Answer:
1.37 rad/s
Explanation:
Given:
Total length of the tape is, 
m
Total time of run is, 
hours
We know, 1 hour = 3600 s
So, 2.1 hours = 2.1 × 3600 = 7560 s
So, total time of run is, 
s
Inner radius is, 
Outer radius is, 
Now, linear speed of the tape is, 
Let the same angular speed be 
.
Now, average radius of the reel is given as the sum of the two radii divided by 2.
So, average radius is, 
Now, common angular speed is given as the ratio of linear speed and average radius of the tape. So,
Therefore, the common angular speed of the reels is 1.37 rad/s.
Answer:
yes, should be
Explanation:
This is a hard yes or no question becuase the amplitudes are the same height but in different beating orders.
Answer:
Magnetic flux through the loop is 1.03 T m²
Explanation:
Given:
Magnetic field, B = 4.35 T
Radius of the circular loop, r = 0.280 m
Angle between circular loop and magnetic field, θ = 15.1⁰
Magnetic flux is determine by the relation:
....(1)
Here A represents area of the circular loop.
Area of circular loop, A = πr²
Hence, the equation (1) becomes:
Substitute the suitable values in the above equation.
= 1.03 T m²
Answer:
I = 21.13 mA ≈ 21 mA
Explanation:
If
I₁ = 5 mA
L₁ = L₂ = L
V₁ = V₂ = V
ρ₁ = 1.68*10⁻⁸ Ohm-m
ρ₂ = 1.59*10⁻⁸ Ohm-m
D₁ = D
D₂ = 2D
S₁ = 0.25*π*D²
S₂ = 0.25*π*(2*D)² = π*D²
If we apply the equation
R = ρ*L / S
where (using Ohm's Law):
R = V / I
we have
V / I = ρ*L / S
If V and L are the same
V / L = ρ*I / S
then
(V / L)₁ = (V / L)₂ ⇒ ρ₁*I₁ / S₁ = ρ₂*I₂ / S₂
If
S₁ = 0.25*π*D² and
S₂ = 0.25*π*(2*D)² = π*D²
we have
ρ₁*I₁ / (0.25*π*D²) = ρ₂*I₂ / (π*D²)
⇒ I₂ = 4*ρ₁*I₁ / ρ₂
⇒ I₂ = 4*1.68*10⁻⁸ Ohm-m*5 mA / 1.59*10⁻⁸ Ohm-m
⇒ I₂ = 21.13 mA
Try this option, the answers are marked with colour.
