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2 years ago

. An electric geyser was used 2 hours a day once a week. The power rating

Physics

1 answer:

iragen [17]2 years ago

3 0

Answer:

Annual energy = 1460000 Joules

Explanation:

Given the following data;

Time = 2 hours

Power = 2000 Watts

To find the energy consumption of the geyser;

Energy = power * time

Energy = 2000 * 2

Energy = 4000 Joules

Next, the annual energy consumption is;

Annual energy = 4000 * 365

Annual energy = 1460000 Joules

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A length of copper wire carries a current of 14 A, uniformly distributed through its cross section. The wire diameter is 2.5 mm,

gavmur [86]

Answer:

a. ρ $_\beta=1.996J/m^3$

b. $U_E=9.445x10^{-15} J/m^3$

Explanation:

a. To find the density of magnetic field given use the gauss law and the equation:

$i=14A$ , $d=2.5mm$ , $R=3.3$ Ω, $l=1 km$ , $E_o=8.85x10^{-12}F/m$ , $u_o=4*x10^{-7}H/m$

ρ $_\beta=\frac{\beta^2}{2*u_o}$

ρ $_\beta=\frac{1}{2*u_o}*(\frac{u_o*i^2}{2\pi *r})^2$

ρ $_\beta=\frac{u_o*i^2}{8\pi*r}=\frac{4\pi *10^{-7}H/m*(14A)^2}{8\pi*(1.25x10^{-3}m)^2}$

ρ $_\beta=1.996J/m^3$

b. The electric field can be find using the equation:

$U_E=\frac{1}{2}*E_o*E^2$

$E=(\frac{i*R}{l})^2$

$U_E=\frac{1}{2}*8.85x10^{-12}*(\frac{14A*3.3}{1000m})^2$

$U_E=9.445x10^{-15} J/m^3$

4 0

3 years ago

How do you find the capacitance in this?

Lostsunrise [7]

Answer:

Explanation:

parallel capacitances add directly

Series capacitances add by reciprocal of sum of reciprocals.

Ceq = [ C ] + [1 / (1/C + 1/C)] + [1 / (1/C + 1/C + 1/C)]

Ceq = [ C ] + [C / 2] + [C / 3]

Ceq = [ 6C/6 ] + [3C / 6] + [2C / 6]

Ceq = 11C/6

3 0

1 year ago

List 3 cases in which potential energy becomes kinetic energy and three cases in which kinetic energy becomes potential energy.

Mrrafil [7]

Answer:

when you open a can of pop

when you jump on your bed

Explanation:

3 0

2 years ago

What is the speed of an object at rest

Alisiya [41]

The speed of an object at rest is 0 m/s. We can use logic to figure this out. Hope this helps!

8 0

2 years ago

Shyam saw a cylindrical copper rod in his store room he realise that it can be used for various purposes it can be used as a flo

Misha Larkins [42]

The error percentage in measuring the volume of the cylindrical box is 6.3 x 10⁻³ %

<h3>Error percentage in measuring the volume of the cylinder</h3>

The total error percentage in measuring the volume of the cylinder is calculated as follows;

Volume of cylinder = πr²h

Volume of cylinder = π(0.03r)²(0.07h)

Volume of cylinder = (6.3 x 10⁻⁵)πr²h

Error in measuring the volume = 6.3 x 10⁻⁵ x 100% = 6.3 x 10⁻³ %

Learn more about percentage error here: brainly.com/question/5493941

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8 0

1 year ago