Answer:
2 x 10⁻³ volts
Explanation:
B = magnetic of magnetic field parallel to the axis of loop = 1 T
= rate of change of area of the loop = 20 cm²/s = 20 x 10⁻⁴ m²
θ = Angle of the magnetic field with the area vector = 0
E = emf induced in the loop
Induced emf is given as
E = B
E = (1) (20 x 10⁻⁴ )
E = 2 x 10⁻³ volts
E = 2 mV
Answer:
a = 18.28 ft/s²
Explanation:
given,
time of force application, t= 10 s
Work = 10 Btu
mass of the object = 15 lb
acceleration, a = ? ft/s²
1 btu = 778.15 ft.lbf
10 btu = 7781.5 ft.lbf
m = 0.466 slug
now,
work done is equal to change in kinetic energy
now, acceleration of object
a = 18.28 ft/s²
constant acceleration of the object is equal to 18.28 ft/s²
Answer:
The distance is 11 m.
Explanation:
Given that,
Friction coefficient = 0.24
Time = 3.0 s
Initial velocity = 0
We need to calculate the acceleration
Using newton's second law
...(I)
Using formula of friction force
....(II)
Put the value of F in the equation (II) from equation (I)
....(III)
Put the value in the equation (III)
We need to calculate the distance,
Using equation of motion
Hence, The distance is 11 m.