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Ksju [112]
3 years ago
14

Several wires of varying thickness are all made of the same material and all have the same length. If the wires are arranged in

order of decreasing thickness, what can be said about the ordering of their resistance?A. The wires will be arranged in order of increasing resistance.B. The wires will be arranged in order of decreasing resistance.C. The ordering of the resistances can not be determined because the resistivity is not given.D. The ordering of the resistances can not be determined because the length is not given.
Physics
1 answer:
aev [14]3 years ago
5 0

Answer:

Option A is correct.

The wires will be arranged in order of increasing resistance.

Explanation:

The resistance of a wire is given by

r = (ρl)/A

where r = resistance of the wire

ρ = resistivity of the wire

L = length of the wire

A = cross sectional area of the wire

Provided all the other parameters are constant, resistance is inversely proportional to cross sectional area

r ∝ (1/A)

And the the cross sectional Area of the wire increases with increase in thickness & decreases with thickness

So, decreasing thickness ----> Decreasing Cross sectional Area ----> Increasing resistance.

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A car accelerates uniformly from rest and
miv72 [106K]

Answer:

29.75 revolutions

Explanation:

The kinematic formula for distance, given a uniform acceleration a and an initial velocity v₀, is

d=v_0t+\frac{1}{2}at^2

This car is starting from rest, so v₀ = 0 m/s. Additionally, we have a = 9.2/9.7 m/s² and t = 9.7 s. Plugging these values into our equation:

d=0t+\frac{1}{2}\left(\frac{9.2}{9.7}\right)(9.7)^2\\d=\frac{1}{2}(9.2)(9.7)\\d=4.6(9.7)\\d=44.62

So, the car has travelled 44.62 m in 9.7 seconds - we want to know how many of the tire's <em>circumferences</em> fit into that distance, so we'll first have to calculate that circumference. The formula for the circumference of a circle given its diameter is c=\pi{d}, which in this case is 47.8π cm, or, using π ≈ 3.14, 47.8(3.14) = 150.092 cm.

Before we divide the distance travelled by the circumference, we need to make sure we're using the same units. 1 m = 100 cm, so 105.092 cm ≈ 1.5 m. Dividing 44.62 m by this value, we find the number of revs is

44.62/1.5\approx29.75 revolutions

7 0
3 years ago
The acceleration of gravity on earth is approximately 10 m/s2 (more precisely, 9.8 m/s2. if you drop a rock from a tall building
VLD [36.1K]
From the equation;
a= -1÷2gT2
a= -5×3(3)
a= -45m/s
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3 years ago
A student is experimenting with some insulated copper wire and a power supply. She winds a single layer of the wire on a tube wi
OverLord2011 [107]

Answer:

P=214.7187\,W

Explanation:

Given that:

Diameter of the solenoid, D=10\,cm=0.1\,m

length of the solenoid, L=90\,cm=0.9\,m

diameter of the wire, d=0.1\,cm=10^{-3}\,m

magnetic field at the center of the solenoid, B=7.4\times 10^{-3}\,T

<u>Now we need the no. of turns incorporated in the length of 90 cm:</u>

N=\frac{Length\,\,of\,\,solenoid}{diameter\,\,of\,\, wire}

N=\frac{L}{d}

N=\frac{0.9}{10^{-3}}

N=900\,\,turns

For solenoids we have:

B=\mu.n.I ...............................(1)

where:

\mu=permeability of the medium

n = no. of turns per unit length

I = current in the coil

So,

n=\frac{900}{0.9}

n=1000\,turns\,.\,m^{-1}

Now putting the respective values in the eq. (1)

7.4\times 10^{-3}=4\pi\times10^{-7}\times 1000\times I

I=5.8887\,A

  • For copper we have resistivity:
  • \rho=1.72\times 10^{-8}\, \Omega.m

We know that resistance is given by:

R=\rho.\frac{l}{a} .....................................(2)

where:

l = length of the conducting wire

a = cross sectional area of the conducting wire

<u>Now we need the length (l) of the wire:</u>

Circumference of the solenoid,

C=\pi.D

C=0.1\pi\,m

\therefore l=C\times N

l=90\pi\,m

&

<u>Cross-sectional area of wire:</u>

a=\pi.\frac{d^2}{4}

a=\pi. \frac{(10^{-3})^2}{4}\,m^2

<u>Resistance from eq. (2):</u>

R=1.72\times 10^{-8}\times \frac{90\pi}{\pi. \frac{(10^{-3})^2}{4}}

R=6.192 \,\Omega

  • For power we have:

P=I^2.R

P=5.8887^2 \times 6.192

P=214.7187\,W

6 0
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They did not affect Florida- never came that far South.
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