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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

ivanzaharov [21]

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

5 0

3 years ago

Starting from rest, a person runs with a constant acceleration, traveling 40 meters in 10 seconds. What is their final velocity?

Assoli18 [71]

Answer:

Final velocity v = 8.944 m/sec

Explanation:

We have given distance S = 40 meters

Time t = 10 sec

As it starts from rest so initial velocity u = 0

From second equation of motion $s=ut+\frac{1}{2}at^2$

$40=0\times 10+\frac{1}{2}a10^2$

$a=0.8944m/sec^2$

Now from first equation of motion $v=u+at$ , here v is final velocity, u is initial velocity, a is acceleration and t is time

So $v=u+at=0+0.8944\times 10=8.944m/sec$

6 0

4 years ago

Unpolarized light with intensity S is incident on a series of polarizing sheets. The first sheet has its transmission axis orien

jeka94

Answer:

Explanation:

Given

Initial Intensity of light is S

when an un-polarized light is Passed through a Polarizer then its intensity reduced to half.

When it is passed through a second Polarizer with its transmission axis $\theat =45^{\circ}$

$S_1=S_0\cos ^2\theta$

here $S_0=\frac{S}{2}$

$S_1=\frac{S}{2}\times \frac{1}{(\sqrt{2})^2}$

$S_1=\frac{S}{4}$

When it is passed through third Polarizer with its axis $90^{\circ}$ to first but $\theta =45^{\circ}$ to second thus $S_2$

$S_2=S_0\cos ^2\theta$

$S_2=\frac{S}{4}\times \frac{1}{2}$

$S_2=\frac{S}{8}$

When middle sheet is absent then Final Intensity will be zero

3 0

3 years ago

Water flows over a section of Niagara Falls at the rate of 1.4 × 106 kg/s and falls 49.8 m. How much power is generated by the f

Ad libitum [116K]

Answer:

Power= 6.84×10⁸ W

Explanation:

Given Data

Niagara falls at rate of=1.4×10⁶ kg/s

falls=49.8 m

To find

Power Generated

Solution

Regarding this problem

GPE (gravitational potential energy) declines each second is given from that you will find much the kinetic energy of the falling water is increasing each second.

So power can be found by follow

Power= dE/dt = d/dt (mgh)

Power= gh dm/dt

Power= 1.4×10⁶ kg/s × 9.81 m/s² × 49.8 m

Power= 6.84×10⁸ W

7 0

3 years ago

Line d represents water. if the atmospheric pressure in a flask is lowered to 70 kpa, water would boil at what temperature?

sp2606 [1]

The diagram is in the picture attached.
Options are:
A) 32 °C
B) 70 °C
C) 92 °C
D) 100 °C

In order to find the value required, you need to look at the diagram and follow these steps:
1) search for the value of 70 kPa on the y-axis;
2) move on a horizontal line towards the right until you reach the line D;
3) move on a vertical line down, towards the x-axis;
4) read at what value of °C you are at.

Doing so, you can see that you are at a value a little bit above 90 °C (see picture).

Hence, the correct answer is C) 92°C.

5 0

3 years ago