To solve this problem we need the concepts of Energy fluency and Intensity from chemical elements.
The energy fluency is given by the equation
Where
The energy fluency
c = Activity of the source
r = distance
E = electric field
In the other hand we have the equation for current in materials, which is given by
Then replacing our values we have that
We can conclude in this part that 1.3*10^7Bq is the activity coming out of the cylinder.
Now the energy fluency would be,
The uncollided flux density at the outer surface of the tank nearest the source is
Answer: the external agent must do work equal to -1.3 × 10⁻⁸ J
Explanation:
Given that;
Mass M1 = 7.0 kg
r = 3.0/2 m = 1.5 m
Mass M2 = 21 kg
we know that G = 6.67 × 10⁻¹¹ N.m²/kg²
work done by an external agent W = -2GM2M1 / r
so we substitute
W = (-2 × 6.67 × 10⁻¹¹ × 21 × 7) / 1.5
W = -1.96098 × 10⁻⁸ / 1.5
W = -1.3 × 10⁻⁸ J
Therefore the external agent must do work equal to -1.3 × 10⁻⁸ J
Answer:
4.7 N
Explanation:
130 g = 0.13 kg
The momentum of the snowball when it's thrown at the wall is
Which is also the impulse. From here we can calculate the magnitude of the average force F knowing the duration of the collision is 0.18 s
Answer:
42.6 m
Explanation:
mass of crate m = 53 kg
coefficient of kinetic friction, μ = 0.36
acceleration due to gravity, g = 9.8 m/s^2
Force, F = 372.098 N
Net force, f = F - friction force
f = 372.098 - μ m x g = 372.098 - 0.36 x 53 x 9.8
f = 185.114 N
acceleration, a = f / m = 185.114 / 53 = 3.49 m/s^2
initial velocity, u = 0
time, t = 4.94 s
s = ut + 1/2 at^2
s = 0 + 1/2 x 3.49 x 4.94 x 4.94
s = 42.6 m
