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2 years ago

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A11) A solenoid of length 18 cm consists of closely spaced coils of wire wrapped tightly around a wooden core. The magnetic fiel

d strength is inside the solenoid near its center when a certain current flows through the coils. If the coils of the solenoid are now pulled apart slightly, stretching it to without appreciably changing the size of the coils, what does the magnetic field become near the center of the solenoid when the same current flows through the coils? (μ0 = 4π × 10-7 T • m/A) A) 1.7 mT B) 3.4 mT C) 0.85 mT D) 2.0 mT

1 answer:

vitfil [10]2 years ago

7 0

Answer:

A

Explanation:

From a Solenoid we know that a magnetic fiel is always inversely proportional to lenght L or BL = constant

$B= frac{\mu_0}{2R}$

As I is constant

$\frac{B2}{B1} = \frac{R1}{R2}$

$B2 = 2mT*\frac{18}{21}$

$B2 = 1.714mT$

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The only force on the system is the mass of the hoop F net = 2.8kg*9.81m/s^2 = 27.468 N The mass equal of the rolling sphere is found by: the sphere rotates around the contact point with the table.
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I = 7/5*mR^2 M = 7/5*m
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2 years ago

A skateboarder with mass ms = 54 kg is standing at the top of a ramp which is hy = 3.3 m above the ground. The skateboarder then

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Answer:

A) $W_{ff} =-744.12J$

B) $F_f=-W_{ff}*sin\theta /hy = 112.75N$

C) $F_{f2}=207.58N$

Explanation:

This question is incomplete. The full question was:

<em>A skateboarder with mass ms = 54 kg is standing at the top of a ramp which is hy = 3.3 m above the ground. The skateboarder then jumps on his skateboard and descends down the ramp. His speed at the bottom of the ramp is vf = 6.2 m/s. </em>

<em>Part (a) Write an expression for the work, Wf, done by the friction force between the ramp and the skateboarder in terms of the variables given in the problem statement. </em>

<em>Part (b) The ramp makes an angle θ with the ground, where θ = 30°. Write an expression for the magnitude of the friction force, fr, between the ramp and the skateboarder. </em>

<em>Part (c) When the skateboarder reaches the bottom of the ramp, he continues moving with the speed vf onto a flat surface covered with grass. The friction between the grass and the skateboarder brings him to a complete stop after 5.00 m. Calculate the magnitude of the friction force, Fgrass in newtons, between the skateboarder and the grass.</em>

For part A), we make a balance of energy to calculate the work done by the friction force:

$W_{ff}=\Delta E$

$W_{ff}=1/2*m*vf^2-m*g*hy$

$W_{ff}=-744.12J$

For part B), we use our previous value for the work:

$W_{ff}=-F_f*(hy/sin\theta)$ Solving for friction force:

$F_f=-W_{ff}*sin\theta /hy$

$F_f=112.75N$

For part C), we first calculate the acceleration by kinematics and then calculate the module of friction force by dynamics:

$Vf^2=Vo^2+2*a*d$

Solving for a:

$a=-3.844m/s^2$

Now, by dynamics:

$|F_f|=|m*a|$

$|F_f|=207.58N$

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2 years ago

A satellite in geostationary orbit is used to transmit data via electromagnetic radiation. The satellite is at a height of 35,00

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Answer:

Explanation:

We have the following relation between power, P and intensity, I

$I = P/(4*pi*r^2)$

= $10^3/(4*pi*(35000*10^3))$

= $6.5*10^-14 W/M^2$

We also have the following relationship between electric field and electromagnetic radiation thus

$I = (ceE^2)/2$

Hence $E = \sqrt{2I/ce}$

substituting the values of I, c and e, we have

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1 year ago

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s344n2d4d5 [400]

Answer:

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Explanation:

For this problem we use energy conservation, the energy initiated is potential and kinetic and the final energy is only potential (infinite r)

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Eo = Ef

½ m1v² - G m1 $M_{earth}$ / R = - G m1 $M_{earth}$ / R

v² = 2G $M_{earth}$ (1 / R - 1 / Rinf)

If we do Rinf = infinity 1 / Rinf = 0

v = √2G $M_{earth}$ / R

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Em = -G m1 $M_{earth}$ / R

Em = - G m1 $M_{earth}$ / R

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1 year ago

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yKpoI14uk [10]

Answer:

2.93 A

Explanation:

From Ohm's law V = IR where V = voltage = 12 V, I = current = ? and R = resistance = 4.1 Ω.

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3 0

1 year ago