Answer
given,
radius of the circular orbit, r = 0.53 x 10⁻¹⁰ m
mass of electron, M = 9.11 x 10⁻³¹ Kg
charge of electron, q₁ = 1.6 x 10⁻¹⁹ C
q₂ = 1.6 x 10⁻¹⁹ C
we know, force between two charges
F = 8.20 x 10⁻⁸ N
b) using newton's second law
F = m a
m a = 8.20 x 10⁻⁸
a = 9 x 10²² m/s²
c) speed of the electron
v² = 4.77 x 10¹²
v = 2.18 x 10⁶ m/s
d) the period of the circular motion.
T = 1.53 x 10⁻¹⁶ s
Answer:
F= 224 N
Explanation:
Given that
mass ,m = 80 kg
Radius ,r= 5 m
speed at the top v= 8 m/s
The force at the top = F
Now by using the Second law of Newton's
Now by putting the values
Take g = 10 m/s²
F= 224 N
Therefore the force exerted by the track at the top position will be 224 N.
Answer:
(a) - 165.032 m/s
(b) 238.37 m/s
Explanation:
initial horizontal velocity, ux = 172 m/s
height, h = 1390 m
g = 9.8 m/s^2
Let it strikes the ground after time t.
Use second equation of motion in vertical direction
-1390 = 0 - 0.5 x 9.8 x t^2
t = 16.84 second
(a) Let vy be the vertical component of velocity as it strikes the ground
Use first equation of motion in vertical direction
vy = uy - gt
vy = 0 - 9.8 x 16.84
vy = - 165.032 m/s
Thus, the vertical component of velocity as it strikes the ground is 165.032 m/s downward direction.
(b)
The horizontal component of velocity remains constant throughout the motion.
vx = 172 m/s
vy = - 165.032 m/s
The resultant velocity is v.
v = 238.37 m/s
Thus, teh velocity with which it hits the ground is 238.37 m/s.