All categories

3 years ago

Which of these is an example of electrical energy being converted to mechanical energy?

Physics

1 answer:

GalinKa [24]3 years ago

7 0

Answer:

Ceiling fan

Explanation:

Ceiling fan is a perfect and typical example of electrical energy being converted to mechanical energy.

In most systems, energy is usually transformed from one form to another. Energy is not created neither is it destroyed. We know this by virtue of law of conservation of energy.

The ceiling fan is powered by electrical energy from an outlet.
The energy from the outlet is used to drive the blades of the fan and set them into motion.
This is mechanical energy.

You might be interested in

If a hot steel tool of 1200°C was put in a bucket to cool and the bucket contained 15L of water of 15°C, and the water temperatu

Mashcka [7]

3.6 kg.

<h3>Explanation</h3>

How much heat does the hot steel tool release?

This value is the same as the amount of heat that the 15 liters of water has absorbed.

Temperature change of water:

$\Delta T = T_2 - T_1= 48\; \textdegree{\text{C}}- 15\; \textdegree{\text{C}} = 33 \; \textdegree{\text{C}}$ .

Volume of water:

$V = 15 \; \text{L} = 15 \; \text{dm}^{3} = 15 \times 10^{3} \; \text{cm}^{3}$ .

Mass of water:

$m = \rho \cdot V = 1.00 \; \text{g} \cdot \text{cm}^{-3} \times 15 \times 10^{3} \; \text{cm}^{3} = 15 \times 10^{3} \; \text{g}$ .

Amount of heat that the 15 L water absorbed:

$Q = c\cdot m \cdot \Delta T = 4.18 \; \text{J} \cdot \text{g}^{-1} \cdot \textdegree{\text{C}}^{-1} \times 15 \times 10^{3} \; \text{g} \times 33 \; \textdegree{\text{C}} = 2.06910 \times 10^{6}\; \text{J}$ .

What's the mass of the hot steel tool?

The specific heat of carbon steel is $0.49 \; \text{J} \cdot \text{g}^{-1} \cdot \textdegree{\text{C}}^{-1}$ .

The amount of heat that the tool has lost is the same as the amount of heat the 15 L of water absorbed. In other words,

$Q(\text{absorbed}) = Q(\text{released}) =2.06910 \times 10^{6}\; \text{J}$ .

$\Delta T = T_2 - T_1 = 1200\; \textdegree{\text{C}} -{\bf 48}\; \textdegree{\text{C}} = 1152\; \textdegree{\text{C}}$ .

$m = \dfrac{Q}{c\cdot \Delta T} = \dfrac{2.06910 \times 10^{6} \; \text{J}}{0.49\; \text{J} \cdot \text{g}^{-1} \cdot \textdegree{\text{C}}^{-1} \times 1152\; \textdegree{\text{C}}} = 3.6 \times 10^{3} \; \text{g} = 3.6 \; \text{kg}$ .

4 0

3 years ago

10. Inertia causes your books on the car seat to continue moving forward even after you brake. TrueFalse 11. Tires with little o

Charra [1.4K]

10. True

11. True

12. True

13. False

14. False

Hope this helps!!

3 0

3 years ago

Two laws are described below:

liubo4ka [24]

Answer:

Explanation:

I can conclude that this means that the law can be broken under certain coditions as long as its not focused on a natural phenomonon

and a pheononmono is a fact or situation that is observed to exist or happen, especially one whose cause or explanation is in question.

3 0

3 years ago

Read 2 more answers

A two-stage rocket is traveling at 1200 mis with respect to the earth when the first stage runs out of fuel. Explosive bolts rel

Aloiza [94]

Answer:

The speed of the second stage after separation is 4905 m/s

Explanation:

Hi there!

Due to conservation of momentum, the momentum of the system first stage - second stage must remain constant before and after the separation. The momentum of the system is calculated by adding the momentums of each object:

initial momentum = final momentum

m₁₊₂ · v = m1 · v1 + m2 · v2

Where:

m₁₊₂ = mass of the two stage rocket

v = velocity of the rocket

m1 = mass of stage 1

v1 = velocity of stage 1

m2 = mass of stage 2

v2 = velocity of stage 2

We have the following data:

m1 = 3 · m2

m₁₊₂ = m1 + m2 = 3 · m2 + m2 = 4 · m2

v = 1200 m/s

v1 = -35 m/s (let´s consider the backward direction as negative)

v2 = ?

Then, replacing these data in the equation of momentum of the system:

m₁₊₂ · v = m1 · v1 + m2 · v2

4 m2 · 1200 m/s = 3 m2 · (-35 m/s) + m2 · v2

Let´s solve the equation for v2:

divide both sides of the equation by m2:

4 · 1200 m/s = 3 · (-35 m/s) + v2

4800 m/s = -105 m/s + v2

v2 = 4800 m/s + 105 m/s = 4905 m/s

The speed of the second stage after separation is 4905 m/s

6 0

3 years ago

Read 2 more answers

Before using a string in a comparison, you can use either the To Upper method or the To Lower method to convert the string to up

diamong [38]

Answer:

True, check attachment for code

Explanation:

To convert java strings of text to upper or lower case, we can use and inbuilt methods To Uppercase and To lower case.

The first two lines of code will set up a String variable to hold the text "text to change", and then we print it out.

The third line sets of a second String variable called result.

The fourth line is where the conversion is done.

We can compare the string

We can compare one string to another. (When comparing, Java will use the hexadecimal values rather than the letters themselves.) For example, if we wanted to compare the word "Fat" with the word "App" to see which should come first, you can use an inbuilt string method called compareTo.

Check attachment for the code

3 0

3 years ago