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3 years ago

Three capacitors with individual capacitances of 2 μF, 5 μF, and 10 μF

Physics

1 answer:

rodikova [14]3 years ago

5 0

Answer:

1.25C

Explanation:

When capacitance is in series we add them like this: 1/Ctotal = 1/C1 +1/C2 + 1/C3.....

1/C = 1/2 + 1/5 + 1/10 = 5 + 2 + 1/10 = 8/10

C = 10/8 = 1.25

Capacitance = Charge/potential difference(Q/V)

1.25 = Charge/12

Total charge = 1.25 ×12 =15coulombs

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A skier leaves the horizontal end of a ramp with a velocity of 25.0 m/s and lands 70.0 m from the base of the ramp. How high is

Valentin [98]

Answer:

Explanation:

<u>Horizontal Motion </u>

When an object is thrown horizontally with an initial speed v and from a height h, it follows a curved path ruled by gravity.

The maximum horizontal distance traveled by the object can be calculated as follows:

$\displaystyle d=v\cdot\sqrt{\frac {2h}{g}}$

If the maximum horizontal distance is known, we can solve the above equation for h:

$\displaystyle h=\frac {d^2g}{2v^2}$

The skier initiates the horizontal motion at v=25 m/s and lands at a distance d=70 m from the base of the ramp. The height is now calculated:

$\displaystyle h=\frac {70^2\cdot 9.8}{2\cdot 25^2}$

$\displaystyle h=\frac {4900\cdot 9.8}{2\cdot 625}$

h= 38.416 m

The end of the ramp is 38.416 m high

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3 years ago

True or false if the temperature does not change but the mass of the object increases the thermal energy in the object decreases

Kay [80]

Answer: im pretty sure its true

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3 years ago

There are devices to put in a light socket that control the current through a lightbulb, thereby increasing its lifetime. Which

Dmitrij [34]

Answer: B

Explanation:

Limiting the maximum current through the bulb. This will help in preserving or improving the bulb's lifetime and also this won't have an effect on the brightness of the bulb as brightness is affected by the average value. Although brightness is a factor of current, reducing the maximum current won't have any bearing on the average current the bulb is getting.

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3 years ago

A point charge with charge q1 = 4.00 μC is held stationary at the origin. A second point charge with charge q2 = -4.40 μC moves

Bezzdna [24]

Answer:

W=0.94J

Explanation:

Electrostatic potential energy is the energy that results from the position of a charge in an electric field. Therefore, the work done to move a charge from point 1 to point 2 will be the change in electrostatic potential energy between point 1 and point 2.

This energy is given by:

$U=\frac{K\left |q_1 \right |\left |q_2 \right |}{r}\\$

So, the work done to move the chargue is:

$W=U_1-U_2\\W=\frac{K\left |q_1 \right |\left |q_2 \right |}{r_1}-\frac{K\left |q_1 \right |\left |q_2 \right |}{r_2}\\r_1=\sqrt{((0.155 m)^2+0 m)^2}=0.115m\\r_2=\sqrt{((0.245 m)^2+(0.270 m)^2}=0.365m\\W=K\left |q_1 \right |\left |q_2 \right |(\frac{1}{r_1}-\frac{1}{r_2})\\W=8.99*10^9\frac{Nm^2}{c^2}(4.00*10^{-6}C)(4.40*10^{-6}C)(\frac{1}{0.115m}-\frac{1}{0.365})\\W=0.94J$

The work is positive since the potential energy in 1 is greater than 2.

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4 years ago

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