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NNADVOKAT
3 months ago
12

A charge of uniform volume density (40 nC/m3) fills a cube with 8.0-cm edges. What is the total electric flux through the surfac

e of this cube?
Physics
1 answer:
Keith_Richards [3.2K]3 months ago
5 0

Answer:

The flux across the cube's surface is 2.314\ Nm^{2}/C.

Solution:

According to the details provided:

Cube edge length, a = 8.0 cm = 8.0\times 10^{- 2}\ m.

Volume charge density, \rho_{v} = 40 nC/m^{3} = 40\times {- 9}\ C/m^{3}.

Now,

To find the electric flux:

\phi = \frac{q}{\epsilon_{o}}

where

\phi = electric flux

\epsilon_{o} = 8.85\times 10^{- 12}\ F/m = permittivity of vacuum.

The volume charge density for this scenario is described by:

\rho_{v} = \frac{Total\ charge, q}{Volume of cube, V}

Cube volume, V = a^{3}.

Thus,

V = (8.0\times 10^{- 2})^{3} = 5.12\times 10^{- 4}\ m^{3}.

The total charge can be derived from equation (2):

q = \rho_{v}V = 40\times {- 9}\times 5.12\times 10^{- 4}.

q = 2.048\times 10^{-11}\ F = 20.48\ pF.

Now, insert the value of 'q' into equation (1):

\phi = \frac{2.048\times 10^{-11}}{8.85\times 10^{- 12}} = 2.314\ Nm^{2}/C.

You might be interested in
Modern wind turbines generate electricity from wind power. The large, massive blades have a large moment of inertia and carry a
Yuliya22 [3333]

Answer:

a)106.48 x 10⁵ kg.m²

b)144.97 x 10⁵ kgm² s⁻¹

Explanation:

a)Given

m = 5500 kg

l = 44 m

The moment of inertia for one blade

I= 1/3 x m l²

where m denotes the mass of the blade

l represents the length of each blade.

Substituting the necessary values, the moment of inertia for one blade is

I= 1/3 x 5500 x 44²

I= 35.49 x 10⁵ kg.m²

Total moment of inertia for 3 blades

I= 3 x 35.49 x 10⁵ kg.m²

I= 106.48 x 10⁵ kg.m²

b) The angular momentum 'L' is calculated using

L =I x ω

where,

I= the moment of inertia of the turbine i.e 106.48 x 10⁵ kg.m²

ω= angular velocity =2π f

f represents the frequency of rotation of the blade i.e 13 rpm

f = 13 rpm=>= 13 / 60 revolutions per second

ω = 2π f => 2π  x  13 / 60 rad / s

L=I x ω =>106.48 x 10⁵ x   2π  x  13 / 60

  = 144.97 x 10⁵  kgm² s⁻¹    

7 0
2 months ago
A 2 000-kg sailboat experiences an eastward force of 3 000 N by the ocean tide and a wind force against its sails with a magnitu
ValentinkaMS [3465]

Respuesta:

La magnitud de la aceleración resultante es 2.2 m/s^2

Explicación:

La masa (m) del velero es 2000 kg

La fuerza que actúa sobre el velero debido a la marea del océano es F_1 = 3000N

Hacia el este significa que se da a lo largo de la dirección positiva del eje x

EntoncesF_{1x} = 3000N y F_{1y}= 0

La fuerza del viento que actúa sobre el velero esF_2 6000N dirigida hacia el noroeste, lo que significa a un ángulo de 45 grados sobre el eje negativo x

Luego

F_{2x} = -(6000N) cos 45 grados = -4242.6 N

F_{2y} = (6000N) cos 45 grados = 4242.6 N

Por lo tanto, la fuerza neta que actúa sobre el velero en la dirección x es

F_x = F_{1x}+ F_{2x}

= - 3000 N + 4242.6 N

= - 3000 N +4242.6 N

= 1242.6N

La fuerza neta que actúa sobre el velero en la dirección y es

= 0+ 4242.6N

= 4242.6N F_y = F_{1y}+ F_{2y}La magnitud de la fuerza resultante =

Usando el teorema de Pitágoras de 1243 N y 4243 N

4420.8 N\sqrt{(1242.6)^2 + (4242.6)^2F = ma

\sqrt{(1544054.76) + (17999654.8)}

\sqrt{(19543709.5)^2}

= 2.2

4 0
2 months ago
A 25kg child sits on one end of a 2m see saw. How far from the pivot point should a rock of 50kg be placed on the other side of
Sav [3153]

Answer:

A rock weighing 50kg should be positioned at a distance of 0.5m from the pivot of the seesaw.

Explanation:

τchild=τrock  

We will utilize the formula for torque:

(F)child(d)child)=(F)rock(d)rock)

The gravitational force acts equally on both objects.

(m)childg(d)child)=(m)rockg(d)rock)

We can eliminate gravity from both sides of the equation for simplification.

 (m)child(d)child)=(m)rock(d)rock)  

Now employing the given masses for the rock and child. The seesaw's total length is 2 meters, with the child sitting at one end, placing them 1 meter from the center of the seesaw.

(25kg)(1m)=(50kg)drock

Solve for the distance where the rock should be positioned in relation to the seesaw's center.

drock=25kg⋅m50kg

drock=0.5m

6 0
3 months ago
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