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2 years ago

What is the primary difference between an ideal emf device and a real emf device?.

Physics

2 answers:

Law Incorporation [45]2 years ago

5 0

Answer:

A real emf device will have internal resistance and the ideal emf will not

Hope this Helps!

EleoNora [17]2 years ago

3 0

Answer:

A real emf device has an internal resistance, but an ideal emf device does not.

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You hear the engine roaring on a race car at the starting line. Predict the changes in the sound as the race starts and the car

Digiron [165]

B. The sound of the engine will get louder and the pitch higher.

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3 years ago

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Consider Compton Scattering with visible light.A photon with wavelength 500nm scatters backward(theta=180degree) from a free ele

JulijaS [17]

Answer: $4.86(10)^{-12}m$

Explanation:

The Compton Shift $\Delta \lambda$ in wavelength when photons are scattered is given by the following equation:

$\Delta \lambda=\lambda' - \lambda_{o}=\lambda_{c}(1-cos\theta)$ (1)

Where:

$\lambda'=500 nm=500(10)^{-9} m$ is the wavelength of the scattered photon

$\lambda_{o}$ is the wavelength of the incident photon

$\lambda_{c}=2.43(10)^{-12} m$ is a constant whose value is given by $\frac{h}{m_{e}.c}$ , being $h=4.136(10)^{-15}eV.s$ the Planck constant, $m_{e}$ the mass of the electron and $c=3(10)^{8}m/s$ the speed of light in vacuum.

$\theta=180\°$ the angle between incident phhoton and the scatered photon.

$\Delta \lambda=2.43(10)^{-12} m (1-cos(180\°))$ (2)

$\Delta \lambda=4.86(10)^{-12}m$ (3) This is the shift in wavelength

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3 years ago

Help help help plzzz. A 15-kilogram wagon originally at rest on a level road is pulled by a rope that exerts a constant horizont

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A. 2m/(s^2)

I hope it will be useful.

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3 years ago

How does heating a gas rigid container change its pressure

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The particles are heating up so they are starting to move faster and closer together which causes them to change the pressure of the container

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3 years ago

A ball was thrown from a projectile building of 30m which moves at a constant velocity of 20m/s and has an angle of 30degrees to

qwelly [4]

Answer:

Time of flight = 4.08seconds

Horizontal component of initial velocity is 17.32m/s

Explanation: complete question( and the horizontal component of the initial velocity.)

The equation for time of flight of a projectile is given as T= 2u/g

T=( 2×20)/9.8

T= 40/9.8= 4.08seconds

Horizontal component of initial velocity Vix= Vi Costheta

Vix= 20× cos 30°

Vix= 20×0.8660

Vix= 17.32m/s

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3 years ago

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