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Rus_ich [418]
3 years ago
10

Suppose a 44-turn coil lies in the plane of the page in a uniform magnetic field that is directed out of the page. The coil orig

inally has an area of 0.350 m2. It is stretched to have no area in 0.100 s. What is the magnitude (in V) and direction (as seen from above) of the average induced emf if the uniform magnetic field has a strength of 1.00 T
Physics
1 answer:
ollegr [7]3 years ago
5 0

Answer:

Explanation:

Dude no one can answer this on this app

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While bats emit a wide variety of sounds, one type emits pulses of sound at a frequency between 39 kHz and 78 kHz. What is the r
sdas [7]

Answer:

The range of wavelengths of the sound is 7692.30 m and 3846.15 m

Explanation:

A bat emits pulses of sound at a frequency between 39 kHz and 78 kHz. It is required to find the range of wavelengths of this sound.

Bat uses ultrasonic waves. It moves with the speed of light.

If f = 39 kHz,

\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{39\times 10^3}\\\\\lambda=7692.30\ m

If f = 78 kHz,

\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{78\times 10^3}\\\\\lambda=3846.15\ m

So, the range of wavelengths of the sound is 7692.30 m and 3846.15 m.

6 0
3 years ago
How does the law of conservation of energy apply to a light bulb? the electrical energy put into the light bulb equals the light
andrew-mc [135]
<span>The law of conservation of energy applies to a light bulb because the energy is being transformed into light and the light bulb is acting as a catalyst. The light bulb itself is not a form of energy, however when in combination with the electrical outlet to the bulb the electricity heats up the metal interior forming it into light. according to the law of conservation energy cannot be created or destroyed, but instead is formed into different kinds of energy. In relation to a light bulb electrical currents are forming heat energy by heating up the metal interior, then the bulb or glass around it allows to radiate light.</span>
7 0
3 years ago
If a nearsighted person has a far point df that is 3.50 m from the eye, what is the focal length f1 of the contact lenses that t
Anastaziya [24]

Answer:

f1 = -3.50 m

Explanation:

For a nearsighted person an object at infinity must be made to  appear  to be at his far point which is 3.50 m away. The image of an object at infinity must be formed on the same side of the lens as the object.

∴ v = -3.5 m

Using mirror formula,

i/f1 = 1/v + 1/u

Where f1 = focal length of the contact lens, v = image distance = -3.5 m, u =         object distance = at infinity(∞) = 1/0

∴ 1/f1 = (1/-3.5) + 1/infinity

  Note that, 1/infinity = 1/(1/0) = 0/1 =0.

∴ 1/f1 = 1/(-3.5) + 0

  1/f1 = 1/(-3.5)

Solving the equation by finding the inverse of both side of the equation.

∴ f1 = -3.50 m

 Therefore a converging lens of focal length  f1 = -3.50 m

would be needed by the person to see an object at infinity clearly

8 0
4 years ago
What initial speed v is required if the blocks m1 =2.5 kg and m2=1.5 kg are to travel a distance d =7.0cm before coming to rest?
ryzh [129]

Answer:

<em>OPTRIMUM</em><em> </em><em>PRIDE</em><em> </em><em>URGH</em><em> </em><em>URGH</em><em> </em><em>URGH</em><em> </em>

Explanation:

AHHAAHAHAHAHA

3 0
3 years ago
A boxed 12.0 kg computer monitor is dragged by friction 5.50 m up along the moving surface of a conveyor belt inclined at an ang
satela [25.4K]

Nah a a yuloure

Explanation:

5 0
3 years ago
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