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GREYUIT [131]
2 years ago
6

A string of 26 identical Christmas tree lights are connected in series to a 120 V source. The string dissipates 73 W. What is th

e equivalent resistance of the light string? Answer in units of Ω.
Physics
1 answer:
spin [16.1K]2 years ago
6 0

To solve this problem we will apply the concepts related to Ohm's law and Electric Power. By Ohm's law we know that resistance is equivalent to,

R_{eq}= \frac{V}{I}

Here,

V = Voltage

I = Current

While the power is equivalent to the product between the current and the voltage, thus solving for the current we have,

P=VI \rightarrow I = \frac{P}{V}

I =0.608 A

Applying Ohm's law

R_{eq} = \frac{120V}{0.608A}

R_{eq} = 197.4\Omega

Therefore the equivalent resistance of the light string is 197.4\Omega

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A 20 kg object is dropped from a very tall building. What is the weight of this objects? After 5 seconds, how has the object fal
prohojiy [21]

1. What is the weight of this objects?

Weight is simply the product of mass and gravitational acceleration. Therefore the weight is:

w = 20 kg * 9.81 m/s^2

w = 196.2 kg m/s^2 = 196.2 N

 

2. After 5 seconds, how has the object fallen and what is its speed at this instant?

We can use the formula:

<span>y = v0 t  + 0.5 g t^2</span>

v = v0 + g t

where v0 = 0 since the object starts from rest, y is the distance it fell, t is time

y = 0 + 0.5 * 9.81 * 5^2 = 122.625 m

<span>v = 0 + 9.81 * 5 = 49.05 m/s</span>

8 0
3 years ago
A satellite of mass M = 270kg is in circular orbit around the Earth at an altitude equal to the earth's mean radius (6370 km). A
zubka84 [21]

To solve this problem we will apply the concepts related to Orbital Speed as a function of the universal gravitational constant, the mass of the planet and the orbital distance of the satellite. From finding the velocity it will be possible to calculate the period of the body and finally the gravitational force acting on the satellite.

PART A)

V_{orbital} = \sqrt{\frac{GM_E}{R}}

Here,

M = Mass of Earth

R = Distance from center to the satellite

Replacing with our values we have,

V_{orbital} = \sqrt{\frac{(6.67*10^{-11})(5.972*10^{24})}{(6370*10^3)+(6370*10^3)}}

V_{orbital} = 5591.62m/s

V_{orbital} = 5.591*10^3m/s

PART B) The period of satellite is given as,

T = 2\pi \sqrt{\frac{r^3}{Gm_E}}

T = \frac{2\pi r}{V_{orbital}}

T = \frac{2\pi (2*6370*10^3)}{5.591*10^3}

T = 238.61min

PART C) The gravitational force on the satellite is given by,

F = ma

F = \frac{1}{4} mg

F = \frac{270*9.8}{4}

F = 661.5N

5 0
3 years ago
Radiant energy comes from the _______.
snow_lady [41]

Answer:

C. Sun

Explanation:

7 0
3 years ago
Read 2 more answers
I need to choose a theme for my physics assignment My experiment is finding g
Kobotan [32]
<h3>Question:</h3>

How to find g (acceleration due to gravity)

<h3>Solution:</h3>

We know,

Acceleration due to gravity (g)

=  \frac{GM}{ {R}^{2} }

where, G = Gravitational constant

= 6.67 \times  {10}^{11} N {m}^{2}/k {g}^{2}  \\

M = Mass of the earth

= 6 \times  {10}^{24} \:  kg

R = Radius of the earth

= 6.4 \times  {10}^{6} m

Putting these values of G, M and R in the above formula, we get

g \:  =  \:  \frac{6.67 \times  {10}^{11} N {m}^{2}/k {g}^{2}   \times \: 6 \times  {10}^{24} \:  kg }{(6.4 \times  {10}^{6}m {)}^{2}  }  \\  = 9.8m/ {s}^{2}

So, the value of acceleration due to gravity is

9.8m/s ^{2}

Hope it helps.

Do comment if you have any query.

5 0
2 years ago
1. The resistance of an electric device is 40,000 microhms. Convert that measurement to ohms.
lidiya [134]

Answer:

The answer would be 0.04ohms.

Explanation:

Hopefully this helps

4 0
2 years ago
Read 2 more answers
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