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Semmy [17]
3 years ago
6

The resistivity of a silver wire with a radius of 5.04 × 10–4 m is 1.59 × 10–8 ω · m. if the length of the wire is 3.00 m, what

is the resistance of the wire?
Physics
1 answer:
Alexxx [7]3 years ago
4 0
<span>5.98 x 10^-2 ohms. Resistance is defined as: R = rl/A where R = resistance in ohms r = resistivity (given as 1.59x10^-8) l = length of wire. A = Cross sectional area of wire. So plugging into the formula, the known values, including the area of a circle being pi*r^2, gives: R = 1.59x10^-8 * 3.00 / (pi * (5.04 x 10^-4)^2) R = (4.77 x 10^-8) / (pi * 2.54016 x 10 ^-7) R = (4.77 x 10^-8) / (7.98015 x 10^-7) R = 5.98 x 10^-2 ohms So that wire has a resistance of 5.98 x 10^-2 ohms.</span>
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Photons are particles of electromagnetic radiation.

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What is the lift (in newtons) due to Bernoulli's principle on a wing of area 76 m2 if the air passes over the top and bottom sur
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So lift will be 30.19632 N

Explanation:

We have given area of the wing a=76m^2

We know that density of air d=1.29kg/m^3

Speed at top surface v_2=290m/sec and speed at bottom surface v_1=150m/sec

According to Bernoulli's principle force is given by

F=A\times d\times \frac{v_2^2-v_1^2}{2}=76\times 1.29\times \frac{290^2-150^2}{2}=3019632N

4 0
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When a deck of cards slips out of your hands and falls to the ground, gravity<br> does work on it
motikmotik

Answer:

yes

Explanation:

3 0
3 years ago
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If an otherwise empty pressure cooker is filled with air of room temperature and then placed on a hot stove, what would be the m
Xelga [282]

Answer:

The magnitude of the net force F₁₂₀ on the lid when the air inside the cooker has been heated to 120 °C is \frac{135.9}{A}N

Explanation:

Here we have

Initial temperature of air T₁ = 20 °C = ‪293.15 K

Final temperature of air T₁ = 120 °C = 393.15 K

Initial pressure P₁ = 1 atm = ‪101325 Pa

Final pressure P₂ = Required

Area = A

Therefore we have for the pressure cooker, the volume is constant that is does not change

By Chales law

P₁/T₁ = P₂/T₂

P₂ = T₂×P₁/T₁ = 393.15 K× (‪101325 Pa/‪293.15 K) = ‭135,889.22 Pa

∴ P₂ = 135.88922 KPa = 135.9 kPa

Where Force = \frac{Pressure}{Area} we have

Force = F_{120}=\frac{135.9}{A}N.

4 0
4 years ago
A convex refracting surface has a radius of 12 cm. Light is incident in air (n = 1) and refracted into a medium with an index of
OLga [1]

Answer:

<h2>36cm from the surface</h2>

Explanation:

Equation of refraction of a lens is expression according to the formula given below;

\dfrac{n_2}{v} = \dfrac{n_1}{u}=  \dfrac{n_2-n_1}{R}

R is the radius of curvature of the convex refracting surface = 12cm

v is the image distance from the refracting surface

u  is the object distance from the refracting surface

n₁ and n₂ are the refractive indices of air and the medium respectively

Given parameters

R = 12 cm

u = \infty (since light incident is parallel to the axis)

n₁  = 1

n₂  = 1.5

Required

<em>focus point of the light that is incident and parallel to the central axis (v)</em>

<em />

Substituting this values into the given formula we will have;

\dfrac{1.5}{v} - \dfrac{1}{\infty}=  \dfrac{1.5-1}{12}\\\\\dfrac{1.5}{v} -0=  \dfrac{0.5}{12}\\\\\dfrac{1.5}{v}=  \dfrac{0.5}{12}\\\\

Cross multiply

1.5*12 = 0.5*v\\ \\18 = 0.5v\\\\v = \frac{18}{0.5}\\ \\v = 36cm

Hence  Light incident parallel to the central axis is focused at a point 36cm from the surface

6 0
3 years ago
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