It’s 25 because I already took the testing
Answer:
Induced emf in the loop is 0.0603 volts.
Explanation:
It is given that,
Radius of the circular loop, r = 1.6 cm = 0.016 m
Magnetic field, B = 0.8 T
When released, the radius of the loop starts to shrink at an instantaneous rate of 75.0 cm/s, 
We need to find the magnitude of induced emf at that instant. Induced emf is given by :

Where
is the magnetic flux, 
, A is the area of cross section




So, the induced emf in the loop is 0.0603 volts. Hence, this is the required solution.
specific gravity of the unknown sample = 2.68
Explanation:
True weight of solid=31.6 N
Apparent weight=19.8 N
loss in weight= 31.6-19.8=11.8 N
Loss in weight= weight of the liquid displaced
so weight of water displaced= 11.8 N
now specific gravity= weight of object/ weight of liquid displaced
specific gravity= 31.6/11.8
specific gravity= 2.68
Answer:
= 922N
Explanation:
m = 64kg
v(t) = (3.0 m/s²)t + (0.20 m/s³)t².
t = 4.0s
differentiation of v(t) so the acceleration is given by

using Newton’s second law
F (net) = ΣFg = ma
⇒R - w = ma
R = m(a + g)
= 64 (9.81 + 4.6)
= 922N