Question

Can anyone help me out with these 3 question(questions 8, 9 & 10)..??

Answer 1

8. electron's initial​ velocity, u = 5×10⁴m/s
acceleration by the electric field, a = 10⁴m/s²

Acceleration in the direction of motion means that you will have to take it as positive during the calculations.

a) final velocity, v =2×u = 10×10⁴ m/s
Applying the first equation of motion, you can easily calculate the time in which the electrons aquire double the initial velocity.
v = u+at
t = (v-u)/a = 2u-u/a= u/a = 5×10⁴/10⁴ = 5 Sec

b) the distance covered can easily be calculated by using the 2nd equation of motion.

s = ut+at²/2
s = 5× 10⁴×5+10⁴×25/2= 75×10⁴/2
=375km

9. Since both the stones are thrown vertically upwards, the acceleration due to gravity acts in vertically downwards direction. Hence, the stones suffer retardation. This means that the stones will have zero final velocity after attaining maximum height.

To calculate maximum height given only acceleration and initial velocity, we can apply the 3rd equation of motion.

v²= u²+2as
s = H
v = 0
a= -g
0²=u²-2gH
H = u²/2g

Hence, for stone 1,
H1 = u1²/2g
and for stone 2,
H2 = u2²/2g

H1:H2=u1²/u2²
Hence proved.

10. The object starts from rest and travels 20m in 2 s.
u = 0, t= 2 and s= 20m
We can calculate​ the acceleration and then the final velocity after these 2 seconds.
s = ut+at²/2
a = 2s/t² = 2 ×20/2² = 10m/s²

v= u+at =0 +10×2 = 20m/s

This velocity = 20m/s becomes the initial velocity of the journey in next 4s.
Now, u = 20m/s, t= 4s, s = 160 m
Let's calculate the acceleration,
s= ut+at²/2
160=20×4+a×4²/2
a = 10m/s²
Final velocity after these 4s,
v = u+at = 20+ 10×4 =60m/s

Hence, we can calculate the final velocity after 1s, since 7 sec from start is equal to 1 sec after the 6 sec journey.
v = u+at = 60+10×1 = 70m/s

Hope you understand!
All the best. Please mark it as brainliest if you find it helpful.