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Luda [366]
3 years ago
7

A parallel-plate capacitor with plates of area 360 cm2 is charged to a potential difference V and is then disconnected from the

voltage source. When the plates are moved 0.8 cm farther apart, the voltage between the plates increases by 100 V. What is the charge Q on the positive plate of the capacitor?
Physics
1 answer:
Softa [21]3 years ago
6 0

Answer:

Q=3.9825\times 10^{-9} C

Explanation:

We are given that a parallel- plate capacitor is charged to a potential difference V and then disconnected from the voltage source.

1 m =100 cm

Surface area =S=\frac{360}{10000}=0.036 m^2

\Delta d=0.8 cm=0.008 m

\Delta V=100 V

We have to find the charge Q on the positive plates of the capacitor.

V=Initial voltage between plates

d=Initial distance between plates

Initial Capacitance of capacitor

C=\frac{\epsilon_0 S}{d}

Capacitance of capacitor after moving plates

C_1=\frac{\epsilon_0 S}{(d+\Delta d)}

V=\frac{Q}{C}

Potential difference between plates after moving

V=\frac{Q}{C_1}

V+\Delta V=\frac{Q}{C_1}

\frac{Qd}{\epsilon_0S}+100=\frac{Q(d+\Delta d)}{\epsilon_0S}

\frac{Q(d+\Delta d)}{\epsilon_0 S}-\frac{Qd}{\epsilon_0S}=100

\frac{Q\Delta d}{\epsilon_0 S}=100

\epsilon_0=8.85\times 10^{-12}

Q=\frac{100\times 8.85\times 10^{-12}\times 0.036}{0.008}

Q=3.9825\times 10^{-9} C

Hence, the charge on positive plate of capacitor=Q=3.9825\times 10^{-9} C

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7 0
3 years ago
A driver in a 2290-kg car car traveling at 42.7 m/s slams on the brakes and skids to a stop. If the coefficient of friction betw
8_murik_8 [283]

Answer:

126.56 m

Explanation:

Applying,

-F = ma............. Equation 1

Where F = frictional force, m = mass of the car, a = acceleration.

Note: Frictional force is negative because it act in opposite direction to motion

But,

F = mgμ.......... Equation 2

Where g = acceleration due to gravity, μ = coefficient of friction

Substitute equation 2 in equation 1

-mgμ = ma

a = -gμ.............. Equation 3

From the question,

Given: μ = 0.735

Constant: 9.8 m/s²

Substitute these values in equation 3

a = -9.8×0.735

a = -7.203 m/s²

Finally,

Applying

v² = u²+2as.............. Equation 4

Where v = final velocity, u = initial velocity, s = distance

From the question,

Given: u = 42.7 m/s, v = 0 m/s (to a stop), a = -7.203 m/s²

Substitute these values into equation 4

0² = 42.7²+2(-7.203)s

-1823.29 = -14.406s

s = -1823.29/-14.406

s = 126.56 m

4 0
3 years ago
Which scientific law describes the relationship between action and reaction force pairs? Newton's third law of motion Kepler's l
Hitman42 [59]

Newton's third law of motion

Explanation:

Newton's third law of motion states that:

<em>"When an object A exerts a force on an object B (action force), then object B exerts an equal and opposite force (reaction force) on object A"</em>

It is important to note that this law is always valid, even when it seems it is not.

Consider for example the gravitational force that the Earth exerts on your body (= your weight). We can say that this is the action force. It may seems that there is no reaction force in this case. However, this is not true: in fact, your body also exerts an equal and opposite force on the Earth, and this is the reaction force. The reason that explains why we don't notice any effect on Earth due to this force is that the mass of the Earth is much larger than your mass, therefore the acceleration produced on the Earth because of the force you apply is negligible.

It is also important to note that the action-reaction pair of forces always act on two different objects, so they never appear in the same free-body diagram.

Learn more about Newton's third law of motion:

brainly.com/question/11411375

#LearnwithBrainly

7 0
3 years ago
A box experiencing a gravitational force of 600 N is being pulled to the right with force of 250 N. A 25 N frictional force acts
bearhunter [10]

Answer: 0 NEWTONS

Explanation:

6 0
3 years ago
The initial height of the water in a sealed container of diameter 100.0 cm is 5.00 m. The air pressure inside the container is 0
katen-ka-za [31]

Answer:

a)  F = 2.66 10⁴ N, b)   h = 1.55 m

Explanation:

For this fluid exercise we use that the pressure at the tap point is

Exterior

          P₂ = P₀ = 1.01 105 Pa

inside

         P₁ = P₀ + ρ g h

the liquid is water with a density of ρ=1000 km / m³

         P₁ = 0.85   1.01 10⁵ + 1000   9.8  5

         P₁ = 85850 + 49000

         P₁ = 1.3485 10⁵ Pa

the net force is

         ΔP = P₁- P₂

         Δp = 1.3485 10⁵ - 1.01 10⁵

         ΔP = 3.385 10⁴ Pa

Let's use the definition of pressure

         P = Fe / A

         F = P A

the area of ​​a circle is

         A = pi r² = [i d ^ 2/4

let's reduce the units to the SI system

         d = 100 cm (1 m / 100 cm) = 1 m

         F = 3.385 104 pi / 4 (1) ²

         F = 2.66 10⁴ N

b) the height for which the pressures are in equilibrium is

        P₁ = P₂

        0.85 P₀ + ρ g h = P₀

        h = \frac{P_o ( 1-0.850)}{\rho \ g}

        h = \frac{1.01 \ 10^5 ( 1 -0.85)}{1000 \ 9.8}

        h = 1.55 m

4 0
2 years ago
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