product of mass of the concerned body with the acceleration of the frame .
For the FBD refer to the attachment . From that ,
<u>Hence</u><u> </u><u>option</u><u> </u><u>d </u><u>is </u><u>correct</u><u> </u><u>choice </u><u>.</u>
<em>I </em><em>hope</em><em> this</em><em> helps</em><em> </em><em>.</em>
<h2>
Speed with which it return to its initial level is 100 m/s</h2>
Explanation:
We have equation of motion v² = u² + 2as
Initial velocity, u = 100 m/s
Acceleration, a = -9.81 m/s²
Final velocity, v = ?
Displacement, s = 0 m
Substituting
v² = u² + 2as
v² = 100² + 2 x -9.81 x 0
v² = 100²
v = ±100 m/s
+100 m/s is initial velocity and -100 m/s is final velocity.
Speed with which it return to its initial level is 100 m/s