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3 years ago

Calculate the current flowing if a charge of 36 kilocoulombs flows in 1 hour.

Physics

1 answer:

o-na [289]3 years ago

6 0

Answer:

Current = 10 Amperes.

Explanation:

Given the following dat;

Quantity of charge, Q = 36 kilocoulombs (KC) = 36 * 1000 = 36000C

Time = 1 hour to seconds = 60*60 = 3600 seconds

To find the current;

Quantity of charge = current * time

Substituting in the equation

36000 = current * 3600

Current = 36000/3600

Current = 10 Amperes.

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A very small source of light that radiates uniformly in all directions produces an electric field amplitude of 8.45 V/m at a poi

Mandarinka [93]

Answer:

4.43 kW

Explanation:

Since Intensity I = P/A = E²/2cμ₀ where P = Power, A = Area = 4πr² where r = distance from source = 61 m and E = electric field amplitude = 8.45 V/m.

P = E²A/2cμ₀ = E²4πr²/2cμ₀ = 2πE²r²/cμ₀

= 2π(8.45 V/m)²(61 m)²/3 × 10⁸ m/s × 4π × 10⁻⁷ Tm/A

= 4428.1 W

= 4.4281 kW ≅ 4.43 kW

6 0

3 years ago

Calculate the energy (in eV/atom) for vacancy formation in some metal, M, given that the equilibrium number of vacancies at 296o

Schach [20]

Explanation:

The given data is as follows.

Temperature of metal = $296^{o}C$ = (296 + 273) K

= 569 K

Density of the metal = 8.85 $g/cm^{3}$ = $8.85 \times 10^{-6} g/m^{3}$ (as $1 cm^{3} = 10^{-6} m^{3}$ )

Atomic mass = 51.40 g/mol

Vacancies = $9.19 \times 10^{23} m^{-3}$

Formula to calculate the number of atomic sites is as follows.

n = $\frac{\rho \times N_{A}}{\text{atomic weight}}$

= $\frac{8.85 \times 10^{-6} \times 6.022 \times 10^{23}}{51.40 g/mol}$

= $1.036 \times 10^{17} atom/m^{3}$

Now, we will calculate the energy as follows.

E = $-KT \times ln (\frac{\text{no. of vacancies}}{\text{no. of atomic sites}})$

where, K = $8.62 \times 10^{-5}$

E = $-8.62 \times 10^{-5} \times 569 K \times ln (\frac{9.19 \times 10^{23}}{1.036 \times 10^{17} atom/m^{3}})$

= $78.46 eV/atom$

Therefore, we can conclude that energy (in eV/atom) for vacancy formation in given metal, M, is $78.46 eV/atom$ .

6 0

4 years ago

A coil of conducting wire carries a current of i(t) = 14.0 sin(1.15 ✕ 103t), where i is in amperes and t is in seconds. A second

Advocard [28]

Answer:

The peak emf in second coil is 1.876 V

Explanation:

Given :

Inductance $L = 130 \times 10^{-6}$ H

The current $I(t) = 14 \sin(1.15\times 10^{3} t)$

We compare above equation with standard equation,

$I(t) = I_{o} \sin (\omega t + \phi)$

From above equation we have,

$\omega = 10^{3}$ and $\phi = 1.15$

Find the inductive resistance,

$X_{L} = \omega L$

$X_{L} = 10^{3} \times 130 \times 10^{-6}$

$X_{L} = 0.134$

The peak emf in second coil is,

$V = I_{o} X_{L}$

$V = 14 \times 0.134$

$V = 1.876$ V

Therefore, the peak emf in second coil is 1.876 V

8 0

4 years ago

After the layers of sediment are deposited in one place, they are

myrzilka [38]

your question doesnt make sense

5 0

4 years ago

Read 2 more answers

What is the ideal banking angle (in degrees) for a gentle turn of 2.00 km radius on a highway with a 125 km/h speed limit (about

nikitadnepr [17]

An "ideal" banking angle assumes no friction is required to keep a car on the road as it turns. Let θ denote the banking angle, and consult the attached free-body diagram for a car making the turn. There are only 2 relevant forces acting on the car,

• the normal force with magnitude n

• the car's weight with magnitude w

and the net force points toward the center of the circle made by the turn, with centripetal acceleration

a = (125 km/h)² / (2.00 km) = 7812.5 km/h² ≈ 0.603 m/s²

Split up the forces into components acting perpendicular (⟂) and parallel (//) to the banked curve, so that by Newton's second law,

∑ F (⟂) = N + W (⟂) = m a (⟂)

and

∑ F (//) = W (//) = m a (//)

Let the direction of N be the positive perpendicular axis, and down the incline and toward the center of the circle the positive parallel axis. The net force vector and acceleration both make an angle θ with the banked curve, and W makes the same angle with the negative perpendicular axis, so that the equations above reduce to

N - m g cos(θ) = m a sin(θ)

and

m g sin(θ) = m a cos(θ)

The second equation is all we need at this point to find the ideal θ. The mass m cancels out, and we can solve for θ to get

tan(θ) = a/g ≈ (0.603 m/s²) / (9.80 m/s²) ≈ 0.0615

→ θ ≈ 3.52°

8 0

4 years ago