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3 years ago

7

A circuit has a voltage drop of 120 V across an 80 resistor that carries a current of 0.75 A. How much power is used by the resi

stor? Use P = VI.
A. 90 W
B. 60 W
C. 9600 W
D. 2 W

1 answer:

Pani-rosa [81]3 years ago

5 0

Answer:

A. 90 W

Explanation:

Just took the quiz

May 28, 2019

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Using Newton's third law of motion, explain what happens when you let an untied balloon go.

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Answer:

balloon pushes you back

Explanation:

3rd Law: Every action has an equal and opposite reaction

So, when you let go of the balloon it's pushed forward so the balloon pushes you back

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Many industries are powered via distant power stations. Calculate the current flowing through a 7,300m long 10. copper power lin

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Answer:

Current, I = 1000 A

Explanation:

It is given that,

Length of the copper wire, l = 7300 m

Resistance of copper line, R = 10 ohms

Magnetic field, B = 0.1 T

$\mu_o=4\pi \times 10^{-7}\ T-m/A$

Resistivity, $\rho=1.72\times 10^{-8}\ \Omega-m$

We need to find the current flowing the copper wire. Firstly, we need to find the radius of he power line using physical dimensions as :

$R=\rho \dfrac{l}{A}$

$R=\rho \dfrac{l}{\pi r^2}$

$r=\sqrt{\dfrac{\rho l}{R\pi}}$

$r=\sqrt{\dfrac{1.72\times 10^{-8}\times 7300}{10\pi}}$

r = 0.00199 m

or

$r=1.99\times 10^{-3}\ m=2\times 10^{-3}\ m$

The magnetic field on a current carrying wire is given by :

$B=\dfrac{\mu_o I}{2\pi r}$

$I=\dfrac{2\pi rB}{\mu_o}$

$I=\dfrac{2\pi \times 0.1\times 2\times 10^{-3}}{4\pi \times 10^{-7}}$

I = 1000 A

So, the current of 1000 A is flowing through the copper wire. Hence, this is the required solution.

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3 years ago

14. The design for a rotating spacecraft below consists of two rings. The outer ring with a radius of 30 m holds the living quar

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Answer:

$T= 11.0003$ s

Explanation:

From the question we are told that

The outer ring with a radius of 30 m

inner Gravity Approximately 9.80 m/s'

Outer Gravity Approximately 5.35 m/s.

Generally the equation for centripetal force is given mathematically as

Centripetal acceleration enables Rotation therefore?

$\omega ^2 r =Angular\ acc$

Considering the outer ring,

$\omega ^2 r = 9.8$

$\omega ^2= \frac{9.8}{30}$

$\omega = \sqrt{\frac{9.8}{30}}$

$\omega= 0.571 rad/s$

Therefore solving for Period T

Generally the equation for solving Period T is mathematically given as

$T= \frac{2\pi}{\omega}$

$T= \frac{2\pi}{0.571 rad/s}$

$T= 11.0003$ s

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3 years ago

Help me to solve it . It’s urgent

Artist 52 [7]

Answer: 0°

Explanation:

Step 1: Squaring the given equation and simplifying it

Let θ be the angle between a and b.

Given: a+b=c

Squaring on both sides:

... (a+b) . (a+b) = c.c

> |a|² + |b|² + 2(a.b) = |c|²

> |a|² + |b|² + 2|a| |b| cos 0 = |c|²

a.b = |a| |b| cos 0]

We are also given;

|a+|b| = |c|

Squaring above equation

> |a|² + |b|² + 2|a| |b| = |c|²

Step 2: Comparing the equations:

Comparing eq( insert: small n)(1) and (2)

We get, cos 0 = 1

> 0 = 0°

Final answer: 0°

[Reminders: every letters in here has an arrow above on it]

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3 years ago

Light with an intensity of 1 kW/m2 falls normally on a surface and is completely absorbed. The radiation pressure is

kobusy [5.1K]

Answer:

The radiation pressure of the light is 3.33 x 10⁻⁶ Pa.

Explanation:

Given;

intensity of light, I = 1 kW/m²

The radiation pressure of light is given as;

$Radiation \ Pressure = \frac{Flux \ density}{Speed \ of \ light}$

I kW = 1000 J/s

The energy flux density = 1000 J/m².s

The speed of light = 3 x 10⁸ m/s

Thus, the radiation pressure of the light is calculated as;

$Radiation \ pressure = \frac{1000}{3*10^{8}} \\\\Radiation \ pressure =3.33*10^{-6} \ Pa$

Therefore, the radiation pressure of the light is 3.33 x 10⁻⁶ Pa.

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