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suter [353]
3 years ago
12

If the cross-section of a wire of fixed length is doubled, how does the resistance of that wire change? (this is for studying fo

r my semester exam. I got the question wrong on one of my chapter tests so I'm trying to find out what the correct answer was)
A. Halved
B. Doubled (I know it's NOT this cause that was the wrong one)
C. Unchanged
D. Quadrupled
Physics
1 answer:
Ymorist [56]3 years ago
6 0
If the cross-section of a wire of fixed length is doubled,  the resistance of that wire change into doubled.We know that <span>the total </span>length<span> of the wires will </span>affect<span> the amount of </span>resistance. <span> The longer the wire, the more </span>resistance<span> that there will be so the answer is doubled.</span>
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A piano tuner sounds two strings simultaneously. One has been previously tuned to vibrate at 293.0 Hz. The tuner hears 3.0 beats
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Answer:

Part a)

f_B = 290 Hz

Part B)

percentage increase is

percentage = 1.38%

Explanation:

Part a)

As we know that the beat frequency is

f_A - f_B = 3

after increasing the tension the beat frequency is decreased and hence the tension in string B will increase

So we have

293 - f_B = 3

f_B = 290 Hz

Part B)

percentage increase in the tension of the string will be given as

f_A - f_B' = 1

f_B' = 292 Hz

now we have

f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}

so we have

T_1 = C (290)^2

T_2 = C(292)^2

so we have

\frac{\Delta T}{T} = \frac{292^2 - 290^2}{290^2}

percentage increase is

percentage = 1.38

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3 years ago
The half life of uranium-235 is 4. 5 billion years. If 0. 5 half-lives have elapsed, how many years have gone by?
igor_vitrenko [27]

Answer:

2.25 billion years

Explanation:

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2 years ago
A toroidal solenoid has an inner radius of 12.0 cm and an outer radius of 15.0 cm . It carries a current of 1.50 A . Part A How
AleksAgata [21]

Answer:

The number of turns, N = 1750

Explanation:

It is given that,

The inner radius of a toroid, r = 12 cm

Outer radius, r' = 15 cm

The magnetic field at points within the coils 14 cm from its center is, B=3.75\ mT=3.75\times 10^{-3}\ T

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The formula for the magnetic field at some distance from its center is given by :

B=\dfrac{\mu_o NI}{2\pi R}

N=\dfrac{2\pi R B}{\mu_o I}

N=\dfrac{2\pi \times 0.14\times 3.75\times 10^{-3}}{4\pi \times 10^{-7}\times 1.5}

N = 1750

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3 years ago
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A rock is sitting at the edge of a flat merry-go-round at a distance of 1.6 meters from the center. The coefficient of static fr
PSYCHO15rus [73]

Answer:

ω = 2.1 rad/sec

Explanation:

  • As the rock is moving along with the merry-go-round, in a circular trajectory, there must be an external force, keeping it on track.
  • This force, that changes the direction of the rock but not its speed, is the centripetal force, and aims always towards the center of the circle.
  • Now, we need to ask ourselves: what supplies this force?
  • In this case, the only force acting on the rock that could do it, is the friction force, more precisely, the static friction force.
  • We know that this force can be expressed as follows:

       f_{frs} = \mu_{s} * F_{n} (1)

      where μs = coefficient of static friction between the rock and the merry-

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  • In this case, as the surface is horizontal, and the rock is not accelerated in the vertical direction, this force in magnitude must be equal to the weight of the rock, as follows:
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  • This static friction force is just the same as the centripetal force.
  • The centripetal force depends on the square of the angular velocity and the radius of the trajectory, as follows:

       F_{c} = m* \omega^{2}*r (3)

  • Since (1) is equal to (3), replacing (2) in (1) and solving for ω, we get:

       \omega = \sqrt{\frac{\mu_{s} * g}{r} } = \sqrt{\frac{0.7*9.8m/s2}{1.6m}} = 2.1 rad/sec

  • This is the minimum angular velocity that would cause the rock to begin sliding off, due to that if it is larger than this value , the centripetal force will be larger that the static friction force, which will become a kinetic friction force, causing the rock to slide off.
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