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miskamm [114]
3 years ago
14

A 594 Ω resistor, an uncharged 1.3 μF capacitor, and a 6.53 V emf are connected in series. What is the current in milliamps afte

r one time constant? You are supposed to use a constant e can you please include what the value should be for this and all other constants.
Physics
1 answer:
ivanzaharov [21]3 years ago
5 0

Answer:

6.88 mA

Explanation:

Given:

Resistance, R = 594 Ω

Capacitance = 1.3 μF

emf, V = 6.53 V

Time, t = 1 time constant

Now,

The initial current, I₀ = \frac{\textup{V}}{\textup{R}}

or

I₀ = \frac{\textup{6.53}}{\textup{594}}

or

I₀ = 0.0109 A

also,

I = I_0[1-e^{-\frac{t}{\tau}}]

here,

τ = time constant

e = 2.717

on substituting the respective values, we get

I = 0.0109[1-e^{-\frac{\tau}{\tau}}]

or

I = 0.0109[1-2.717^{-1}]

or

I = 0.00688 A

or

I = 6.88 mA

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A stone with a weight of 5.30 N is launched vertically from ground level with an initial speed of 23.0 m/s, and the air drag on
frozen [14]

Answer:

a) h=25.7\ m

b) v'=21.8733\ m.s^{-1}

Explanation:

Given:

  • weight of the stone, w=5.3\ N
  • initial velocity of vertical projection, u=23\ m.s^{-1}
  • air drag acting opposite to the motion of the stone, D=0.266\ N

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m=\frac{w}{g}

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m=0.5408\ kg

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D=m.d

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d = deceleration

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<u>In course of going up the net acceleration on the stone will be:</u>

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a)

Now using the equation of motion:

v^2=u^2-2 g'.h

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v= final velocity when the stone reaches at the top of the projectile = 0

h = height attained by the stone before starting to fall down

0^2=23^2-2\times 10.2918\times h

h=25.7\ m

b)

during the course of descend from the top height of the projectile:

initial velocity, v=0\ m.s^{-1}

The acceleration will be:

g"=g-d

g"=9.8-0.4918

g"=9.3082\ m.s^{-2}

here the gravity still acts downwards but the drag acceleration acts in the direction opposite to the motion of the stone, now the stone is falling down hence the drag  acts upwards.

Using equation of  motion:

v'^2=v^2+2g".h (+ve acceleration because it acts in the direction of motion)

v'^2=0^2+2\times 9.3082\times 25.7

v'=21.8733\ m.s^{-1}

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