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3 years ago

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A circular loop of radius 0.10 m is rotating in a uniform external magnetic field of 0.20 t. (a) find the magnetic flux through

the loop due to the external field when the plane of the loop and the magnetic field vector are parallel.

1 answer:

solniwko [45]3 years ago

6 0

Your answer is zero. :)

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A photovoltaic array of (solar cells) is 10.0% efficient in gathering solar energy and converting it to electricity. If the aver

Natasha_Volkova [10]

Answer:

$A_{area}=0.243m^{2}$

Explanation:

Given data

Average intensity of sunlight on one day I=700 W/m²

Power P=170 W

To find

Area A

Solution

As we know that:

$I_{intensity}=\frac{P_{power}}{A_{area}}\\ A_{area}=\frac{P_{power}}{I_{intensity}} \\ A_{area}=\frac{170W}{700W/m^{2} } \\ A_{area}=0.243m^{2}$

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4 years ago

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Two charges 5 c and 15 c are seperated by 10 cm. what is the electric force between them?

icang [17]

Answer:

6.75×10^13N

Explanation:

The electric force between the charges can be determined using coulombs law which states that 'the force of attraction between two charges is directly proportional to the product of the charges and inversely proportional to the square of their distance between them.

Mathematically, F = kq1q2/r² where;

q1 and q2 are the charges

r is the distance between the charges

F is the force of attraction

k is the coulombs constant

Given q1 = 5C q2 = 15C r = 10cm = 0.1m k = 9×10^9Nm²/C²

Substituting the given values in the formula we have;

F =9×10^9×5×15/0.1²

F = 6.75×10^11/0.01

F = 6.75×10^13N

Therefore the electric force between them is 6.75×10^13N

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3 years ago

The mass of a basketball is three times greater then the mass of a softball. Compare the momentum’s of a softball and a basketba

vagabundo [1.1K]

Let m = the mass of the softball.
Then the mass of the basketball is 3m.

By definition, momentum = mass * velocity.
If both the softball and basketball move at the same velocity, v, then
Momentum of the softball = mv
Momentum of the basketball = 3mv

Answer: The momentum of the basketball is 3 times that of the softball.

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4 years ago

During the internal examination, the pathologist drains the intestines, removes any undigested food and feces,and examines the c

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During the internal examination, the pathologist drains the intestines, removes any undigested food and feces, and examines the contents of the stomach. This examination could give the pathologist clues of the time of death, and the location of death. The process of digesting and defecation vary from person to person, the entire process is generally considered to take approximately 40 hours in adults

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3 years ago

An object has the acceleration graph shown in (Figure 1). Its velocity at t=0s is vx=2.0m/s. Draw the object's velocity graph fo

timama [110]

Answer:

Explanation:

We may notice that change in velocity can be obtained by calculating areas between acceleration lines and horizontal axis ("Time"). Mathematically, we know that:

$v_{b}-v_{a} = \int\limits^{t_{b}}_{t_{a}} {a(t)} \, dt$

$v_{b} = v_{a}+ \int\limits^{t_{b}}_{t_{a}} {a(t)} \, dt$

Where:

$v_{a}$ , $v_{b}$ - Initial and final velocities, measured in meters per second.

$t_{a}$ , $t_{b}$ - Initial and final times, measured in seconds.

$a(t)$ - Acceleration, measured in meters per square second.

Acceleration is the slope of velocity, as we know that each line is an horizontal one, then, velocity curves are lines with slopes different of zero. There are three region where velocities should be found:

Region I (t = 0 s to t = 4 s)

$v_{4} = 2\,\frac{m}{s} +\int\limits^{4\,s}_{0\,s} {\left(-2\,\frac{m}{s^{2}} \right)} \, dt$

$v_{4} = 2\,\frac{m}{s}+\left(-2\,\frac{m}{s^{2}} \right) \cdot (4\,s-0\,s)$

$v_{4} = -6\,\frac{m}{s}$

Region II (t = 4 s to t = 6 s)

$v_{6} = -6\,\frac{m}{s} +\int\limits^{6\,s}_{4\,s} {\left(1\,\frac{m}{s^{2}} \right)} \, dt$

$v_{6} = -6\,\frac{m}{s}+\left(1\,\frac{m}{s^{2}} \right) \cdot (6\,s-4\,s)$

$v_{6} = -4\,\frac{m}{s}$

Region III (t = 6 s to t = 10 s)

$v_{10} = -4\,\frac{m}{s} +\int\limits^{10\,s}_{6\,s} {\left(2\,\frac{m}{s^{2}} \right)} \, dt$

$v_{10} = -4\,\frac{m}{s}+\left(2\,\frac{m}{s^{2}} \right) \cdot (10\,s-6\,s)$

$v_{10} = 4\,\frac{m}{s}$

Finally, we draw the object's velocity graph as follows. Graphic is attached below.

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4 years ago