Which device helps safely transmit electricity from the power plant to your home?

A firework is ignited, and explodes with a flash and a loud bang as it is blown apart. The system consists of: the firework, the

butalik [34]

The complete question is:

Study the scenario. A firework is ignited, and explodes with a flash and a loud bang as it is blown apart. The system consists of: the firework, the earth, and the air. which choice best describes how energy is transformed in the system?

A) When the firework is ignited, a chemical reaction absorbs energy from the surrounding environment. the energy is in several forms including sound and light, and mechanical energy of the fragments of the firecracker that are launched through the air. Eventually all the energy released is transformed into thermal energy.

B) When the firework is ignited, a chemical reaction releases energy in several forms, including sound, light, and the mechanical energy of the fragments being launched through the air. Eventually all the energy released is transformed into mechanical energy.

C) When the firework is ignited, a chemical reaction releases energy in several forms, including sound, light, and the mechanical energy of the fragments being launched through the air. Eventually all the energy released is transformed into thermal or mechanical energy.

D) When the firework is ignited, a chemical reaction absorbs energy from the surrounding environment. The energy is taken in in several forms including sound and light, and mechanical energy of the fragments being launched through the air. Eventually all the energy is transformed into thermal energy.

Answer:

C) When the firework is ignited, a chemical reaction releases energy in several forms, including sound, light, and the mechanical energy of the fragments being launched through the air. Eventually all the energy released is transformed into thermal or mechanical energy.

Explanation:

Energy is released from the system, not absorbed.

All the sound, light and movement of the debris is as a result of energy transformation from the chemical energy.

Eventually, most of the energy are finally wasted away as heat energy.

Some of the energy is used up by the flying particles from the fireworks.

8 0

2 years ago

(a) What is the potential between two points situated 10 cm and 20 cm from a 3.0-μC point charge? (b) To what location should th

julia-pushkina [17]

Answer:

(a) 135 kV

(b) The charge chould be moved to infinity

Explanation:

(a)

The potential at a distance of r from a point charge, Q, is given by

$V = -\dfrac{kQ}{r}$

where $k = 9\times 10^9 \text{ F/m}$

Difference in potential between the points is

$kQ\left[-\dfrac{1}{0.2\text{ m}} -\left( -\dfrac{1}{0.1\text{ m}}\right)\right] = \dfrac{kQ}{0.2\text{ m}} = \dfrac{9\times10^9\text{ F/m}\times3\times10^{-6}\text{ C}}{0.2\text{ m}}$

$PD = 135\times 10^3\text{ V} = 135\text{ kV}$

(b)

If this potential difference is increased by a factor of 2, then the new pd = 135 kV × 2 = 270 kV. Let the distance of the new location be x.

$270\times10^3 = kQ\left[-\dfrac{1}{x}-\left(-\dfrac{1}{0.1\text{ m}}\right)\right]$

$10 - \dfrac{1}{x} = \dfrac{270000}{9\times10^9\times3\times10^{-6}} = 10$

$\dfrac{1}{x} = 0$

$x = \infty$

The charge chould be moved to infinity

7 0

2 years ago

Two charges, each of 2.9 microC are placed at two corners of a square 50cm on a side, If the charges are on one side of the squa

anyanavicka [17]

Answer:

The magnitude of the electric field and direction of electric field are $146.03\times10^{3}\ N/C$ and 75.36°.

Explanation:

Given that,

First charge $q_{1}= 2.9\mu C$

Second charge $q_{2}= 2.9\mu C$

Distance between two corners r= 50 cm

We need to calculate the electric field due to other charges at one corner

For E₁

Using formula of electric field

$E_{1}=\dfrac{kq}{r'^2}$

Put the value into the formula

$E_{1}=\dfrac{9\times10^{9}\times2.9\times10^{-6}}{(50\sqrt{2}\times10^{-2})^2}$

$E_{1}=52200=52.2\times10^{3}\ N/C$

For E₂,

Using formula of electric field

$E_{1}=\dfrac{kq}{r^2}$

Put the value into the formula

$E_{2}=\dfrac{9\times10^{9}\times2.9\times10^{-6}}{(50\times10^{-2})^2}$

$E_{2}=104400=104.4\times10^{3}\ N/C$

We need to calculate the horizontal electric field

$E_{x}=E_{1}\cos\theta$

$E_{x}=52.2\times10^{3}\times\cos45$

$E_{x}=36910.97=36.9\times10^{3}\ N/C$

We need to calculate the vertical electric field

$E_{y}=E_{2}+E_{1}\sin\theta$

$E_{y}=104.4\times10^{3}+52.2\times10^{3}\sin45$

$E_{y}=141310.97=141.3\times10^{3}\ N/C$

We need to calculate the net electric field

$E_{net}=\sqrt{E_{x}^2+E_{y}^2}$

Put the value into the formula

$E_{net}=\sqrt{(36.9\times10^{3})^2+(141.3\times10^{3})^2}$

$E_{net}=146038.69\ N/C$

$E_{net}=146.03\times10^{3}\ N/C$

We need to calculate the direction of electric field

Using formula of direction

$\tan\theta=\dfrac{141.3\times10^{3}}{36.9\times10^{3}}$

$\theta=\tan^{-1}(\dfrac{141.3\times10^{3}}{36.9\times10^{3}})$

$\theta=75.36^{\circ}$

Hence, The magnitude of the electric field and direction of electric field are $146.03\times10^{3}\ N/C$ and 75.36°.

4 0

2 years ago

The electric motor in the car is powered by a battery.

Vinil7 [7]

Answer:

I = 30 A.

Explanation:

Given that,

The voltage of the battery, V = 230 V

Power used to charge the battery, P = 6.9 kW

We need to find the current used to charge the battery. The formula for the power is given by :

P = VI

Where

I is current

So,

So, the required current is 30 A.

6 0

1 year ago

Which best explains why the shape of a liquid can change?

topjm [15]

The particles in a liquid are extremely fast, not faster than a gas, but faster than a solid. The way the particles move allow you to get volume but not area for it doesn't have a defiant shape

3 0

2 years ago

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