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xxTIMURxx [149]
3 years ago
13

A constant voltage is applied across a circuit element. If the resistance of the element is doubled, what is the effect on the p

ower dissipated by this element? The power dissipated is reduced by a factor of 2. The power dissipated remains constant. The power dissipated is doubled. The power dissipated is quadrupled. The power dissipated is reduced by a factor of 4.
Physics
1 answer:
miv72 [106K]3 years ago
5 0

Answer:

The power dissipated is reduced by a factor of 2

Explanation:

The power dissipated by a resistor is given by:

P=I^2 R

where

I is the current

R is the resistance

by using Ohm's law, I=\frac{V}{R}, we can rewrite the previous equation in terms of the voltage applied across the resistor (V):

P=(\frac{V}{R})^2R=\frac{V^2}{R}

In this problem, the resistance of the element is doubled, while the voltage is kept constant. So we have R'=2R while V remains the same; substituting into the formula, we have:

P'=\frac{V^2}{2R}=\frac{1}{2}\frac{V^2}{R}=\frac{P}{2}

so, the power dissipated is reduced by a factor 2.

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A friction force of 280 N exists between a cart and the path. If the force of reaction is 766 N, what is the minimum action forc
MatroZZZ [7]

force of friction on the cart is given as

F_f = 280 N

here we also know the reaction force due to surface

R = 766 N

so we can say reaction force is given as

R = \sqrt{F_n^2 + F_f^2}

766 = \sqrt{F_n^2 + 280^2}

F_n^2 = 766^2 - 280^2

F_n = 713 N

now by force balance we will say

F_y + F_n = mg

F_y = mg - F_n

F_x = F_f

also we know that

F_f = \mu * F_n

280 = \mu * 713

\mu = 0.39

now minimum force required to set this into motion

F_{min} = \frac{\mu mg}{\sqrt{1 + \mu^2}}

here we know that

mg = F_n = 713 N

F_{min} = \frac{0.39* 713}{\sqrt{1 + 0.39^2}}

F_{min} = 259 N

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3 years ago
The most common isotope of hydrogen contains a proton and an electron 'separated by about -11-27 5.0 x 10 m. The mass of proton
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Answer:

A)   F_g = 4.05 10⁻⁴⁷ N, B)   F_e = 9.2 10⁻⁸N, C)    \frac{F_e}{F_g} = 2.3 10³⁹

Explanation:

A) It is asked to find the force of attraction due to the masses of the particles

Let's use the law of universal attraction

            F = G \frac{m_1m_2}{r^2}

let's calculate

            F = 6.67 \ 10^{-11} \ \frac{9.1 \ 10^{-31} \ 1.67 \ 10 ^{-27} }{(5 \ 10^{-11})^2 }

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B) in this part it is asked to calculate the electric force

Let's use Coulomb's law

            F = k \  \frac{q_1q_2}{r^2}

let's calculate

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C) It is asked to find the relationship between these forces

        \frac{F_e}{F_g} = \frac{9.2 \ 10^{-8} }{4.05 \ 10^{-47} }

        = 2.3 10³⁹

therefore the electric force is much greater than the gravitational force

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