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Murljashka [212]
3 years ago
10

A wire 6.60 m long with diameter of 2.05 mm has a resistance of 0.0310 Ω.

Physics
1 answer:
Alex73 [517]3 years ago
6 0

Answer:

1.551×10^-8 Ωm

Explanation:

Resistivity of a material is expressed as shown;.

Resistivity = RA/l

R is the resistance of the material

A is the cross sectional area

l is the length of the wire.

Given;

R = 0.0310 Ω

A = πd²/4

A = π(2.05×10^-3)²/4

A = 0.000013204255/4

A = 0.00000330106375

A = 3.30×10^-6m

l = 6.60m

Substituting this values into the formula for calculating resistivity.

rho = 0.0310× 3.30×10^-6/6.60

rho = 1.023×10^-7/6.60

rho = 1.551×10^-8 Ωm

Hence the resistivity of the material is 1.551×10^-8 Ωm

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2. A bus drove 8 meters East, then turned to drive 8 meters North, then 2 meters
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2 years ago
A5.0 kg TNT explosive, initially at rest, explodes into two pieces. One of the pieces weighing 2.0 kg flies off to
nordsb [41]

Answer:

v = 24 m/s, rightwards

Explanation:

Given that,

The mass of TBT explosive = 5 kg

It explodes into two pieces.

One of the pieces weighing 2.0 kg flies off to  the left at 36 m/s. Let left be negative and right be positive.

The law of conservation of momentum holds here. Let v be the final speed of the remaining piece. So,

5\times 0=2\times (-36)+3\times v\\\\-72=-3v\\\\v=24\ m/s

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4 0
3 years ago
A photographer wants to determine the color of light he can use in the darkroom that will not expose the films he is processing.
Sonbull [250]

For a photographer that wishes to determine the color of light that he can use in a dark room that will not expose the films he is processing, having used a Blue Incandescent bulb, he should proceed to use a Red Incandescent bulb for the next trial.

The photographer in question is performing an experiment. For these kinds of experiments it is important to identify the variables present, which can be of three kinds:

  1. Control variables
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  3. Independent variables

For this experiment, the dependent variable is the exposure of the light onto the films, given that this is what we wish to measure. The independent variable will be the color of the light being used which is what will affect the dependent variable.

The remaining variable must be the control variable. Unlike the previous variables, we can have more than one of these. The control variable is there to make sure that only the dependent variable is affecting the outcome. We do this by keeping the control variable the same through each trial, which is why the photographer should not change the type of bulb in the second experiment, changing only the color of the light.

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brainly.com/question/1549017?referrer=searchResults

4 0
3 years ago
g determine what frequency is required of a source powering a 100 uf capacitor a 500 ohm resistor and a s50 mH inductor in serie
polet [3.4K]

Answer: 71.16\ Hz

Explanation:

Given

Capacitance C=100\ \mu F

Resistance R=500\ \Omega

Inductance L=50\ mH

In LCR circuit, current is maximum at resonance frequency i.e.

X_L=X_C\ \text{and}\ \omega_o=\dfrac{1}{\sqrt{LC}}

Insert the values

\Rightarrow \omega_o=\dfrac{1}{\sqrt{50\times 10^{-3}\times 100\times 10^{-6}}}\\\\\Rightarrow \omega_o=\dfrac{1}{\sqrt{5}\times 10^{-3}}\\\\\Rightarrow \omega_o=0.447\times 10^{3}

Also, frequency is given by

\Rightarrow 2\pi f=\omega_o\\\\\Rightarrow f=\frac{\omega_o}{2\pi}

\Rightarrow f=\dfrac{1}{2\pi}\times 0.447\times 10^3\\\\\Rightarrow f=71.16\ Hz

8 0
3 years ago
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