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3 years ago

What is the current in a 10V Circuit if the resistance is 2 ohms

Physics

1 answer:

VladimirAG [237]3 years ago

8 0

V=IR can be changed to V/R=I so 10V/2 ohms = 5amps so 5 amps is your answer boss

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You illuminate the surface of some metal with green laser and observe the photocurrent. If you decrease the intensity of laser,

gregori [183]

Remain the same.

Explanation:

The kinetic energy of the ejected electrons, and thus the resulting photocurrent, does not depend on the intensity of the incident radiation. Instead, it depends on the frequency. So since the stopping potential is used to reduce the photocurrent to zero and the photocurrent does not depend on the intensity, the stopping potential remains the same.

7 0

2 years ago

A house is 54.0 ft long and 48 ft wide, and has 8.0-ft-high ceilings. What is the volume of the interior of the house in cubic m

pshichka [43]

Explanation:

Length of house, l = 54 ft = 16.45 m

Breadth of house, b = 48 ft = 14.63 m

Height of house, h = 8 ft = 2.43 m

We need to find the volume of the interior of the house. The house is in the form of cuboid. The volume of cuboid is given by :

$V=l\times b\times h$

$V=16.45\ m\times 14.63\ m\times 2.43\ m$

$V=584.81\ m^3$

Since, $1\ m^3=1000000\ cm^3$

or the volume of the interior of the house, $V=5.84\times 10^8\ cm^3$

Hence, this is the required solution.

7 0

3 years ago

A particle of charge Q is fixed at the origin of an xy coordinate system. At t = 0 a particle (m = 0.959 g, q = 5.84 µC is locat

NemiM [27]

Answer:

$8.66\times 10^{-6}\ C$ or $8.66\ \mu C$ .

Explanation:

<u>Given:</u>

Charge on the particle at origin = Q.
Mass of the moving charged particle, $\rm m = 0.959\ g = 0.959\times 10^{-3}\ kg.$
Charge on the moving charged particle, $\rm q = 5.84\ \mu C = 5.84\times 10^{-6}\ C.$
Distance of the moving charged particle from first at t = 0 time, $\rm r=20.7\ cm = 0.207\ m.$
Speed of the moving particle, $\rm v = 47.9\ m/s.$

For the moving particle to circular motion, the electrostatic force between the two must be balanced by the centripetal force on the moving particle.

The electrostatic force on the moving particle due to the charge Q at origin is given by Coulomb's law as:

$\rm F_e = \dfrac{kqQ}{r^2}.$

where, $\rm k$ is the Coulomb's constant having value $\rm 9\times 10^9\ Nm^2/C^2.$

The centripetal force on the moving particle due to particle at origin is given as:

$\rm F_c = \dfrac{mv^2}{r}.$

For the two forces to be balanced,

$\rm F_e = F_c\\\dfrac{kqQ}{r^2}=\dfrac{mv^2}{r}\\\Rightarrow Q = \dfrac{mv^2}{r}\times \dfrac{r^2}{kq}\\=\dfrac{mv^2r}{kq}\\=\dfrac{(0.959\times 10^{-3})\times (47.9)^2\times (0.207)}{(9\times 10^9)\times (5.84\times 10^{-6})}\\=8.66\times 10^{-6}\ C\\=8.66\ \mu C.$

6 0

3 years ago

What is kinematics ???<br>need a legendary answer -,-" <br>xD

Alenkasestr [34]

Explanation:

the features or properties of motion in an object

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Aloiza [94]

Answer:

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