Answer:
Option B
Explanation:
Magnification of Microscope is
Mo= Magnification of objective lens and
Me= magnification of the eyepiece.
Both magnifications( of objective and eyepiece) are inversely proportional to the focal length.
Magnification,
when the focal length is less magnification will be high and when the magnification is the low focal length of the microscope will be more.
Thus. Magnification will increase by decreasing the focal length.
The correct answer is Option B
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