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olganol [36]
3 years ago
13

Considerable scientific work is currently under way to determine whether weak oscillating magnetic fields such as those found ne

ar outdoor electric power lines can affect human health. One study indicated that a magnetic field of magnitude 1.40 ✕ 10^−3 T, oscillating at 60 Hz, might stimulate red blood cells to become cancerous. If the diameter of a red blood cell is 7.8 μm, determine the maximum emf that can be generated around the perimeter of the cell.
Physics
1 answer:
VLD [36.1K]3 years ago
5 0

Answer:

The the maximum emf is 2.52\times10^{-11}\ V

Explanation:

Given that,

Magnetic field B = 1.40\times10^{-3}\ T

Frequency = 60 Hz

Diameter = 7.8 μm

We need to calculate the maximum emf

Using formula of emf

\epsilon=NBA\omega

Where, N = number of turns

B= magnetic field

A = area

Put the value in to the formula

\epsilon=1\times1.40\times10^{-3}\times\pi\times(\dfrac{7.8\times10^{-6}}{2})^2\times2\times\pi\times60

\epsilon=2.52\times10^{-11}\ V

Hence, The the maximum emf is 2.52\times10^{-11}\ V

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kykrilka [37]

Answer:

Potential energy =mass* acceleration due to gravity * height

mass*acceleration due to gravity =weight

hence potential energy of the puppy= weight * height

=18*2

=36 joule

6 0
3 years ago
What is sound energy? List three examples of sound energy that you experienced today.
Helga [31]
1. A broom swishing against the floor
2. a bee buzzing
3. a car engine


hope this helps!
4 0
4 years ago
What speed would a fly with a mass of 0.55g need in order to have a kinetic energy of 7.6 •10^4 j?
masya89 [10]

Answer:

16613 m/s

Explanation:

Given that

mass of the fly, m = 0.55 g = 0.55*10^-3 kg

Kinetic Energy of the fly, E = 7.6*10^4 J

Speed of the fly, v = ? m/s

We know that the Kinetic Energy is that energy that an object, in this case, the fly, possesses due to its motion.

The Kinetic Energy, KE of any object is represented by the formula

KE = 1/2 * m * v²

If we substitute the values in the relation, we have,

7.6*10^4 = 1/2 * 0.55*10^-3 * v²

v² = (15.2*10^4) / 0.55*10^-3

v² = 2.76*10^8

v = √2.76*10^8

v = 16613 m/s

Thus, the fly would need a speed of 16.6 km/s in order to have a Kinetic Energy of 7.6*10^4 J

7 0
3 years ago
At 5.0 minutes, the temperature of the water reaches 100 °C. The volume of the water in the urn
serious [3.7K]

Answer:

fdvevddvevkejokef0jeovdlvkjeuiyv

Explanation:

ddf4edscd

4 0
3 years ago
An astronaut inside a spacecraft, which protects her from harmful radiation, is orbiting a black hole at a distance of 120 km fr
mestny [16]

An astronaut inside a spacecraft, which protects her from harmful radiation, is orbiting a black hole at a distance of 120 km from its center. The black hole is 5.00 times the mass of the sun and has a Schwarzschild radius of 15.0 km. The astronaut is positioned inside the spaceship such that one of her 0.030 kg ears is 6.0 cm farther from the black hole than the center of mass of the spacecraft and the other ear is 6.0 cm closer.

What is the tension between her ears?

Would the astronaut find it difficult to keep from being torn apart by the gravitational forces?

Answer:

The tension between the ears = 2.07 KN

The astronaut will find it difficult to keep and will eventually be in trouble because the tension is now greater compared to the tension in the human tissues.

Explanation:

Given that:

Orbital radius of the spacecraft (R) = 120 Km = 120 × 10³ m

Mass of the black hole (m) = 5 \ * (M \ _{sun})

where : M_{sun} = 1.99*10^{33} \ kg

Then; we have:

 m = 5*(1.99*10^{30} \ kg ) \\ = 9.95*10^{30} kg

Schwarzchild radius of the black hole

r - 15.0 km

Mass of each ear m_{ear} = 0.030 \ kg

Farther distance between one ear and the black hole (d) = 6.0 cm

= 0.06 m

Closer distance between the other ear and the black home is (d) 6.0 cm

= 0.6 cm

NOW, If we assume that the tension force should be T; then definitely the two ears will posses the same angular velocity .

The net force on the ear closer to the black hole will be:

\frac{GMm_{ear} }{(R-d)}- T = m_{ear} (R -  d) \omega^2

\frac{GMm_{ear} }{(R-d)^2}- \frac{T}{(R-d)} = m_{ear} \omega^2 \ ----> \ (1)

The net force on the ear farther to the black hole is :

\frac{GMm_{ear} }{(R+d)}- T = m_{ear} (R +  d) \omega^2

\frac{GMm_{ear} }{(R+d)^2}- \frac{T}{(R+d)} = m_{ear} \omega^2 \ ----> \ (2)

Equating equation (1) and (2) & therefore making (T) the subject of the formula; we have:

T = \frac{3GMm_{ear}d}{R^3}

T = \frac{3(6.67*10^{-11}N.m^2/kg^2)(1.95*10^{30}kg)(0.03kg)(0.06m)}{(120*10^3m)^3}

T = 2073.9 N\\T = 2.07 KN

The tension between the ears = 2.07 KN

The astronaut will find it difficult to keep and will eventually be in trouble because the tension is now greater compared to the tension in the human tissues.

3 0
4 years ago
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