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Delvig [45]
2 years ago
14

If a current is two amps and the resistance is 3 ohms, how much voltage was needed?

Physics
1 answer:
Hunter-Best [27]2 years ago
7 0

Answer:

6 V

Explanation:

We can solve the problem by using Ohm's law:

V=RI

where

V is the voltage in the circuit

R is the resistance

I is the current

In this problem, we know the current, I=2 A, and the resistance, R=3 \Omega, therefore we can find the voltage in the circuit:

V=RI=(3 \Omega )(2 A)=6 V

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A(n) 55.5 g ball is dropped from a height of 53.6 cm above a spring of negligible mass. The ball compresses the spring to a maxi
Serggg [28]

Answer:

The spring force constant is  k=243\ \frac{N}{m} .

Explanation:

We are told the mass of the ball is m=0.0555\ kg, the height above the spring where the ball is dropped is h=0.536\ m,  the length the ball compresses the spring is d=0.04897\ m and the acceleration of gravity is 9.8\ \frac{m}{s^{2}} .

We will consider the initial moment to be when the ball is dropped and the final moment to be when the ball stops, compressing the spring. We supose that there is no friction so the initial mechanical energy E_{mi} is equal to the final mechanical energy E_{mf} :

                                                    E_{mf}=E_{mi}

Initially there is only gravitational potential energy because the force of the spring isn't present and the speed is zero. In the final moment there is only elastic potential energy because the height is zero and the ball has stopped. So we have that:

                                                   \frac{1}{2}kd^{2}=mgh

If we manipulate the equation we have that:

                                                    k=\frac{2mgh}{d^{2} }

                                         k=\frac{2\ 0.0555\ kg\ 9.8\frac{m}{s^{2}}\ 0.536\ m}{(0.04897)^{2}m^{2}}

                                              k=\frac{0.58306\ \frac{kgm^{2}}{s^{2}}}{2.398x10^{-3}m^{2}}

                                                     k=243\ \frac{N}{m}

                                                   

                             

5 0
3 years ago
A point charge q is located at the center of a spherical shell of radius a that has a charge −q uniformly distributed on its sur
muminat

Answer:

a) E = 0

b) E =  \dfrac{k_e \cdot q}{ r^2 }

Explanation:

The electric field for all points outside the spherical shell is given as follows;

a) \phi_E = \oint E \cdot  dA =  \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}

From which we have;

E \cdot  A =  \dfrac{{\Sigma Q}}{\varepsilon _{0}} = \dfrac{+q + (-q)}{\varepsilon _{0}}  = \dfrac{0}{\varepsilon _{0}} = 0

E = 0/A = 0

E = 0

b) \phi_E = \oint E \cdot  dA =  \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}

E \cdot  A  = \dfrac{+q }{\varepsilon _{0}}

E  = \dfrac{+q }{\varepsilon _{0} \cdot A} = \dfrac{+q }{\varepsilon _{0} \cdot 4 \cdot \pi \cdot r^2}

By Gauss theorem, we have;

E\oint dS =  \dfrac{q}{\varepsilon _{0}}

Therefore, we get;

E \cdot (4 \cdot \pi \cdot r^2) =  \dfrac{q}{\varepsilon _{0}}

The electrical field outside the spherical shell

E =  \dfrac{q}{\varepsilon _{0} \cdot (4 \cdot \pi \cdot r^2) }= \dfrac{q}{4 \cdot \pi \cdot \varepsilon _{0} \cdot r^2 }=  \dfrac{q}{(4 \cdot \pi \cdot \varepsilon _{0} )\cdot r^2 }

k_e=  \dfrac{1}{(4 \cdot \pi \cdot \varepsilon _{0} ) }

Therefore, we have;

E =  \dfrac{k_e \cdot q}{ r^2 }

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2 years ago
Contrast the behavior of a water wave that travel by a stone barrier to a sound wave that travels through a door
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Explanation:

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An ideal parallel-plate capacitor consists of a set of two parallel plates of area Separated by a very small distance 푑. This ca
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Answer:

doubled the initial value

Explanation:

Let the area of plates be A and the separation between them is d.

Let V be the potential difference of the battery.

The energy stored in the capacitor is given by

U = Q^2/2C   ...(1)

Now the battery is disconnected, it means the charge is constant.

the separation between the plates is doubled.

The capacitance of the parallel plate capacitor is inversely proportional to the distance between the plates.

C' = C/2

the new energy stored

U' = Q^2 /  2C'

U' = Q^2/C = 2 U

The energy stored in the capacitor is doubled the initial amount.

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3 years ago
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Hello There!

It's a Property of matter where atoms in an object are aligned into domains

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3 years ago
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