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3 years ago

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An electric generator has an 18-cmcm-diameter, 120-turn coil that rotates at 60 HzHz in a uniform magnetic field that is perpend

icular to the rotation axis. Part A What magnetic field strength is needed to generate a peak voltage of 330 VV

1 answer:

kow [346]3 years ago

7 0

Explanation:

<h2>OK ok nosepo </h2><h2> $2825.55$ </h2>

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A father uses a rope to pull a child on a sled at a constant speed. of the ones listed which forces are present?

Amanda [17]

All the forces may be present.

The child on the sled has a weight- which is due to the force of gravity on the child and the sled.

The sled and the child exert a force on the ground equal to the combined weight of the sled and the child. The ground exerts a normal force on the sled.

The force used by the father to pull the sled is the applied force.

The sled slides on the ground and as a result, the force of friction exists between the ground and the sled, directed opposite to the direction of the motion of the sled.

The father pulls the sled using a rope. As a result, the rope is under Tension.

As the sled moves it also experiences a force of air resistance, which is dependent on the sled's speed.

However, since the father pulls the sled along with constant speed, the sum of all the forces acting on the sled is zero.

Since there is no movement in the upward or downward directions, the weight of the child and the sled is equal to the normal force acted upon the sled by the ground.

The force applied by the father on the rope is equal to the tension in the rope.

Friction and air resistance act opposite to the direction of motion of the sled. for the sled to move at constant speed, the tension in the rope must be equal to the sum of the forces due to friction and air resistance.

4 0

3 years ago

For a vertical spring-mass oscillator that is moving up and down, which of the following statements are true? (more than one sta

omeli [17]

Answer:

At the lowest point in the oscillation, the momentum is zero.

At the lowest point in the oscillation, $mg < ks_s$

Explanation:

Since spring block system is performing to and fro motion along straight line

So here we can say at the lowest position of its path the velocity will become zero.

So we can say that momentum of the spring block system is given as

$P = mv$

$P = 0$

Also we know that after reaching the lowest point the block will again go up towards its mean position

So at the lowest point of the spring block system the block will move upwards again

So this will accelerate upwards hence

$F_{spring} > mg$

$Ks > mg$

6 0

3 years ago

A photon of wavelength 2.78 pm scatters at an angle of 147° from an initially stationary, unbound electron. What is the de Brogl

Elena-2011 [213]

Answer:

2.07 pm

Explanation:

The problem given here is the very well known Compton effect which is expressed as

$\lambda^{'}-\lambda=\frac{h}{m_e c}(1-cos\theta)$

here, $\lambda$ is the initial photon wavelength, $\lambda^{'}$ is the scattered photon wavelength, h is he Planck's constant, $m_e$ is the free electron mass, c is the velocity of light, $\theta$ is the angle of scattering.

Given that, the scattering angle is, $\theta=147^{\circ}$

Putting the respective values, we get

$\lambda^{'}-\lambda=\frac{6.626\times 10^{-34} }{9.11\times 10^{-31}\times 3\times 10^{8} } (1-cos147^\circ ) m\\\lambda^{'}-\lambda=2.42\times 10^{-12} (1-cos147^\circ ) m.\\\lambda^{'}-\lambda=2.42(1-cos147^\circ ) p.m.\\\lambda^{'}-\lambda=4.45 p.m.$

Here, the photon's incident wavelength is $\lamda=2.78pm$

Therefore,

$\lambda^{'}=2.78+4.45=7.23 pm$

From the conservation of momentum,

$\vec{P_\lambda}=\vec{P_{\lambda^{'}}}+\vec{P_e}$

where, $\vec{P_\lambda}$ is the initial photon momentum, $\vec{P_{\lambda^{'}}}$ is the final photon momentum and $\vec{P_e}$ is the scattered electron momentum.

Expanding the vector sum, we get

$P^2_{e}=P^2_{\lambda}+P^2_{\lambda^{'}}-2P_\lambda P_{\lambda^{'}}cos\theta$

Now expressing the momentum in terms of De-Broglie wavelength

$P=h/\lambda,$

and putting it in the above equation we get,

$\lambda_{e}=\frac{\lambda \lambda^{'}}{\sqrt{\lambda^{2}+\lambda^{2}_{'}-2\lambda \lambda^{'} cos\theta}}$

Therefore,

$\lambda_{e}=\frac{2.78\times 7.23}{\sqrt{2.78^{2}+7.23^{2}-2\times 2.78\times 7.23\times cos147^\circ }} pm\\\lambda_{e}=\frac{20.0994}{9.68} = 2.07 pm$

This is the de Broglie wavelength of the electron after scattering.

6 0

3 years ago

An object has an acceleration of 6.0 m/s/s. If the net force was doubled and the mass was one-third the original value, then the

alexandr402 [8]

Hahahahaha. Okay.

So basically , force is equal to mass into acceleration.

F=ma

so when F=ma , we get acceleration=6m/s/s

Force is doubled.

Mass is 1/3 times original.

2F=1/3ma

Now , we rearrange , and we get 6F=ma

So , now for 6 times the original force , we get 6 times the initial acceleration.

So new acceleration = 6*6= 36m/s/s

5 0

3 years ago

What is the relationship between sunspots and geomagnetic storms?

shutvik [7]

Sunspots are spot like structure on the surface of sun, they are created due to geomagnetic flux which inhibits the convection of heat waves thus reducing the temperature of spot with some factor

8 0

3 years ago