Assume that the body's muscle mechanism can be approximated by a spring with a uniform continuous mass distribution that follows
1 answer:
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Answer:
d) 1.2 mT
Explanation:
Here we want to find the magnitude of the magnetic field at a distance of 2.5 mm from the axis of the coaxial cable.
First of all, we observe that:
- The internal cylindrical conductor of radius 2 mm can be treated as a conductive wire placed at the axis of the cable, since here we are analyzing the field outside the radius of the conductor. The current flowing in this conductor is
I = 15 A
- The external conductor, of radius between 3 mm and 3.5 mm, does not contribute to the field at r = 2.5 mm, since 2.5 mm is situated before the inner shell of the conductor (at 3 mm).
Therefore, the net magnetic field is just given by the internal conductor. The magnetic field produced by a wire is given by
where
is the vacuum permeability
I = 15 A is the current in the conductor
r = 2.5 mm = 0.0025 m is the distance from the axis at which we want to calculate the field
Substituting, we find:
Answer:
A) 1.55
B) 1.55
C) 12.92
D) 34.08
E) 57.82
Explanation:
The free body diagram attached, R is the radius of the wheel
Block B is lighter than block A so block A will move upward while A downward with the same acceleration. Since no snipping will occur, the wheel rotates in clockwise direction.
At the centre of the whee, torque due to B is given by
Similarly, torque due to A is given by
The sum of torque at the pivot is given by
Replacing and
by
and
respectively yields
Substituting for
in the equation
The angular acceleration of the wheel is given by
where a is the linear acceleration
Substituting for
into equation
we obtain
Net force on block A is
Net force on block B is
Where g is acceleration due to gravity
Substituting and
for
and
respectively into equation
and making a the subject we obtain
Since = 3kg and
= 7kg
g=9.81 and R=0.12m, I=0.22
Substituting these we obtain
Therefore, the linear acceleration of block A is 1.55
(B)
For block B
Therefore, the acceleration of both blocks A and B are same
1.55
(C)
The angular acceleration is
(D)
Tension on left side of cord is calculated using
(E)
Tension on right side of cord is calculated using
