Ask question

Ask question

All categories

BabaBlast [244]

2 years ago

7

A capacitor with C = 6.00 μF is fully charged by connecting it to a battery that has emf 50.0 V. The capacitor is disconnected f

rom the battery. A resistor of resistance R = 185 Ω is connected across the capacitor, and the capacitor discharges through the resistor.
What is the charge q on the capacitor when the current in the resistor is 0.180 A?

1 answer:

Arte-miy333 [17]2 years ago

4 0

Answer:

1.99×10^-4coulombs

Explanation:

The charge (Q) across the resistor the directly proportional to the voltage (V) where capacitance of the capacitor(C) is the proportionality constant. Mathematically, Q = CV

If V is the voltage across the resistor, V = IR (according to ohm's law) where I is the current in the resistor and R is the resistance.

We need to calculate the voltage on the resistor first when 0.18A current is passed through it.

V = 0.18 × 185

V = 33.3Volts

The charge Q on the resistor will be;

Q = CV

Were C = 6.00 μF, V = 33.3

Q= 6×10^-6 ×33.3

Q = 0.0001998

Q= 1.99×10^-4Coulombs

You might be interested in

An alpha particle (α), which is the same as a helium-4 nucleus, is momentarily at rest in a region of space occupied by an elect

beks73 [17]

Answer:

Speed of the alpha particle is $v=1.8180\times 10^3m/sec$

Explanation:

We have given charge on alpha particle $q=3.2\times 10^{-19}C$

Mass of the alpha particle $m=6.68\times 10^{-27}kg$

Potential difference $V=-3.45\times 10^{-3}volt$

We have to find the speed of the alpha particle

From energy conservation we know that

$\frac{1}{2}mv^2=qV$

$\frac{1}{2}\times 6.68\times 10^{-27}\times v^2=3.2\times 10^{-19}\times 3.45\times 10^{-3}$

$v=1.8180\times 10^3m/sec$

4 0

3 years ago

Before the start of a trip, air in a tire is at 320 kPa gage pressure and 27°C temperature. At the end of the trip the tire pres

tia_tia [17]

Answer:

The temperature of air in the tire is 55.57 ºC

Explanation:

Please look at the solution in the attached Word file

7 0

3 years ago

g An airplane is flying through a thundercloud at a height of 1500 m. (This is a very dangerous thing to do because of updrafts,

Stolb23 [73]

Answer:

$523269.9\ \text{N/m}$

Explanation:

q = Charge

r = Distance

$q_1=25\ \text{C}$

$r_1=3000\ \text{m}$

$q_2=40\ \text{C}$

$r_2=850\ \text{m}$

The electric field is given by

$E=E_1+E_1\\\Rightarrow E=k(\dfrac{q_1}{r_1^2}+\dfrac{q_2}{r_2^2})\\\Rightarrow E=9\times 10^9\times (\dfrac{25}{3000^2}+\dfrac{40}{850^2})\\\Rightarrow E=523269.9\ \text{N/m}$

The electric field at the aircraft is $523269.9\ \text{N/m}$

4 0

3 years ago

3. Maverick and Goose are flying a training mission in their F-14. They are

Elanso [62]

Answer:

A. The bomb will take <em>17.5 seconds </em>to hit the ground

B. The bomb will land <em>12040 meters </em>on the ground ahead from where they released it

Explanation:

Maverick and Goose are flying at an initial height of $y_0=1500m$ , and their speed is v=688 m/s

When they release the bomb, it will initially have the same height and speed as the plane. Then it will describe a free fall horizontal movement

The equation for the height y with respect to ground in a horizontal movement (no friction) is

$y=y_0 - \frac{gt^2}{2}$ [1]

With g equal to the acceleration of gravity of our planet and t the time measured with respect to the moment the bomb was released

The height will be zero when the bomb lands on ground, so if we set y=0 we can find the flight time

The range (horizontal displacement) of the bomb x is

$x = v.t$ [2]

Since the bomb won't have any friction, its horizontal component of the speed won't change. We need to find t from the equation [1] and replace it in equation [2]:

Setting y=0 and isolating t we get

$t=\sqrt{\frac{2y_0}{g}}$

Since we have $y_0=1500m$

$t=\sqrt{\frac{2(1500)}{9.8}}$

$t=17.5 sec$

Replacing in [2]

$x = 688\ m/sec \ (17.5sec)$

$x = 12040\ m$

A. The bomb will take 17.5 seconds to hit the ground

B. The bomb will land 12040 meters on the ground ahead from where they released it

6 0

2 years ago

Which planet is the farthest?

matrenka [14]

The planet that is the farthest is Neptune, pls make me brainliest:)

3 0

2 years ago