Answer:
200 , 0 , 133.33333
Explanation:
velocity = change of X / change of T
so
400/2 = 200
0/2 = 0
400/3 = 133.33333
Answer: Anurag Mishra - Problems in Physics - Electricity and Magnetism ... Between two infinitely long wires having linear charge densities}. and -].·there are two points A ... the ratio of the electric force between them to t:-:c grav:tadonal force between them? (a) 10 8 ... d between the first two charges on the straight line at a distance
Explanation:
MORE POWER
Answer:
391.5 J
Explanation:
The amount of work done can be calculated using the formula:
- W = F║d
- where the force is parallel to the displacement
Looking at the formula, we can see that the mass of the object does not affect the work done on it.
Substitute the force applied and the displacement of the object into the equation.
- W = (87 N)(4.5 m)
- W = 391.5 J
The amount of work done on the object is 391.5 J in order to move it 4.5 meters with an applied force of 87 Newtons.
It would be Joules.
Workdone is measured in Joules.
Workdone = Force * distance
Force = mass * acceleration
= kg * ms⁻²
= kgms⁻²
Distance = m
So, Force * distance
kgms⁻² * m
Apply laws of indices that says
x² * x³ = x²⁺³ = x⁵
Therefore, It would be kgm²s⁻²
m¹ * m¹ = m¹⁺¹ = m²
s⁻² is also = s / 2
Answer:
Explanation:
q = 2e = 3.2 x 10^-19 C
mass, m = 6.68 x 10^-27 kg
Kinetic energy, K = 22 MeV
Current, i = 0.27 micro Ampere = 0.27 x 10^-6 A
(a) time, t = 2.8 s
Let N be the alpha particles strike the surface.
N x 2e = q
N x 3.2 x 10^-19 = i t
N x 3.2 x 10^-19 = 0.27 x 10^-6 x 2.8
N = 2.36 x 10^12
(b) Length, L = 16 cm = 0.16 m
Let N be the alpha particles
K = 0.5 x mv²
22 x 1.6 x 10^-13 = 0.5 x 6.68 x 10^-27 x v²
v² = 1.054 x 10^15
v = 3.25 x 10^7 m/s
So, N x 2e = i x t
N x 2e = i x L / v
N x 3.2 x 10^-19 = 2.7 x 10^-7 x 0.16 / (3.25 x 10^7)
N = 4153.85
(c) Us ethe conservation of energy
Kinetic energy = Potential energy
K = q x V
22 x 1.6 x 10^-13 = 2 x 1.5 x 10^-19 x V
V = 1.17 x 10^7 V