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Aliun [14]
3 years ago
5

The heating coils in a hair dryer are 0.800 cm in diameter, have a combined length of 1.00 m, and a total of 400 turns. (a) What

is their total self-inductance assuming they act like a single solenoid?
Physics
2 answers:
VashaNatasha [74]3 years ago
7 0

Answer: 13.1 μH

Explanation:

Given

length of heating coil, l = 1 m

Diameter of heating coil, d = 0.8 cm = 8*10^-3 m

No of loops, N = 400

L = μN²A / l

where

μ = 4π*10^-7 = 1.26*10^-6 T

A = πd²/4 = (π * .008 * .008) / 4 = 6.4*10^-5 m²

L = μN²A / l

L = [1.26*10^-6 * 400 * 400* 6.5*10^-5] / 1

L = 1.26*10^-6 * 1.6*10^5 * 6.5*10^-5

L = 1.31*10^-5

L = 13.1 μH

Thus, from the calculations above, we can say that the total self inductance of the solenoid is 13.1 μH

lisov135 [29]3 years ago
4 0

Answer:

Their total self-inductance assuming they act like a single solenoid is 10.11 μH

Explanation:

Given;

diameter of the heating coil, d = 0.800 cm

combined length of heating coil and hair dryer, l = 1.0 m

number of turns, N = 400 turns

Formula for self-inductance is given as;

L = \frac{\mu_oN^2A}{l}

where

μ₀ is constant = 4π x 10⁻⁷ T.m/A

A is the area of the coil:

A = πd²/4

A = π (0.8 x 10⁻²)²/4

A = 5.027 x 10⁻⁵ m²

L = \frac{\mu_oN^2 A}{l } = \frac{4\pi *10^{-7}(400)^2 *5.027*10^{-5}}{1 } \\\\L =1.011 *10^{-5} \ H\\\\L = 10.11 \mu H

Therefore, their total self-inductance assuming they act like a single solenoid is 10.11 μH

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suter [353]

Answer:

5.82812 rad/s

Explanation:

L = Length of meter stick = 1 m = 100 cm

m_c = The center of mass of the stick = \frac{L}{2}-0.22=0.5-0.22=0.28\ m

\omega = Angular velocity

Moment of inertia of the system is given by

I=I_c+mr^2\\\Rightarrow I=\frac{mL^2}{12}+mr^2\\\Rightarrow I=\frac{m1^2}{12}+m0.28^2\\\Rightarrow I=m(\frac{1}{12}+0.0784)

As the energy in the system is conserved

mgh=I\frac{\omega^2}{2}\\\Rightarrow mgh=m(\frac{1}{12}+0.0784)\frac{\omega^2}{2}\\\Rightarrow gh=(\frac{1}{12}+0.0784)\frac{\omega^2}{2}\\\Rightarrow \omega=\sqrt{\frac{2gh}{\frac{1}{12}+0.0784}}\\\Rightarrow \omega=\sqrt{\frac{2\times 9.81\times 0.28}{\frac{1}{12}+0.0784}}\\\Rightarrow \omega=5.82812\ rad/s

The maximum angular velocity is 5.82812 rad/s

4 0
3 years ago
A tank of liquid (SG=0.80) that is 1 ft in diameter and 1.0 ft high is rigidly fixed to a rotating arm having a 2 ft radius. The
IgorLugansk [536]

Answer:

the pressure at B is 527psf

Explanation:

Angular velocity, ω = v / r

ω = 20 /1.5

= 13.333 rad/s

Flow equation from point A to B

P_A+rz_A-\frac{1}{2} Pr_A^2w^2=P_B+rz_B-\frac{1}{2} pr^2_Bw^2\\\\P_B = P_A + r(z_A-z_B)+\frac{1}{2} pw^2[(r_B^2)-(r_A)^2]\\\\P_B = [25 +(0.8+62.4)(0-1)+\frac{1}{2}(0.8\times1.94)\times(13.333)^2[2.5^2-1.5^2]  ]\\\\P_B = 25 - 49.92+551.79\\\\P_B = 526.87psf\\\approx527psf

the pressure at B is 527psf

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3 years ago
Select all of the statements that are true.
Strike441 [17]

I think its all four of them could be wrong but try all four !!!!!!

7 0
3 years ago
A block weighting 400kg rests on a horizontal surface and support on top of it another block of weight 100kg placed on the top o
masha68 [24]

The horizontal force applied to the block is approximately 1,420.84 N

The known parameters;

The mass of the block, w₁ = 400 kg

The orientation of the surface on which the block rest, w₁ = Horizontal

The mass of the block placed on top of the 400 kg block, w₂ = 100 kg

The length of the string to which the block w₂ is attached, l = 6 m

The coefficient of friction between the surface, μ = 0.25

The state of the system of blocks and applied force = Equilibrium

Strategy;

Calculate the forces acting on the blocks and string

The weight of the block, W₁ = 400 kg × 9.81 m/s² = 3,924 N

The weight of the block, W₂ = 100 kg × 9.81 m/s² = 981 N

Let <em>T</em> represent the tension in the string

The upward force from the string = T × sin(θ)

sin(θ) = √(6² - 5²)/6

Therefore;

The upward force from the string = T×√(6² - 5²)/6

The frictional force = (W₂ - The upward force from the string) × μ

The frictional force, F_{f2} = (981 - T×√(6² - 5²)/6) × 0.25

The tension in the string, T = F_{f2} × cos(θ)

∴ T = (981 - T×√(6² - 5²)/6) × 0.25 × 5/6

Solving, we get;

T = \dfrac{5886}{\sqrt{6^2 - 5^2} + 28.8} \approx 183.27

Frictional \ force, F_{f2} = \left (981 -  \dfrac{5886}{\sqrt{6^2 - 5^2} + 28.8}  \times \dfrac{\sqrt{6^2 - 5^2} }{6} \times  0.25 \right) \approx 219.92

The frictional force on the block W₂, F_{f2} ≈ 219.92 N

Therefore;

The force acting the block w₁, due to w₂ F_{w2} = 219.92/0.25 ≈ 879.68

The total normal force acting on the ground, N = W₁ + \mathbf{F_{w2}}

The frictional force from the ground, \mathbf{F_{f1}} = N×μ + \mathbf{F_{f2}} = P

Where;

P = The horizontal force applied to the block

P = (W₁ + \mathbf{F_{w2}}) × μ + \mathbf{F_{f2}}

Therefore;

P = (3,924 + 879.68) × 0.25 + 219.92 ≈ 1,420.84

The horizontal force applied to the block, P ≈ 1,420.84 N

Learn more about friction force here;

brainly.com/question/18038995

3 0
3 years ago
If we drop two iron balls, one ball is bigger than another one, from
loris [4]

Answer:

The bigger one. Ignoring air resistance, they will fall at the same speed, but the bigger one will hit first because it sticks out lower.

Explanation:

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