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Natali5045456 [20]

3 years ago

13

An uncharged capacitor is connected to the terminals of a 4.0 V battery, and 6.0 μC flows to the positive plate. The 4.0 V batte

ry is then disconnected and replaced with a 7.0 V battery, with the positive and negative terminals connected in the same manner as before.

1 answer:

Ugo [173]3 years ago

5 0

Answer:

1 μC extra charge will be flow here

Explanation:

Given data

battery V1 = 4.0 V

flows Q1 = 6.0 μC

replace battery V2 = 7.0 V

to find out

what happen if we replace battery

solution

we apply here principal of capacitor

that is Q directly proportional voltage

so we say Q2/Q1 = V2/ V1

put all value here

Q2/Q1 = V2/ V1

Q2/6 = 7/ 6

Q2 = 7

so we see here 7 μC will be flow

and Q = Q2 - Q1 = 7 - 6 = 1 μC

so we also say that 1 μC extra charge will be flow here

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Oksanka [162]

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6 0

3 years ago

Read 2 more answers

Two cars are traveling along a straight line in the same direction, the lead car at 25 m/s and the other car at 35 m/s. At the m

Phoenix [80]

Answer:

a. $t_1=12.5\ s$

b. $a_2=-13.61\ m.s^{-2}$ must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping

c. $t_2=2.5714\ s$ is the time taken to stop after braking

Explanation:

Given:

speed of leading car, $u_1=25\ m.s^{-1}$

speed of lagging car, $u_{2}=35\ m.s^{-1}$

distance between the cars, $\Delta s=45\ m$

deceleration of the leading car after braking, $a_1=-2\ m.s^{-2}$

a.

Time taken by the car to stop:

$v_1=u_1+a_1.t_1$

where:

$v_1=0$ , final velocity after braking

$t_1=$ time taken

$0=25-2\times t_1$

$t_1=12.5\ s$

b.

using the eq. of motion for the given condition:

$v_2^2=u_2^2+2.a_2.\Delta s$

where:

$v_2=$ final velocity of the chasing car after braking = 0

$a_2=$ acceleration of the chasing car after braking

$0^2=35^2+2\times a_2\times 45$

$a_2=-13.61\ m.s^{-2}$ must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping

c.

time taken by the chasing car to stop:

$v_2=u_2+a_2.t_2$

$0=35-13.61\times t_2$

$t_2=2.5714\ s$ is the time taken to stop after braking

7 0

3 years ago

If we were to illuminate them only with light from the Balmer transition considered above, would the solar panels produce a curr

Ugo [173]

Answer:

No

Explanation:

The reason why no current is produced are basically that, the wavelengths of light in the Balmer transition are reflected, not absorbed in solar panels, hence no current is produced.

The Balmer series consists of lines in the visible spectrum. It corresponds to emission of a photon of light when electrons descend from higher energy levels to the n=2 level in the hydrogen spectrum. The various wavelengths in the Balmer series can be separated by a prism since they are all in the visible region of the electromagnetic spectrum.

In solar panels, light corresponding to the wavelengths in the Balmer series is merely reflected by the panel and not absorbed. Since light is not absorbed, no current can be produced when the panel is irradiated with light corresponding to the wavelengths in the Balmer series.

6 0

2 years ago

Convert 500ml to how many liters

Mashutka [201]

500 ml = 0.5 liters. that's what i'm getting

hope it helps

7 0

2 years ago

Read 2 more answers

A solenoid with an inductance of 8 mH is connected in series with a resistance of 5 Ω and an EMF forming a series RL circuit. A

monitta

Answer:

induced EMF = 240 V

and by the lenz's law direction of induced EMF is opposite to the applied EMF

Explanation:

given data

inductance = 8 mH

resistance = 5 Ω

current = 4.0 A

time t = 0

current grow = 4.0 A to 10.0 A

to find out

value and the direction of the induced EMF

solution

we get here induced EMF of induction is express as

E = - L $\frac{dI}{dt}$ ...................1

so E = - L $\frac{I2 - I1}{dt}$

put here value we get

E = - 8 × $10^{-3}$ $\frac{10 - 4}{0.2*10^{-3}}$

E = -40 × 6

E = -240

take magnitude

induced EMF = 240 V

and by the lenz's law we get direction of induced EMF is opposite to the applied EMF

5 0

3 years ago