Answer:
(a) 0.81 V
(b) 0.52 V
Explanation:
Number of turns, N = 150
Radius, r = 2.5 cm = 0.025 m
Magnetic field, B = 0.060 T
f = 440 rev/min = 440 / 60 = 7.33 rps
A.
The maximum emf is given by
e = N x B x A x 2 x π x f
e = 150 x 0.060 x 3.14 x 0.025 x 0.025 x 2 x 3.14 x 7.33
e = 0.81 V
B.
The back emf is given by
e' = 2e / π = 2 x 0.81 / 3.14 = 0.52 V
Explanation:
2.
4.
In only the above cases (i.e 1,2,4,5,6,8 ) the object possibly moves at a constant velocity of
You should have noticed that the sets of forces applied to the object are the same asthe ones in the prevous question. Newton's 1st law (and the 2nd law, too) makes nodistinction between the state of re st and the state of moving at a constant velocity(even a high velocity).
In both cases, the net force applied to the object must equal zero.
Answer: The beaker containing pure water has decreased more.
Explanation:
In both cases, the decrease of water level is due to evaporation. We know that evaporation is a surface phenomenon. In the case of salt water, the salt molecules somewhat hinders the evaporation process of the water molecules and hence the salt water evaporates at a slower rate than pure water.
Hence, pure water level falls more.
Answer:
longitudinal engineering strain = 624.16
true strain is 6.44
Explanation:
given data
diameter d1 = 0.5 mm
diameter d2 = 25 mm
to find out
longitudinal engineering and true strains
solution
we know both the volume is same
so
volume 1 = volume 2
A×L(1) = A×L(2)
( π/4 × d1² )×L(1) = ( π/4 × d2² )×L(2)
( π/4 × 0.5² )×L(1) = ( π/4 × 25² )×L(2)
0.1963 ×L(1) = 122.71 ×L(2)
L(1) / L(2) = 122.71 / 0.1963 = 625.16
and we know longitudinal engineering strain is
longitudinal engineering strain = L(1) / L(2) - 1
longitudinal engineering strain = 625.16 - 1
longitudinal engineering strain = 624.16
and
true strain is
true strain = ln ( L(1) / L(2))
true strain = ln ( 625.16)
true strain is 6.44
Answer:
Banked
Explanation:
Banked curves are formed when the inner edge is below the outer edge.
It is done in order to ensure the reliability of the frictional force as it varies when the road is wet wet or oily. Thus in order to avoid these problems the curved roads are banked.
Banking of the curve provides the necessary centripetal force, i.e., the horizontal component of the normal reaction force to keep the vehicle i motion and thus helps in reducing the effect of the forward motion force on the vehicle.