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ad-work [718]
2 years ago
6

To practice Problem-Solving Strategy 29.1: Faraday's Law. A metal detector uses a changing magnetic field to detect metallic obj

ects. Suppose a metal detector that generates a uniform magnetic field perpendicular to its surface is held stationary at an angle of 15.0∘∘ to the ground, while just below the surface there lies a silver bracelet consisting of 6 circular loops of radius 5.00 cmcm with the plane of the loops parallel to the ground. If the magnetic field increases at a constant rate of 0.0250 T/sT/s, what is the induced emf EEEMF? Take the magnetic flux through an area to be positive when B⃗ B→B_vec crosses the area from top to bottom.
Physics
1 answer:
ExtremeBDS [4]2 years ago
4 0

Answer:

1.138\times 10^{-3}V

Explanation:

Apply Faraday's Newmann Lenz law to determine the induced emf in the loop:

\epsilon=\frac{d\phi}{dt}

where:

d\Phi-variation of the magnetic flux

dt-is the variation of time

#The magnetic flux through the coil is expressed as:

\Phi=NBA \ Cos \theta

Where:

N- number of circular loops

A-is the Area of each loop(A=\pi r^2=\pi \times 5^2=78.5398)

B-is the magnetic strength of the field.

\theta=15\textdegree- is the angle between the direction of the magnetic field and the normal to the area of the coil.

\epsilon=-\frac{d(78.5398\times 10^{-3}NB \ Cos \theta)}{dt}\\\\=-(78.5398\times 10^{-3}N\ Cos \theta)}{\frac{dB}{dt}

\frac{dB}{dt}-=0.0250T/s is given as rate at which the magnetic field increases.

#Substitute in the emf equation:

=-(78.5398\times 10^{-3} m^2 \times 6\ Cos 15 \textdegree)\times 0.0250T/s\\\\=1.138\times 10^{-3}V

Hence, the induced emf is 1.138\times 10^{-3}V

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The sun transfers energy to the earth by radiation at a rate of approximately 1.00 kW per square meter of surface.
Mashutka [201]

Answer:

1320336992.2512 m²

1320.33 kilometers or 509.79 miles

Explanation:

Energy transferred by the sun

W=0.24\times 1\times 10^3=240\ W/m^2

Energy required by the United States is 1\times 10^{19}\ J/yr (assumed)

Power

P=\frac{E}{t}\\\Rightarrow P=\frac{1\times 10^{19}}{365.25\times 24\times 3600}\\\Rightarrow P=316880878140.2895\ W

Area

A=\frac{P}{W}\\\Rightarrow A=\frac{316880878140.2895}{240}\\\Rightarrow A=132033699.2512\ m^2

Area of the solar collector would be 1320336992.2512 m²

Converting to km²

1\ m^2=\frac{1}{1000\times 1000}\ km^2

1320336992.2512\ m^2=1320336992.2512\times \frac{1}{1000\times 1000}\ km^2=1320.33\ km^2

Converting to mi²

1\ m^2=\frac{1}{1609.34\times 1609.34}\ mi^2

1320336992.2512\ m^2=1320336992.2512\times \frac{1}{1609.34\times 1609.34}\ mi^2=509.79\ mi^2

Each side of the square would be 1320.33 kilometers or 509.79 miles

4 0
2 years ago
Whale sharks swim forward while ascending or descending. They swim along a straight-line path at a shallow angle as they move fr
Aneli [31]
You did not include the quesetion, but I can help you to understand the problem and how to find the relevant information.

1) The angle of 13° with which the shark ascends meets this:

Vertical ascending velocity = 0.85m/s * sin(13°)

Horizontal velocity = 0.85m/s * cos(13°)

2) The length swan by the shark ascending meets this

Vertical ascending length = 50 m

Horizontal length, y:

\frac{y}{50} = \frac{0.85sin(13)}{0.85cos(13)}

From that y = 50 * tan(13°)

=> y = 11.54 m.

3) Conclusions:

1) The shark run 50 m vertically upward and 11.54 m horizontally.

2) The length of the path run by the shark may be calculated using Pythagoras' theorem:

hypotenuse^2 = (50m)^2 + (11.54m)^2 = 2633.25m^2

hypotenuse = 51.35m

So, the shark swan 51.35 m to reach the surface.

4 0
2 years ago
The late news reports the story of a shooting in the city. Investigators think that they have recovered the weapon and they run
Gemiola [76]

Answer:

50000 N

Explanation:

From the question given above, the following data were obtained:

Mass (m) of bullet = 0.050 kg

velocity (v) = 400 m/s

Distance (s) = 0.080 m

Force (F) =?

Next, we shall determine the acceleration of the bullet. This can be obtained as follow:

Initial velocity (u) = 0 m/s

Final velocity (v) = 400 m/s

Distance (s) = 0.080 m

Acceleration (a) =?

v² = u² + 2as

400² = 0 + (2 × a × 0.08)

160000 = 0 + 0.16a

160000 = 0.16a

Divide both side by 0.16

a = 160000 / 0.16

a = 1×10⁶ m/s²

Finally, we shall determine the force exerted by the bullet on the target. This can be obtained as follow:

Mass (m) of bullet = 0.050 kg

Acceleration (a) of bullet = 1×10⁶ m/s²

Force (F) =?

F = ma

F = 0.050 × 1×10⁶

F = 50000 N

Thus, the bullet exerted a force of 50000 N on the target.

7 0
2 years ago
A container of volume 0.6 m^3 contains 5.3 mol of argon gas at 24°C. Assuming argon behaves as an ideal gas, find the total inte
Vitek1552 [10]

Answer:

the internal energy of the gas is 433089.52 J

Explanation:

let n be the number of moles, R be the gas constant and T be the temperature in Kelvins.

the internal energy of an ideal gas is given by:

Ein = 3/2×n×R×T

     = 3/2×(5.3)×(8.31451)×(24 + 273)

     = 433089.52 J

Therefore, the internal energy of this gas is 433089.52 J.

5 0
2 years ago
A 50.0 kg crate is being pulled along a horizontal, smooth surface. The pulling force is 10.0 N and is directed 20.0 degree abov
allsm [11]

Explanation:

It is given that,

Mass of the crate, m = 50 kg

Force acting on the crate, F = 10 N

Angle with horizontal, \theta=20^{\circ}

Let N is the normal force acting on the crate. Using the free body diagram of the crate. It is clear that,

N=mg-F\ sin\theta

N=50\times 9.8-10\ sin(20)

N = 486.57 N

or

N = 487 N

If a is the acceleration of the crate. The horizontal component of force is balanced by the applied forces as :

ma=F\ cos\theta

a=\dfrac{F\ cos\theta}{m}

a=\dfrac{10\times \ cos(20)}{50}

a=0.1879\ m/s^2

or

a=0.188\ m/s^2

So, the normal force the crate and the magnitude of the acceleration of the crate is 487 N and 0.188\ m/s^2 respectively.

4 0
2 years ago
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