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tensa zangetsu [6.8K]
3 years ago
13

Two identical air-filled parallel-plate capacitors C1 and C2 are connected in series to a battery that has voltage V. The charge

on each capacitor is Q0. While the two capacitors remain connected to the battery, a dielectric with dielectric constant K>1 is inserted between the plates of capacitor C1, completely filling the space between them.
Physics
1 answer:
Llana [10]3 years ago
5 0

Complete question:

Two identical air-filled parallel-plate capacitors C₁ and C₂ are connected in series to a battery that has voltage V. The charge on each capacitor is Qo. While the two capacitors remain connected to the battery, a dielectric with dielectric constant K > 1 is inserted between the plates of capacitor C1, completely filling the space between them. What is the charge on capacitor C1 after the dielectric is inserted? Express your answer in terms of K and Qo.

Answer:

The charge on capacitor after the dielectric is inserted is Qo( k + 1)

Explanation:

Given;

two identical capacitor, C₁ and C₂

charge on each capacitor, Q = Qo

Voltage of the battery, = V

dielectric constant K > 1

After inserting dielectric constant K, the equivalent capacitance in parallel becomes;

C.equ. = KC₁ + C₂

Charge, on capacitor Qo = C.equ.V

Q = V(KC₁ + C₂)

Q = KC₁V + C₂V

since, the charge on each capacitor is Qo, then

Q = KQo + Qo

Q = Qo( k + 1)

Therefore, the charge on capacitor after the dielectric is inserted is Qo( k + 1), in terms of K and Qo.

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dalvyx [7]

Answer:

The velocity is 19.39 m/s

Solution:

As per the question:

Mass, m = 75 kg

Radius, R = 19.2 m

Now,

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F_{C} = F_{G}

\frac{mv^{2}}{R} = mg

where

v = velocity

g = acceleration due to gravity

v = \sqrt{2gR} = \sqrt{2\times 9.8\times 19.2} = 19.39\ m/s

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3 years ago
What are gears and what is a lever ?​
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3 years ago
Mike has a mass of 97 kg. He jumps out of a perfectly good airplane that is 2000 m above the ground. After he falls 1000 m, when
Ne4ueva [31]

Answer:

a)  fr = 224.3 N , b)   fr = 224.3 N , c)   v = 198.0  m/s

Explanation:

a) For this exercise let's start by calculating the acceleration in the fall

             v² = v₀² - 2 a (y-y₀)

When it jumps the initial vertical speed is zero

             a = -v² / 2 (y-y₀)

             a = -68 2/2 (1000-2000)

             a = 2,312 m / s²

Let's use the second net law to enter the average friction force

            fr = m a

            fr = 97 2,312

            fr = 224.3 N

b) let's look for acceleration

            v² = v₀² - 2 a y

            a = (v² –v₀²) / 2 (y-y₀)

            a = (4² - 68²) / 2 (0-1000)

            a = 2,304 m / s²

            fr = m a

            fr = 97 2,304

            fr = 223.5 N

c) the speed of the wallet is searched with kinematics

           v² = v₀² - 2 g (y-y₀)

           v = √ (0-2 9.8 (0-2000))

           v = 198.0  m/s

4 0
3 years ago
A small meteorite with mass of 1 g strikes the outer wall of a communication satellite with a speed of 2Okm/s (relative to the s
strojnjashka [21]

Answer:

The energy coverted to heat is 200 kilojoules.

Explanation:

GIven the absence of external forces exerted both on the small meteorite and on the communication satellite, the Principle of Linear Momentum is considered and let suppose that collision is completely inelastic and that satellite is initially at rest. Hence, the expression for the satellite-meteorite system:

m_{M}\cdot v_{M} + m_{S}\cdot v_{S} = (m_{M}+m_{S})\cdot v

Where:

m_{M}, m_{S} - Masses of the small meteorite and the communication satellite, measured in kilograms.

v_{M}, v_{S} - Speeds of the small meteorite and the communication satellite, measured in meters per second.

v - Final speed of the satellite-meteorite system, measured in meters per second.

The final speed of the satellite-meteorite system is cleared:

v = \frac{m_{M}\cdot v_{M}+m_{S}\cdot v_{S}}{m_{M}+m_{S}}

If m_{M} = 1\times 10^{-3}\,kg, m_{S} = 200\,kg, v_{M} = 20000\,\frac{m}{s} and v_{S} = 0\,\frac{m}{s}, the final speed is now calculated:

v = \frac{(1\times 10^{-3}\,kg)\cdot \left(20000\,\frac{m}{s} \right)+(200\,kg)\cdot \left(0\,\frac{m}{s} \right)}{1\times 10^{-3}\,kg+200\,kg}

v = 0.1\,\frac{m}{s}

Which means that the new system remains stationary and all mechanical energy from meteorite is dissipated in the form of heat. According to the Principle of Energy Conservation and the Work-Energy Theorem, the change in the kinetic energy is equal to the dissipated energy in the form of heat:

K_{S} + K_{M} - K - Q_{disp} = 0

Q_{disp} = K_{S}+K_{M}-K

Where:

K_{S}, K_{M} - Initial translational kinetic energies of the communication satellite and small meteorite, measured in joules.

K - Kinetic energy of the satellite-meteorite system, measured in joules.

Q_{disp} - Dissipated heat, measured in joules.

The previous expression is expanded by using the definition for the translational kinetic energy:

Q_{disp} = \frac{1}{2}\cdot [m_{M}\cdot v_{M}^{2}+m_{S}\cdot v_{S}^{2}-(m_{M}+m_{S})\cdot v^{2}]

Given that m_{M} = 1\times 10^{-3}\,kg, m_{S} = 200\,kg, v_{M} = 20000\,\frac{m}{s}, v_{S} = 0\,\frac{m}{s} and v = 0.1\,\frac{m}{s}, the dissipated heat is:

Q_{disp} = \frac{1}{2}\cdot \left[(1\times 10^{-3}\,kg)\cdot \left(20000\,\frac{m}{s} \right)^{2}+(200\,kg)\cdot \left(0\,\frac{m}{s} \right)^{2}-(200.001\,kg)\cdot \left(0.001\,\frac{m}{s} \right)^{2}\right]Q_{disp} = 200000\,J

Q_{disp} = 200\,kJ

The energy coverted to heat is 200 kilojoules.

4 0
3 years ago
Sallys physical education teacher timed her run and recorded the time and distance in the table below. What is her average speed
Fudgin [204]
Answer: B) 2.5 m/s

Explanation: Find the average of the time and distance, and see how far they go in only 1 second.


1 + 2 + 3 + 4 + 5 = 15
15 divided by 5 = 3

3 seconds

2 + 5 + 7 + 10 + 12 = 36
36 divided by 5 = 7.2


7.2m per 3 seconds.

7.2 divided by 3 = 2.4

Therefore, the answer is technically 2.4m/s
4 0
4 years ago
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