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Juliette [100K]
3 years ago
13

A cube with sides of area 22 cm2 contains a 21.2 nanoCoulomb charge. Find the flux of the electric field through the surface of

the cube in unis of Nm2/C. Enter a number with one digit behind the decimal point.
Physics
1 answer:
VladimirAG [237]3 years ago
3 0

Answer:

Explanation:

According to Gausses theorem, electric flux coming out of  charge   on all sides q is equal to

q / ε₀

q is charge , ε₀  is permittivity of air whose value is given by

8.85 x 10⁻¹²; q = 21.2 x 10⁻⁹ C

Putting the values we get

FLUX = \frac{21.2\times10^{-9}}{8.85\times10^{-12}}

2.4 x 10³ Nm²/C

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When the area over which pressure is applied is increased,
Tema [17]

Answer:

A. the pressure decrease

Explanation:

pressure decreases when the surface area over which a force is applied increases. pressure increases when the surface area over which force is applied decreases.

3 0
3 years ago
Estás cuentas son ingresos fijos o variables o egresos fijos o variables?
Brut [27]
¿El salario es un costo fijo o variable?
Los salarios anuales son costos fijos, pero otros tipos de compensación, como comisiones o horas extraordinarias, son costos variables.
8 0
3 years ago
You have been assigned to investigate a traffic accident. The masses of car A and car B are 1300 kg and 1200 kg, respectively. C
jarptica [38.1K]

Answer:

The velocity of A before impact = 17.90 m/s

Explanation:

Coefficient of restitution = (speed of seperation)/(speed of approach)

= (v₁ - v₂)/(u₂ - u₁)

where v₁ = velocity of the car A after the impact = ?

v₂ = velocity of the car B after the impact = ?

u₂ = velocity of the car B before the impact = 0 m/s (it was initially at rest)

u₁ = velocity of car A before the impact = ?

First of, we can solve for v₂, the velocity of car B after the impact, from some of the information given in the question.

- Skid marks indicate car B slid 10 m after the impact

- The coefficient of kinetic friction the tires and road is 0.8.

According to the work energy theorem, the work done by frictional force in stopping the car B is equal to the change in kinetic energy of the car B. (All after collision)

W = ΔK.E

ΔK.E = (1/2)(1200)(v₂²) - 0 (final kinetic energy is 0 since the car comes to stop eventually)

ΔK.E = (600v₂²) J

W = F × d

where F = frictional force = μmg = 0.8×1300×9.8 = 10,192 N

d = distance the car skids over before stopping = 10 m

W = 10,192 × 10 = 101,920 J

W = ΔK.E

101,920 = 600v₂²

v₂² = (101920/600) = 169.867

v₂ = 13.03 m/s

But recall,

Coefficient of restitution = (v₁ - v₂)/(u₂ - u₁)

For the sake of convention, we take the direction of car A's initial velocity to be the positive direction.

u₁ = ?

u₂ = 0 m/s

v₁ = ?

v₂ = +13.03 m/s

Coefficient of restitution = 0.4

0.4 = (v₁ - 13.03)/(0 - u₁)

-0.4u₁ = v₁ - 13.03

v₁ = 13.03 - 0.4u₁

But this is a collision. In a collision, the linear momentum is usually conserved.

Momentum before collision = Momentum after collision

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

1300u₁ + (1200×0) = 1300v₁ + (1200×13.03)

1300u₁ + 0 = 1300v₁ + 15639.95

1300u₁ = 1300v₁ + 15639.95

But recall, from the coefficient of restitution relation,

v₁ = 13.03 - 0.4u₁

Substituting this into the momentum balance equation.

1300u₁ = 1300v₁ + 15639.95

1300u₁ = 1300(13.03 - 0.4u₁) + 15639.95

1300u₁ = 16943.28 - 520u₁ + 15639.95

1820u₁ = 32,583.23

u₁ = (32,583.23/1820)

u₁ = 17.90 m/s

Therefore, the velocity of A before impact = 17.90 m/s

Hope this Helps!!!

4 0
4 years ago
The intensity of the Sun's light in the vicinity of the Earth is about 1000W/m^2. Imagine a spacecraft with a mirrored square sa
NeX [460]

Explanation:

Formula to represent thrust is as follows.

             F = \frac{dP}{dt}

                = \frac{2p}{dt}

or,           p = \frac{E}{c}

         \frac{p}{dt} = \frac{W}{c}

                  F = \frac{2IA}{c}

                     = \frac{2 (1000 W/m^{2})(5.5 \times 10^{3} m)^{2}}{3 \times 10^{8} m/s}

                     = 201.67 N

Thus, we can conclude that the thrust is 201.67 N.

8 0
4 years ago
Put the following objects in order by their
maksim [4K]

Answer:

I Think its ABC

Explanation:

No air is the lightest

Helium is lighter than regular air that's why it goes up so then the regular air would be heaviest.

4 0
3 years ago
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