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GREYUIT [131]
3 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]3 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 city government received a $1 million grant to build swimming pools and skating rinks for youth. based on the data provided in
ivolga24 [154]

We have that for the Question "A city government received a $1 million grant to build swimming pools and skating rinks for youth. based on the data provided in the graph, what is the opportunity cost of building one swimming pool?" it can be said that

OC=2

From the question we are told

A city government received a $1 million grant to build swimming pools and skating rinks for youth. based on the data provided in the graph, what is the opportunity cost of building one swimming pool?

Generally the equation for the Opportunity cost  is mathematically given as

OC=\frac{y_1-y_2}{x_1-x_2}\\\\OC=\frac{2-0}{1-0}\\\\OC=2

Therefore

From the graph of the  question we can ascertain that Opportunity cost

OC=2

For more information on this visit

brainly.com/question/18670421

Graph is attached below

6 0
2 years ago
What is the speed at 5s?
Masteriza [31]
25 km/hr I hope this helps;)
6 0
2 years ago
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A magnetic field is uniform over a flat, horizontal circular region with a radius of 1.50 mm, and the field varies with time. In
Daniel [21]

Answer:81.57\mu V

Explanation:

Given

radius of circular region r=1.50 mm

A=\pi r^2=7.069\times 10^{-6}\ m^2

Magnetic Field B=1.50\ T

time t=130 ms

Flux is given by

\phi =B\cdot A

change in Flux d\phi =(B_f-B_i)A

Emf induced is e=\frac{\mathrm{d} \phi}{\mathrm{d} t}

e=\frac{(1.5)\cdot 7.069\times 10^{-6}}{130\times 10^{-3}}

e=81.57 \mu V

3 0
3 years ago
Suppose that you are standing on a train accelerating at 0.20g (where g is the acceleration due to gravity). What minimum coeffi
prisoha [69]

Answer:

0.2

Explanation:

The given parameters are;

The acceleration of the train, a = 0.2·g

The mass of the person standing on the train = m

Let μ represent the coefficient of static friction, we have;

The force acting on the person, F = m × a = m × 0.2·g

The force of friction acting between the feet and the floor, F_f = m·g·μ

For the person not to slide we have;

The force acting on the person = The force of friction acting between the feet and the floor

F = F_f

∴ m × 0.2·g = m·g·μ

From which we get;

0.2 = μ

The coefficient of static friction that must exist between the feet and the floor if the person is not to slide, μ = 0.2.

7 0
3 years ago
A mass is oscillating with amplitude A at the end of a spring.
Dmitry_Shevchenko [17]

A) x=\pm \frac{A}{2\sqrt{2}}

The total energy of the system is equal to the maximum elastic potential energy, that is achieved when the displacement is equal to the amplitude (x=A):

E=\frac{1}{2}kA^2 (1)

where k is the spring constant.

The total energy, which is conserved, at any other point of the motion is the sum of elastic potential energy and kinetic energy:

E=U+K=\frac{1}{2}kx^2+\frac{1}{2}mv^2 (2)

where x is the displacement, m the mass, and v the speed.

We want to know the displacement x at which the elastic potential energy is 1/3 of the kinetic energy:

U=\frac{1}{3}K

Using (2) we can rewrite this as

U=\frac{1}{3}(E-U)=\frac{1}{3}E-\frac{1}{3}U\\U=\frac{E}{4}

And using (1), we find

U=\frac{E}{4}=\frac{\frac{1}{2}kA^2}{4}=\frac{1}{8}kA^2

Substituting U=\frac{1}{2}kx^2 into the last equation, we find the value of x:

\frac{1}{2}kx^2=\frac{1}{8}kA^2\\x=\pm \frac{A}{2\sqrt{2}}

B) x=\pm \frac{3}{\sqrt{10}}A

In this case, the kinetic energy is 1/10 of the total energy:

K=\frac{1}{10}E

Since we have

K=E-U

we can write

E-U=\frac{1}{10}E\\U=\frac{9}{10}E

And so we find:

\frac{1}{2}kx^2 = \frac{9}{10}(\frac{1}{2}kA^2)=\frac{9}{20}kA^2\\x^2 = \frac{9}{10}A^2\\x=\pm \frac{3}{\sqrt{10}}A

3 0
3 years ago
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