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motikmotik
3 years ago
7

One wire has a cross-sectional area of 1,250 cmil and a resistance of 7 ohms. A second piece of wire, identical except for cross

-sectional area, has a resistance of 10 ohms. Determine what the cross-sectional area of this second wire is. (Cross-sectional area and resistance are inversely proportional.)
Physics
1 answer:
Kruka [31]3 years ago
8 0

Answer:

875 cmil

Explanation:

Cross section area of wire=A_1=1250 cmil

Resistance of wire,R_1=7\Omega

R_2=10\Omega

We have to find the cross sectional area of second wire

We know that

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

According to question

l_1=l_2=l,\rho_1=\rho_2=\rho

R_1=\frac{\rho l}{1250}

7=\frac{\rho l}{1250}....(1)

10=\frac{\rho l}{A}...(2)

Equation (1) divided by equation (2) then, we get

\frac{7}{10}=\frac{A}{1250}

A=\frac{7}{10}\times 1250=875cmil

Hence, the cross- sectional area of second wire=875 cmil

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In a pig caller can produce a sound intensity level of 107 dB. How many pig callers would be needed to generate an intensity lev
myrzilka [38]

Answer:

20 pig callers

Explanation:

Given that:

A pig caller produced intensity level of  a sound = 107 dB

To find how many pig callers required to generate an intensity level of 120 dB;

we have:

120 dB - 107 dB = 13 dB

Taking the logarithm function;

10 \ log \bigg(\dfrac{I}{I_o} \bigg) = 13 \ dB

where;

I_o = initial intensity

log \bigg(\dfrac{I}{I_o} \bigg) = 1.3

\dfrac{I}{I_o}=  10^{1.3 }

I = 19.95I_o

I ≅ 20 pig callers

6 0
3 years ago
Experiment with the battery voltage set to 15 volts, measure the current in a parallel circuit with 1,2,3, and 4 light bulbs. (I
koban [17]

Answer:

1) current = I

2) Resistance = V/I

3) current = 2I

4) resistance = V/2I

5) current = 3I

6) Resistance = V/3I

7) Current = 4I

8) Resistance = V/4I

Explanation:

When one bulb is connected across the battery then let say the current is given as I

Then resistance is given as

R = \frac{V}{I}

When two bulbs are in parallel with the battery then

total current becomes twice of initial current

so we have

current = 2I

Resistance of the circuit is now

R = \frac{V}{2I}

When three bulbs are in parallel with the battery then

total current becomes three times of initial current

so we have

current = 3I

Resistance of the circuit is now

R = \frac{V}{3I}

When four bulbs are in parallel with the battery then

total current becomes four times of initial current

so we have

current = 4I

Resistance of the circuit is now

R = \frac{V}{4I}

3 0
3 years ago
A 62 kg skier is moving at 6.5 m/s on frictionless horizontal snow-covered plateau when she encounters a rough patch 3.50 m long
Degger [83]

Answer:

(A). The work done by friction in crossing the patch is -637.98 J.

(B). The speed of skier is 10.57 m/s.

Explanation:

Given that,

Mass of skier = 62 kg

Speed = 6.5 m/s

Length = 3.50 m

Coefficient kinetic friction = 0.30

Height = 2.5 m

(A) we need to calculate the work done by friction in crossing the patch

Using formula of work done

W=-\mu mg\times l

Put the value into the formula

W=-0.30\times62\times9.8\times3.50

W=-637.98\ J

The work done by friction in crossing the patch is -637.98 J.

(B) we need to calculate the speed of skier

Using conservation of energy

K.E_{i}+U_{i}-W_{friction}=K.E_{f}+U_{f}

\dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2+U_{f}

Final potential energy is zero

So, \dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2

\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}v_{1}^2+gh-\mu gl

Put the value into the formula

\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}\times6.5^2+9.8\times2.5+0.30\times9.8\times3.50

v_{2}=\sqrt{2\times55.915}

v_{2}=10.57\ m/s

The speed of skier is 10.57 m/s.

Hence,  (A).The work done by friction in crossing the patch is -637.98 J.

(B).The speed of skier is 10.57 m/s.

6 0
3 years ago
A beam of light strikes one face of a window pane with an angle of incidence of 30.0°. The index of refraction of the glass is 1
natka813 [3]

Answer:

(a). The angle of refraction is 19.26°.

(b). That is proved that the rays in air on either side of the glass are parallel to each other

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Given that,

Angle of incidence = 30.0°

Index of reflection of glass = 1.52

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Using snell's law

\dfrac{\sin i}{\sin r}=\mu

\sin r=\dfrac{\sin i}{\mu}

Put the value into the formula

\sin r=\dfrac{\sin 30}{1.52}

r=\sin^{-1}(0.329)

r=19.26^{\circ}

(b). We know that,

The incident ray and emerging ray  is equal then the ray will be parallel.

We need to prove that the rays in air on either side of the glass are parallel to each other

Using formula for emerging ray

\dfrac{\sin e}{\sin r}=\mu

\sin e=\sin r\times \mu

Put the value into the formula

\sin e=0.3289\times 1.52

e=\sin^{-1}(0.499)

e=29.9\approx 30^{\circ}

So, \sin i=\sin e

This is proved.

Hence, (a). The angle of refraction is 19.26°.

(b). That is proved that the rays in air on either side of the glass are parallel to each other

6 0
3 years ago
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Mkey [24]

When it rains, dust particles and oil residues float on the water and this reduces the traction of tires.

<h3>What is traction?</h3>

This concept refers to a force between the tires and road that causes the movement of the wheels or vehicle is slower.

<h3>What happens with traction when it rains?</h3>

It is well-known more accidents occur when it rains, which is caused by cars slidding on the road. This is because when it rains traction or the grip of the wheel drastically reduces.

Learn more about traction in: brainly.com/question/14525337

8 0
2 years ago
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