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olga nikolaevna [1]
3 years ago
11

What is the relationship between an object’s temperature and its heat

Physics
1 answer:
Sedaia [141]3 years ago
8 0

Heat is the total energy of the motion of the molecules inside the object or particle, where is temperature is merely a measure of this energy. So our relationship could be, the more heated an object is there higher the temperature the object will have.

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How many ways can motion change
meriva

Answer:

There are four main ways of doing that :-

  • Velocity
  • Acceleration
  • Momentum
  • Kinetic energy

Hope it helps!

7 0
4 years ago
Read 2 more answers
Guys this is urgent and I don’t know please help me I will mark brainliest!
VLD [36.1K]

Answer:

Not acted upon there is no motion

Forces acted upon there is motion

You can refer to the definition stated below:

Newton's first law states that every object will remain at rest or in uniform motion in a straight line unless compelled to change its state by the action of an external force. This is normally taken as the definition of inertia. ... If that velocity is zero, then the object remains at rest.

8 0
3 years ago
An energy plant produces an output potential of 1500 kV and serves a city 143 km away. A high-voltage transmission line carries
jekas [21]

Answer:

2123.55 $/hr

Explanation:

Given parameters are:

V_{plant} = 1500 KV

L = 143 km

I = 500 A

\rho = 2.4 \Omega / km

So, we will find the voltage potential provided for the city as:

V_{wire} =IR = I\rho L = 1500*2.4*143 = 514.8 kV

V_{city} = V_{plant}- V_{wire} = 1500-514.8 = 985.2 kV

Then, we will find dissipated power because of the resistive loss on the transmission line as:

P = I^2R = I^2\rho L=500^2*2.4*143 = 8.58*10^7 W

Since the charge of plant is not given for electric energy, let's assume it randomly as x =  \frac{\dollar 0.081}{kW.hr}

Then, we will find the price of energy transmitted to the city as:

Cost = P * x = 8.58*10^7 * 0.081 * 0.001 = 6949.8 $/hr

To calculate money per hour saved by increasing the electric potential of the power plant:

Finally,

I_{new} = P/V_{new} = I/1.2\\P_{new} = I_{new}^2R_{wire}\\Cost = P_{new}/1.44=6949.8/1.44 = 4826.25 $/hr

The amount of money saved per hour = 6949.8 - 6949.8/1.44 = 2123.55 $/hr

Note: For different value of the price of energy, it just can be substituted in the equations above, and proper result can be found accordingly.

3 0
3 years ago
A series of parallel linear water wave fronts are traveling directly toward the shore at 15.5 cm/s on an otherwise placid lake.
nata0808 [166]

Find the given attachment

7 0
3 years ago
(a) What magnitude point charge creates a 10,000 N/C electric field at a distance of 0.250 m? (b) How large is the field at 10.0
Sergio039 [100]
<h2>Answer:</h2>

(a) 6.95 x 10⁻⁸ C

(b) 6.25N/C

<h2>Explanation:</h2>

The electric field (E) on a point charge, Q, is given by;

E = k x Q / r²              ---------------(i)

Where;

k = constant = 8.99 x 10⁹ N m²/C²

r = distance of the charge from a reference point.

Given from the question;

E = 10000N/C

r = 0.250m

Substitute these values into equation(i) as follows;

10000 = 8.99 x 10⁹ x Q / (0.25)²

10000 = 8.99 x 10⁹ x Q / (0.0625)

10000 = 143.84 x 10⁹ x Q

Solve for Q;

Q = 10000/(143.84 x 10⁹)

Q = 0.00695 x 10⁻⁵C

Q = 6.95 x 10⁻⁸ C

The magnitude of the charge is 6.95 x 10⁻⁸ C

(b) To get how large the field (E) will be at r = 10.0m, substitute these values including Q = 6.95 x 10⁻⁸ C into equation (i) as follows;

E = k x Q / r²

E = 8.99 x 10⁹ x 6.95 x 10⁻⁸ / 10²

E = 8.99 x 10⁹ x 6.95 x 10⁻⁸ / 100

E = 6.25N/C

Therefore, at 10.0m, the electric field will be just 6.25N/C

3 0
3 years ago
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