Answer:
The speed of Susan is 2.37 m/s
Explanation:
To visualize better this problem, we need to draw a free body diagram.
the work is defined as:
here we have the work done by Paul and the friction force, so:
Now the change of energy is:
<span>You are given a QL = -26 μC charge that is placed on the x-axis at x = - 0.2 m and a QR = 26 μC charge that is placed at x = +0.2 m. The answers are:
The x-component of the electric field at x = 0 m and y = 0.2 m is 3.
The y-component of the electric field at x = 0 m and y = 0.2 m is 2.
</span>
Answer:
4.42 x 10⁷ W/m²
Explanation:
A = energy absorbed = 500 J
η = efficiency = 0.90
E = Total energy
Total energy is given as
E = A/η
E = 500/0.90
E = 555.55 J
t = time = 4.00 s
Power of the beam is given as
P = E /t
P = 555.55/4.00
P = 138.88 Watt
d = diameter of the circular spot = 2.00 mm = 2 x 10⁻³ m
Area of the circular spot is given as
A = (0.25) πd²
A = (0.25) (3.14) (2 x 10⁻³)²
A = 3.14 x 10⁻⁶ m²
Intensity of the beam is given as
I = P /A
I = 138.88 / (3.14 x 10⁻⁶)
I = 4.42 x 10⁷ W/m²
Top of the U ramp: potential energy is the highest, while kinetic energy is the lowest
Bottom of the U ramp(aka the curve part): potential energy is the lowest and the kinetic energy is the highest
THEREFORE, PE and KE have an INVERSE RELATIONSHIP.
