Ask question

Ask question

All categories

2 years ago

9

Wire B has twice the length and twice the radius of wire A. both wires are made from the same material. if wire A has a resistan

ce R, what is the resistance of wire B?

1 answer:

otez555 [7]2 years ago

3 0

Answer:

Hello There!!

Explanation:

1/2R because wire B has twice the radius,it has four times the cross-sectional area of wire A.

hope this helps,have a great day!!

~Pinky~

You might be interested in

A segment of wire carries a current of 25 A along the x axis from x = −2.0 m to x = 0 and then along the y axis from y = 0 to y

Rashid [163]

Answer: F = 2N

Explanation: If a current i is flowing in a wire of length L lying in a region of magnetic field B, then the magnetic force acting on the wire is given by

F = BIL

Please find the attached file for the solution

8 0

3 years ago

URGENT PLEASE ANSWER THIS ASAP I WILL MARK YOU THE BRAINLIEST !!!

kykrilka [37]

Answer:

C would be the answer

Explanation:

Hope this helps! :)

5 0

2 years ago

Read 2 more answers

When two ions form a bond, the overall charge of that compound will ALWAYS become..

Maslowich

Answer:

C. Neutral

Explanation:

Ions will combine in a way that the overall ionic compound will always be neutral.

8 0

3 years ago

A disk rotates about its central axis starting from rest and accelerates with constant angular acceleration. At one time it is r

aivan3 [116]

Answer:

Explanation:

Given that,

Initial angular velocity is 0

ωo=0rad/s

It has angular velocity of 11rev/sec

ωi=11rev/sec

1rev=2πrad

Then, wi=11rev/sec ×2πrad

wi=22πrad/sec

And after 30 revolution

θ=30revolution

θ=30×2πrad

θ=60πrad

Final angular velocity is

ωf=18rev/sec

ωf=18×2πrad/sec

ωf=36πrad/sec

a. Angular acceleration(α)

Then, angular acceleration is given as

wf²=wi²+2αθ

(36π)²=(22π)²+2α×60π

(36π)²-(22π)²=120πα

Then, 120πα = 8014.119

α=8014.119/120π

α=21.26 rad/s²

Let. convert to revolution /sec²

α=21.26/2π

α=3.38rev/sec

b. Time Taken to complete 30revolution

θ=60πrad

∆θ= ½(wf+wi)•t

60π=½(36π+22π)t

60π×2=58πt

Then, t=120π/58π

t=2.07seconds

c. Time to reach 11rev/sec

wf=wo+αt

22π=0+21.26t

22π=21.26t

Then, t=22π/21.26

t=3.251seconds

d. Number of revolution to get to 11rev/s

∆θ= ½(wf+wo)•t

∆θ= ½(0+11)•3.251

∆θ= ½(11)•3.251

∆θ= 17.88rev.

5 0

3 years ago

Read 2 more answers

A 265 g mass attached to a horizontal spring oscillates at a frequency of 3.40 Hz . At t =0s, the mass is at x= 6.20 cm and has

lara [203]

Answer:

The phase constant is 7.25 degree

Explanation:

given data

mass = 265 g

frequency = 3.40 Hz

time t = 0 s

x = 6.20 cm

vx = - 35.0 cm/s

solution

as phase constant is express as

y = A cosФ ..............1

here A is amplitude that is = $\sqrt{(\frac{v_x}{\omega })^2+y^2 }$ = $\sqrt{(\frac{35}{2\times \pi \times y})^2+6.2^2 }$ = 6.25 cm

put value in equation 1

6.20 = 6.25 cosФ

cosФ = 0.992

Ф = 7.25 degree

so the phase constant is 7.25 degree

5 0

3 years ago