Question

An electron of mass 9.11×10−31 kg leaves one end of a TV picture tube with zero initial speed and travels in a straight line to the accelerating grid, which is a distance 2.20 cm away. It reaches the grid with a speed of 2.70×106 m/s . The accelerating force is constant.

Answer 1

Answer:

$F=1.509\times 10^{-16}N$

Explanation:

We have given that mass of electron $m=9.11\times 10^{-31}kg$

It is given that initial velocity u=0 m/sec

Final velocity v = $2.7\times 10^6m/sec$

Distance S = 2.2 cm = 0.022 m

According to third law of motion $v^2=u^2+2as$

$(2.7\times 10^6)^2=0^2+2\times a\times 0.022$

$a=165.68\times 10^{12}m/sec^2$

Accelerating force F is given by F = ma

So accelerating force $F=9.11\times 10^{-31}\times 165.68\times 10^{12}=1.509\times 10^{-16}N$