Answer:
Range, 
Explanation:
The question deals with the projectile motion of a particle mass M with charge Q, having an initial speed V in a direction opposite to that of a uniform electric field.
Since we are dealing with projectile motion in an electric field, the unknown variable here, would be the range, R of the projectile. We note that the electric field opposes the motion of the particle thereby reducing its kinetic energy. The particle stops when it loses all its kinetic energy due to the work done on it in opposing its motion by the electric field. From work-kinetic energy principles, work done on charge by electric field = loss in kinetic energy of mass.
So, [tex]QER = MV²/2{/tex} where R is the distance (range) the mass moves before it stops
Therefore {tex}R = MV²/2QE{/tex}
Answer:
Option A
Explanation:
Velocity is expressed as distance covered per unit time, with respect to direction. Therefore, v=d/t
Given distance west as 60 km and time as 1.33 then velocity will be
V=60/1.33=45.112781954887 km/h
Rounded off as 45.11 km/h West
Velocity in East will also be given by substituting 40 km for d and 0.67 h for h hence
V=40/0.67=59.701492537313 km/h rounded off as 59.70 km/h East
Taking East as positive then West as negative, the sum of two velocities will be (59.70+-45.11)/2=7.295 km/h East
Approximately 10 km/h East since it is positive
Answer:
F = 4.8 N
Explanation:
Given that,
Mass of the curtain, m = 1.2 kg
Acceleration of the pull, a = 4 m/s²
We need to find the force she exerts on the curtain.
We know that,
F = ma
Put al the values,
F = 1.2 kg × 4 m/s²
= 4.8 N
So, the force is 4.8 N.
Answer:

Explanation:
Equation of the rocket is,

Here, v' is the relative velocity of rocket.
In space F is zero.
So,

Now the momentum can be obtained by multiply by m on both sides.

Now for maxima, 

Now,

Therefore, the mass of the rocket while having maximum momentum is 
<span>A concave mirror and a converging lens will only produce a real image if the object is located beyond the focal point (i.e., more than one focal length away). The image of an object is found to be upright and reduced in size.</span>