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gregori [183]
2 years ago
13

At the same temperature, two wires made of pure copper have different resistances. The same voltage is applied at the ends of ea

ch wire. The wires may differ in Check all that apply. View Available Hint(s) Check all that apply.(A) length.(B) cross-sectional area.(C) resistivity.(D) amount of electric current passing through them.
Physics
2 answers:
vazorg [7]2 years ago
7 0

Answer:

A and B true, C and D false.

Explanation:

We only know that they have the same resistance and the same voltage is applied to them. We need to remember the formula of resistance of a wire:

R=\rho \frac{L}{A}

Where \rho is the resistivity (a property of the material), L its length and A its cross-sectional area. We could change L and A and still obtain the same R. From this we see that:

A) True. They could have any length.

B) True. They could have any cross-sectional area.

C) False. Resistivity is a property of the material, in this case, copper.

D) False. I=V/R, and both are the same for both wires.

Anastasy [175]2 years ago
3 0

Answer:

(A) length.(B) cross-sectional area.

Explanation:

The two options that they could have different are the length and the cross sectional area, and that can make the have the different resitances the longer the wire the more resitance it will have, since they are made from the same material and the same voltage is applied ot them, neither C or D could be because the resistivity depends solely on the material with which things are made of and both are pure copper wires.

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A charge Q is distributed uniformly along the x axis from x1 to x2. What would be the magnitude of the electric field at x0 on t
Lena [83]

Answer:

  E =  k Q    1 / (x₀-x₂) (x₀-x₁)

Explanation:

The electric field is given by

              dE = k dq / r²

In this case as we have a continuous load distribution we can use the concept of linear density

              λ= Q / x = dq / dx

              dq = λ dx

We substitute in the equation

           ∫ dE = k ∫ λ dx / x²

We integrate

           E = k λ (-1 / x)

We evaluate between the lower limits x = x₀- x₂ and higher x = x₀-x₁

           E = k λ (-1 / x₀-x₁ + 1 / x₀-x₂)

           E = k λ  (x₂ -x₁) / (x₀-x₂) (x₀-x₁)

We replace the density

             E = k (Q / (x₂-x₁)) [(x₂-x₁) / (x₀-x₂) (x₀-x₁)]

             E =  k Q    1 / (x₀-x₂) (x₀-x₁)

8 0
2 years ago
Charge q1 is distance r from a positive point charge q. charge q2=q1/3 is distance 2r from q. what is the ratio u1/u2 of their p
makvit [3.9K]
We need the power law for the change in potential energy (due to the Coulomb force) in bringing a charge q from infinity to distance r from charge Q. We are only interested in the ratio U₁/U₂, so I'm not going to bother with constants (like the permittivity of space). 

<span>The potential energy of charge q is proportional to </span>
<span>∫[s=r to ∞] qQs⁻²ds = -qQs⁻¹|[s=r to ∞] = qQr⁻¹, </span>

<span>so if r₂ = 3r₁ and q₂ = q₁/4, then </span>
<span>U₁/U₂ = q₁Qr₂/(r₁q₂Q) = (q₁/q₂)(r₂/r₁) </span>
<span>= 4•3 = 12.</span>
5 0
2 years ago
A child of mass 27 kg swings at the end of an elastic cord. At the bottom of the swing, the child's velocity is horizontal, and
snow_tiger [21]

Answer:

The magnitude of the rate of change of the child's momentum is 794.11 N.

Explanation:

Given that,

Mass of child = 27 kg

Speed of child in horizontal = 10 m/s

Length = 3.40 m

There is a rate of change of the perpendicular component of momentum.

Centripetal force acts always towards the center.

We need to calculate the magnitude of the rate of change of the child's momentum

Using formula of momentum

\dfrac{dp}{dt}=F

\dfrac{dP}{dt}=\dfrac{mv^2}{r}

Put the value into the formula

\dfrac{dP}{dt}=\dfrac{27\times10^2}{3.40}

\dfrac{dP}{dt}=794.11\ N

Hence, The magnitude of the rate of change of the child's momentum is 794.11 N.

7 0
2 years ago
A flywheel of mass M is rotating about a vertical axis with angular velocity ω0. A second flywheel of mass M/5 is not rotating a
Contact [7]

Answer:

0.83 ω

Explanation:

mass of flywheel, m = M

initial angular velocity of the flywheel, ω = ωo

mass of another flywheel, m' = M/5

radius of both the flywheels = R

let the final angular velocity of the system is ω'

Moment of inertia of the first flywheel , I = 0.5 MR²

Moment of inertia of the second flywheel, I' = 0.5 x M/5 x R² = 0.1 MR²

use the conservation of angular momentum as no external torque is applied on the system.

I x ω = ( I + I') x ω'

0.5 x MR² x ωo = (0.5 MR² + 0.1 MR²) x ω'

0.5 x MR² x ωo = 0.6 MR² x ω'

ω' = 0.83 ω

Thus, the final angular velocity of the system of flywheels is 0.83 ω.

5 0
2 years ago
A 0.200 kg plastic ball moves with a velocity of 0.30 m/s. It collides with a second plastic ball of mass 0.100 kg, which is mov
zzz [600]

Answer:

0.22m/s

Explanation:

The total momentum of the System is conserved. Total momentum of the system before the collision is equal to the total momentum of the system after collision. The total momentum is the sum of individual momentum of all the objects in that system.

momentum of an object = mass* velocity

Total Momentum before collision = 0.2*0.3 + 0.1*0.1= 0.07 kg⋅m/s;

Total momentum after collision = 0.1*0.26 + 0.2*x = 0.07;

Solve for x.

4 0
2 years ago
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