Correct question:
Consider the motion of a 4.00-kg particle that moves with potential energy given by
a) Suppose the particle is moving with a speed of 3.00 m/s when it is located at x = 1.00 m. What is the speed of the object when it is located at x = 5.00 m?
b) What is the magnitude of the force on the 4.00-kg particle when it is located at x = 5.00 m?
Answer:
a) 3.33 m/s
b) 0.016 N
Explanation:
a) given:
V = 3.00 m/s
x1 = 1.00 m
x = 5.00
At x = 1.00 m
= 4J
Kinetic energy = (1/2)mv²
= 18J
Total energy will be =
4J + 18J = 22J
At x = 5
= -0.24J
Kinetic energy =
= 2Vf²
Total energy =
2Vf² - 0.024
Using conservation of energy,
Initial total energy = final total energy
22 = 2Vf² - 0.24
Vf² = (22+0.24) / 2
= 3.33 m/s
b) magnitude of force when x = 5.0m
At x = 5.0 m
= 0.016N
Explanation:
a)
Sum of moments = 0 (Equilibrium)
T . cos (Q)*L = m*g*L/2
b) If the String is shorter the Q increases; hence, Cos Q decreases which in turn increases Tension in the string due to inverse relationship!
c)
Answer:
0.012 N
Explanation:
The formula to apply is that adopted from the Ampere law which is;
F= μ* I₁*I₂*l /2πd where
F is force that one conductor exerts on the other.
μ = magnetic permeability of free space = 4π×10⁻⁷ T. m/A
I₁ = current in conductor one=10 A
I₂ = current in conductor two= 10 A
l= length of conductor= 3 m
d= distance between the conductors = 5 mm = 0.005 m
Applying the values in the equation
F= 4π×10⁻⁷ *10*10*3 / 2π*0.005
F= 6 * 10⁻⁵ / 0.005
F=0.012 N
Answer: The reflected ray will be 44 degrees.
Explanation: The Law of Reflection states the angle of incidence ( the incoming light ray) equals the angle of reflection ( the light ray being reflected) Therefore, the angle of reflection will be 44 degrees.
Hope that helped!
