Explanation:
The electric field at a distance r from the charged particle is given by :
k is electrostatic constant
if r = 2 m, electric field is given by :
If r = 1 m, electric field is given by :
Dividing equation (1) and (2) we get :
So, at a point 1 m from the particle, the electric field is 4 times of the electric field at a point 2 m.
Answer:
sorry I don't know I am too bad a t this coz I am only at class 7
mass of the ball m = 0.63 kg
initial height h = 1.8 m
final height h ' = 3.03 m
initial speed v = 7.09 m / s
final speed v ' = 4.21 m / s
Let the work done on the ball by air resistance W = ?
we know from law of conservation of energy ,
total energy at height h + work done by air = total energy at height h '
mgh + ( 1/ 2) mv^ 2 + W = mgh ' + ( 1/ 2) mv'^ 2
0.630*9.8*1.8 + 0.63*7.09^2 + W = mgh ' + ( 1/ 2) mv'^ 2
From there you can find W
if there is negative sign indicates it work opposite direction to motion
Answer: d₂ = 170 mGya
Explanation:
the relationship between absonbed 'd' and exposure 'E' is given as;
D(Gv) = F . x (AS/xB)
F is a conversion coefficient depending on medium
so we can simply write
d₁/d₂ = x₁/x₂
Given that;
our x₁ = 60 mAs, x₂ = 120 mAs, d₁ = 85 mGya, d₂ = ?
from the given formula,
d₂ = (x₂d₁ / x₁)
now we substitute
d₂ = (120 × 85) / 60
d₂ = 170 mGya
∴ if 120 mAa is used, the new exposure will be 170 mGya
Answer:
Ep = 0.6095 [J]
Explanation:
As defined in the problem statement, potential energy is defined as the product of mass by gravity by height. But first we must convert all the values given to measures of the international system (SI)
g = gravity = 10 [m/s^2]
h = elevation = 40 [ft] = 12.19 [m]
m = mass = 5 [g] = 0.005 [kg]
Ep = potential energy [J]
Ep = 0.005*10*12.19 = 0.6095 [J]
