Answer:
Inductance as calculated is 13.12 mH
Solution:
As per the question:
Length of the coil, l = 12 cm = 0.12 m
Diameter, d = 1.7 cm = 0.017 m
No. of turns, N = 235
Now,
Area of cross-section of the wire, A = 
We know that the inductance of the coil is given by the formula:

Answer:
The angular acceleration is zero
Explanation:
When an object is in rotational motion, it has a certain angular velocity, which is the rate of displacement of its angular position.
This angular velocity can change or remain constant - this is given by the angular acceleration, which is:

where
is the change in angular velocity
is the time elapsed
Therefore, the angular acceleration is the rate of change of angular velocity.
In this problem, the bicycle rotates at a constant angular velocity of

This means that the change in angular velocity is zero:

And so, that the angular acceleration is zero:

Answer:
Such limitations are given below.
Explanation:
- Each pn junction provides limited measurements of maximum forwarding current, highest possible inversion voltage as well as the maximum output level.
- If controlled within certain adsorption conditions, the pn junction could very well offer satisfying performance.
- In connector operation, the maximum inversion voltage seems to be of significant importance.
After the collision the magnitude of the momentum of the system is Mv
Given:
mass of 1st object = M
speed of 1st object = v
mass of 2nd object = M
speed of 2nd object = 0
To Find:
magnitude of the momentum after collision
Solution: Product of the mass of a particle and its velocity. Momentum is a vector quantity; i.e., it has both magnitude and direction. Isaac Newton's second law of motion states that the time rate of change of momentum is equal to the force acting on the particle.
Applying conservation of linear momentum
Mv + M(0) = 2MV
Mv = 2MV
V = v/2
So, after collision momentum is
p = 2MV = 2xMxv/2 = Mv
So, after collision momentum is Mv
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Answer:
592.92 x 10³ Pa
Explanation:
Mole of ammonia required = 10 g / 17 =0 .588 moles
We shall have to find pressure of .588 moles of ammonia at 30 degree having volume of 2.5 x 10⁻³ m³. We can calculate it as follows .
From the relation
PV = nRT
P x 2.5 x 10⁻³ = .588 x 8.32 x ( 273 + 30 )
P = 592.92 x 10³ Pa