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butalik [34]
3 years ago
10

A 591 μF capacitor is discharged through a resistor, whereby its potential difference decreases from its initial value of 88.5 V

to 11.9 V in 3.09 s. Find the resistance of the resistor in kilohms.
Physics
1 answer:
Stels [109]3 years ago
8 0

Answer:

2.6 kilo Ohm

Explanation:

Capacitance, C = 591 μF = 591 x 10^-6 F

Vo = 88.5 V

V = 11.9 V

t = 3.09 s

Let the resistance is R.

V = V_{0}e^{\frac{-t}{RC}}

\frac{11.9}{88.5} = e^{\frac{-t}{RC}}

0.135 = e^{\frac{-t}{RC}}

Take natural log on oth the sides

ln 0.135 = - 3.09 / RC

RC = 1.545

R = 1.545 / ( 591 x 10^-6)

R = 2614.2 ohm

R = 2.6 kilo Ohm

Thus the resistance is  2.6 kilo Ohm.

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An ideal gas in a cylinder occupies a volume of 0.065 m3 at room temperature (T = 293 K). The gas is confined by a piston with a
slamgirl [31]

Answer:

zero

Explanation:

There are three forces acting on the piston

1. force due to atmospheric pressure = F1 downward

2. force due to gaseous pressure = F2 upward

3. force due to the weight placed on the piston = F3 = mg downward

As the piston is in equilibrium condition, so the net force on the piston is zero.

8 0
4 years ago
PLEASE HELP MEE THIS IS DUE IN 45 MINS
guajiro [1.7K]

Answer:

The distance travelled does not depend on the mass of the vehicle. Therefore, s = d

Explanation:

This deceleration situation can be analyzed by means of Work-Energy Theorem, where change in translational kinetic energy is equal to the work done by friction:

\frac{1}{2}\cdot m\cdot v^{2}-\mu\cdot m\cdot g \cdot s = 0 (1)

Where:

m - Mass of the car, in kilogram.

v - Initial velocity, in meters per second.

\mu - Coefficient of friction, no unit.

s - Travelled distance, in meters.

Then we derive an expression for the distance travelled by the vehicle:

\frac{1}{2}\cdot v^{2} = \mu \cdot g \cdot s

s = \frac{v^{2}}{\mu\cdot g}

As we notice, the distance travelled does not depend on the mass of the vehicle. Therefore, s = d

3 0
3 years ago
The pressure exerted by a gas is 2.0 atm while it has a volume of 350 mL. What would be the volume of this sample of gas at stan
34kurt

Answer:

700 mL or 0.0007 m³

Explanation:

P₁ = Initial pressure = 2 atm

V₁ = Initial volume = 350 mL

P₂ = Final pressure = 1 atm

V₂ = Final volume

Here the temperature remains constant. So, Boyle's law can be applied here.

P₁V₁ = P₂V₂

\frac{P_1V_1}{P_2}=V_2\\\Rightarrow V_2=\frac{2\times 350}{1}\\\Rightarrow V_2=700\ mL

So, volume of this sample of gas at standard atmospheric pressure would be 700 mL or 0.0007 m³

7 0
3 years ago
Party hearing. As the number of people at a party increases, you must raise your voice for a listener to hear you against the ba
dusya [7]

Answer:

\frac{r_i}{1.77} m

Explanation:

Given that

At starting separated = 1.20m

And the increase in background noise by Δβ = 5 dB, due to which the level of sound also rises

Based on the above information, the separation rf that is needed is shown below:

As we know that

I_f = I_o \times 10^{\frac{\beta}{10} }\\\\I_f = I_i \times 10^{0.5}\\\\I_f = 3.16 \times  I_i\\\\I \alpha \frac{1}{r^2} \\\\\frac{r_i^2}{r_f^2} = 3.16\\\\r_f = \frac{\sqrt{r_i^2}}{{\sqrt3.16}} \\\\= \frac{r_i}{1.77} m

Hence, the separation r_f i.e. required is  \frac{r_i}{1.77} m

We simply applied the above equation so that the correct separation could come

6 0
3 years ago
You decide to visit Santa Claus at the north pole to put in a good word about your splendid behavior throughout the year. While
ANTONII [103]

To solve this problem it is necessary to apply the concepts related to the Rotational Force described from the equilibrium and Newton's second law.

When there is equilibrium, the Force generated by the tension is equivalent to the Force of the Weight. However in rotation, the Weight must be equivalent to the Centrifugal Force and the tension, in other words:

W = F_T + m\omega^2r_E

Where

\omega = \frac{2\pi}{T} \rightarrow Angular velocity is equal to the Period, at this case Earth's period

r_E = 6.371*10^6m \rightarrow Radius of the Earth

m = mass

F_T= Force of Tension

W = mg \rightarrow Newton's second law

Replacing and re-arrange to find the Tension we have,

F_T = W- \frac{W}{g} (\frac{2\pi}{T})^2r_E

F_T = W(1-(\frac{2\pi}{T})^2\frac{r_E}{g})

F_T = (505)(1-(\frac{2\pi}{24hours})^2\frac{6.371*10^6}{9.8})

F_T = (505)(1-(\frac{2\pi}{24hours(\frac{3600s}{1hour})})^2\frac{6.371*10^6}{9.8})

F_T = (505)(1-(\frac{2\pi}{86400})^2\frac{6.371*10^6}{9.8})

F_T = 503.26N

Therefore when Sneezy is on the equator he is in a circular orbit with a Force of tension of 503.26N

7 0
3 years ago
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