<h2>
Answer:</h2>
143μH
<h2>
Explanation:</h2>
The inductance (L) of a coil wire (e.g solenoid) is given by;
L = μ₀N²A / l --------------(i)
Where;
l = the length of the solenoid
A = cross-sectional area of the solenoid
N= number of turns of the solenoid
μ₀ = permeability of free space = 4π x 10⁻⁷ N/A²
<em>From the question;</em>
N = 183 turns
l = 2.09cm = 0.0209m
diameter, d = 9.49mm = 0.00949m
<em>But;</em>
A = π d² / 4 [Take π = 3.142 and substitute d = 0.00949m]
A = 3.142 x 0.00949² / 4
A = 7.1 x 10⁻⁵m²
<em>Substitute these values into equation (i) as follows;</em>
L = 4π x 10⁻⁷ x 183² x 7.1 x 10⁻⁵ / 0.0209 [Take π = 3.142]
L = 4(3.142) x 10⁻⁷ x 183² x 7.1 x 10⁻⁵ / 0.0209
L = 143 x 10⁻⁶ H
L = 143 μH
Therefore the inductance in microhenrys of the Tarik's solenoid is 143
Ohm's Law states V = IR
So,
I = V/R
The answer is B. 10/5=2 amps
Answer:2.47
Explanation:
So, the beaker weighs 1.40N when filled with water, brine of density weighs about 1.7N, you add the density + water. Have a good day!
Answer:
The first part can be solved via conservation of energy.

For the second part,
the free body diagram of the car should be as follows:
- weight in the downwards direction
- normal force of the track to the car in the downwards direction
The total force should be equal to the centripetal force by Newton's Second Law.

where
because we are looking for the case where the car loses contact.

Now we know the minimum velocity that the car should have. Using the energy conservation found in the first part, we can calculate the minimum height.

Explanation:
The point that might confuse you in this question is the direction of the normal force at the top of the loop.
We usually use the normal force opposite to the weight. However, normal force is the force that the road exerts on us. Imagine that the car goes through the loop very very fast. Its tires will feel a great amount of normal force, if its velocity is quite high. By the same logic, if its velocity is too low, it might not feel a normal force at all, which means losing contact with the track.
The air pressure is most likely lower.