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asambeis [7]
2 years ago
13

Suppose that a rectangular toroid has 2,000 windings and a self-inductance of 0.060 H. If the height of the rectangular toroid i

s h = 0.60 m, what is the current (in A) flowing through the rectangular toroid when the energy in its magnetic field is 4.0 ✕ 10−6 J?
Physics
1 answer:
andrey2020 [161]2 years ago
3 0

Answer:

0.01154 A

Explanation:

We have given the energy in the magnetic field U=4\times 10^{-6}J

Value of inductance L =0.060 H

Energy stored in magnetic field is given by U=\frac{1}{2}Li^2

i=\sqrt{\frac{2U}{L}}

i=\sqrt{\frac{2\times 4\times 10^{-6}}{0.06}}=0.01154\ A

So the current flowing through rectangular toroid will be 0.01154 A

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The heater element of a particular 120-V toaster is a 8.9-m length of nichrome wire, whose diameter is 0.86 mm. The resistivity
AleksAgata [21]

Answer:

Power, P = 722.96 watts

Explanation:

It is given that,

Voltage, V = 120 V

Length of nichrome wire, l = 8.9 m

Diameter of wire, d = 0.86 mm

Radius of wire, r = 0.43 mm = 0.00043 m

Resistivity of wire, \rho=1.3\times 10^{-6}\ \Omega-m

We need to find the power drawn by this heater. Power is given by :

P=\dfrac{V^2}{R}

And, R=\rho\dfrac{l}{A}

P=\dfrac{V^2\times A}{\rho\times l}

P=\dfrac{120^2\times \pi (0.00043)^2}{1.3\times 10^{-6}\times 8.9}

P = 722.96 watts

So, the power drawn by this heater element is 722.96 watts. Hence, this is the required solution.    

3 0
2 years ago
The free-body diagram of a crate is shown. What is the net force acting on the crate? 352 N to the left 176 N to the left 528 N
Umnica [9.8K]

As per given conditions there are two directions along which forces are acting

1. Net force along left direction is given as

F_{left} = 352N + 176 N = 528 N

2. Net force towards right direction is given as

F_{right} = 528N + 440 N = 968 N

now since the two forces here in opposite direction so here we will have net force given as

F_{net} = F_{right} - F_{left}

F_{net} = 968 - 528

F_{net} = 440 N

so here net forces must be 440 N towards right

7 0
2 years ago
Read 2 more answers
Water runs through a plumbing with a flow of 0.750m3/s and arrives to every exit of a fountain. At what speed will the water com
Lubov Fominskaja [6]

Divide the flow rate (0.750 m³/s) by the cross-sectional area of each pipe:

diameter = 40 mm   ==>   area = <em>π</em> (0.04 m)² ≈ 0.00503 m²

diameter = 120 mm   ==>   area = <em>π</em> (0.12 m)² ≈ 0.0452 m²

Then the speed at the end of the 40 mm pipe is

(0.750 m³/s) / (0.00503 m²) ≈ 149.208 m/s ≈ 149 m/s

(0.750 m³/s) / (0.0452 m²) ≈ 16.579 m/s ≈ 16.6 m/s

7 0
1 year ago
If you find an igneous rock which has 450 radioactive isotopes and 3,150 stable daughter isotopes, how many half-lifes of this i
slavikrds [6]

Answer:

3t_{1/2}  

Explanation:

To find the half-lifes of the isotope we need to use the following equation:

N_{t} = N_{0}2^{-\frac{t}{t_{1/2}}}     (1)

<em>where Nt: is the amount of the isotope that has not yet decayed after a time t, N₀: is the initial amount of the isotope, t: is the time and </em>t_{1/2}<em>: is the half-lifes.</em>

By solving equation (1) for t we have:

\frac{t}{t_{1/2}} = - \frac{Ln(Nt/N_{0})}{Ln(2)}

<u>Having that:</u>

Nt = 450

N₀ = 3150 + 450 = 3600,

The half-lifes of the isotope is:

t = - \frac{Ln(450/3600)}{Ln(2)} \cdot t_{1/2} = 3t_{1/2}

Therefore, 3 half-lives of the isotope passed since the rock was formed.

I hope it helps you!

3 0
2 years ago
A cylinder has 500 cm3 of water. After a mass of 100 grams of sand is poured into the cylinder and all air bubbles are removed b
Nina [5.8K]

Answer: SG = 2.67

Specific gravity of the sand is 2.67

Explanation:

Specific gravity = density of material/density of water

Given;

Mass of sand m = 100g

Volume of sand = volume of water displaced

Vs = 537.5cm^3 - 500 cm^3

Vs = 37.5cm^3

Density of sand = m/Vs = 100g/37.5 cm^3

Ds = 2.67g/cm^3

Density of water Dw = 1.00 g/cm^3

Therefore, the specific gravity of sand is

SG = Ds/Dw

SG = (2.67g/cm^3)/(1.00g/cm^3)

SG = 2.67

Specific gravity of the sand is 2.67

3 0
2 years ago
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