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1 year ago

In a circuit of 2 lamps in parallel if the current in one lamp is 2a the current in the other lamp is?

Physics

1 answer:

ioda1 year ago

4 0

In a circuit having 2 lamps are connected in parallel to a battery

then the two lamps will be having the same potential as the battery

i.e

$V_{1} = V_{2} = V_{battery}$

As per Ohm's law,

$I_{1} = \frac{V_{1}}{R_{1}}$ and $I_{2} = \frac{V}{R_{2} }$

In other words, each lamp's current is inversely related to its individual resistance. We only know the current in one of the bulbs in this specific instance. We would therefore need further information in order to calculate the current in the other light. Therefore, there isn't enough data to make a statement.

Under the assumption that all physical parameters, including temperature, remain constant, Ohm's law asserts that "the voltage across a conductor is directly proportional to the current flowing through it".

Learn more about Ohm's law here

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The side length of a square is modeled by s=A, where A is the area of the square. Graph the function. What is the side length of

Naya [18.7K]

Answer:

√35 square units

Explanation:

Area of square = 35square units

Formula of area of square is a =s*s where a is area and s is side

So,

s*s = 35

s^2= 35

s = √35

tell me if this is the answer u r looking for

3 0

2 years ago

Ccording to coulomb's law, which pair of charged particles has the lowest potential energy? according to coulomb's law, which pa

Sladkaya [172]

Coulombs law says that the force between any two charges depends on the amount of charges and distance between them. This force is directly proportional to the magnitude of the two charges and inversely proportional to the distance between them.

$F=k\frac{|q_1| |q_2|}{r^2}$

where $q_1\hspace{1mm}and\hspace{1mm}q_2$ are charges, $r$ is the distance between them and k is the coulomb constant.

case 1:

$q_1=-e\\ q_2=+3e\\ r=100pm\\ \Rightarrow F=k\frac{|-e||3e|}{(100pm)^2}=3ke^2\times10^8$

case 2

$q_1=-e\\ q_2=+2e\\ r=100pm\\ \Rightarrow F=k\frac{|-e||2e|}{(100pm)^2}=2ke^2\times10^8$

case 3:

$q_1=-e\\ q_2=+e\\ r=100pm\\ \Rightarrow F=k\frac{|-e||e|}{(200pm)^2}=0.25ke^2\times10^8$

Comparing the 3 cases:

The maximum potential force according to coulombs law is between -1 charge and +3 charge separated by a distance of 100 pm.

3 0

3 years ago

Read 2 more answers

A man-made satellite of mass 6105 kg is in orbit around the earth, making one revolution in 430 minutes. What is the magnitude o

blondinia [14]

Answer:

A gravitational force of 6841.905 newtons is exerted on the satellite by the Earth.

Explanation:

At first we assume that Earth is represented by an uniform sphere, such that the man-made satellite rotates in a circular orbit around the planet. Hence, the following condition must be satisfied:

$\left(\frac{4\pi^{2}}{T^{2}} \right)\cdot r = \frac{G\cdot M}{r^{2}}$ (1)

Where:

$T$ - Period of rotation of the satellite, measured in seconds.

$r$ - Distance of the satellite with respect to the center of the planet, measured in meters.

$G$ - Gravitational constant, measured in newton-square meters per square kilogram.

$M$ - Mass of the Earth, measured in kilograms.

Now we clear the distance of the satellite with respect to the center of the planet:

$r^{3} = \frac{G\cdot M\cdot T^{2}}{4\pi^{2}}$

$r = \sqrt[3]{\frac{G\cdot M\cdot T^{2}}{4\pi^{2}} }$ (2)

If we know that $G = 6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}$ , $M = 6.0\times 10^{24}\,kg$ and $T = 25800\,s$ , then the distance of the satellite is:

$r = \sqrt[3]{\frac{\left(6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (6.0\times 10^{24}\,kg)\cdot (25800\,s)^{2}}{4\pi^{2}} }$

$r \approx 18.897\times 10^{6}\,m$

The gravitational force exerted on the satellite by the Earth is determined by the Newton's Law of Gravitation:

$F = \frac{G\cdot m\cdot M}{r^{2}}$ (3)

Where:

$m$ - Mass of the satellite, measured in kilograms.

$F$ - Force exerted on the satellite by the Earth, measured in newtons.

If we know that $G = 6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}$ , $M = 6.0\times 10^{24}\,kg$ , $m = 6105\,kg$ and $r \approx 18.897\times 10^{6}\,m$ , then the gravitational force is:

$F = \frac{\left(6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (6105\,kg)\cdot (6\times 10^{24}\,kg)}{(18.897\times 10^{6}\,m)^{2}}$

$F = 6841.905\,N$

A gravitational force of 6841.905 newtons is exerted on the satellite by the Earth.

4 0

3 years ago

Water is formed when two hydrogen atoms bond to an oxygen atom. The hydrogen and the oxygen in this example are different

Goryan [66]

They both are two different elements.SO the answer is Element

8 0

3 years ago

Kevin is refinishing his rusty wheelbarrow. He moves his sandpaper back and forth 45 times over a rusty area, each time moving a

dmitriy555 [2]

W = _|....F*dx*cos(a)........With F=force, x=distance over which force acts on object,
.......0.............................and a=angle between force and direction of travel.

Since the force is constant in this case we don't need the equation to be an integral expression, and since the force in question - the force of friction - is always precisely opposite the direction of travel (which makes (a) equal to 180 deg, and cos(a) equal to -1) the equation can be rewritted like so:

W = F*x*(-1) ............ or ............. W = -F*x

The force of friction is given by the equation: Ffriction = Fnormal*(coeff of friction)

Also, note that the total work is the sum of all 45 passes by the sandpaper. So our final equation, when Ffriction is substituted, is:

W = (-45)(Fnormal)(coeff of friction)(distance)
W = (-45)...(1.8N).........(0.92).........(0.15m)
W = ................-11.178 Joules

5 0

3 years ago