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2 years ago

A process generates 250 watts of useful energy and 600 watts of waste energy. WhT is the efficiency of the process?

Physics

1 answer:

Cloud [144]2 years ago

3 0

The efficiency of the process is 42%

From the question given above, the following data were obtained:

Useful energy = 250 watt

Wasteful energy = 600 watt

<h3>Efficiency =? </h3>

The efficiency of the process can be obtained as illustrated below:

$efficiency \: = \frac{useful \: energy}{wasteful \: energy} \times 100 \\ \\ efficiency \: = \frac{250}{600} \times 100 \\$

<h3>Efficiency = 42%</h3>

Therefore, the efficiency of the process is 42%

Learn more: brainly.com/question/2300041

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A car with a momentum of 3,200 kg-m/s moves foward at a rate of 2 m/s. What is the mass of the vehicle?

natulia [17]

Answer:

1600 kg

Explanation:

use the formula p=mv. p=3200, v=2. Plug in and rearrange.

3200=(m)(2)

m= 3200/2

m=1600

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2 years ago

How does free energy challenge the scientific definition of energy?

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3 years ago

Dependent variable is the _______ that happens because of the UI independent variable. (No answer choices provided)

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Answer: One is called the dependent variable and the other the independent variable. The independent variable is the variable the experimenter changes or controls and is assumed to have a direct effect on the dependent variable.

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3 years ago

The masses of blocks A and B are 20kg and 10kg, respectively. The blocks are initially at rest on the floor and are connecte by

Goryan [66]

Answer

mass of the block A = 20 Kg

mass of block B = 10 Kg

acceleration of Block A = ?

acceleration of Block B = ?

Assuming the magnitude of force = 124 N

Applying newton's second law

F = 2 T

T = F/2

now, Tension in the string =

T = 124/2 = 62 N

Weight of the block A

W = m₁ g

W = 20 x 9.8 = 196 N

Weight of the block B

W = m₂ g

W = 10 x 9.8

W = 98 N

Weight of blocks is greater than force applied by the pulley so, the blocks will not move.

Hence, acceleration of the block A and B = 0 m/s²

If the Force magnitude is increased to 294 N

T = F/2 = 294/2 = 147 N

Since this Force is less than Weight A acceleration of the block A = 0 m/s².

For Block B

Tension is more than Weight hence block will move

Net Force = 147 - 98 = 49 N

acceleration = $\dfrac{F}{m}$

$a = \dfrac{49}{10}$

a = 4.9 m/s²

8 0

3 years ago

O professor Hosney, levou os alunos da segunda série ao laboratório para realizar um experimento. Pegou um recipiente de capacid

givi [52]

Answer:

The temperature beyond which the substance overflows the container is 86.23°C.

Explanation:

English Translation

Professor Hosney took the second grade students to the laboratory to perform an experiment. He took a 1000ml capacity container at a temperature of 68oF and poured 980 ml of a substance at 20oC into it. While placing the set to heat, he consulted a table where he found the volumetric expansion coefficient of the substance, 4 x 10-4 ºC-1 and the linear expansion coefficient of the container material, 3 x 10-5 ºC-1. Hosney then asked students to determine the temperature from which the substance would overflow. A student then asked, what is the melting temperature of the substance, and the teacher answered promptly 290.8 K. What is the temperature from which the substance will overflow?

Solution

The change in volume of a substance is given as

ΔV = γV₀(ΔT)

where

γ = coefficient of volume expansion

V₀ = Initial volume

(ΔT) = change in temperature.

At the temperature where the substance will overflow, the volume of the substance and the container will both be the same.

Let this temperature be T.

For the substance,

γ = coefficient of volume expansion = (4 × 10⁻⁴) °C⁻¹

V₀ = Initial volume = 980 mL

(ΔT) = change in temperature = (T - 20)

We will still leave ΔT as ΔT

ΔV₁ = (4 × 10⁻⁴) × 980 × ΔT

ΔV₁ = 0.392 ΔT

New volume of the substance at that temperature = V₀ + ΔV₁ = 980 + 0.392ΔT

For the container

γ = coefficient of volume expansion = 3 × coefficient of linear expansion = 3 × (3 × 10⁻⁵) °C⁻¹ = (9 × 10⁻⁵) °C⁻¹

V₀ = Initial volume = 1000 mL

(ΔT) = change in temperature = (T - 20) (note that 68°F = 20°C)

We will still leave ΔT as ΔT

ΔV₂ = (9 × 10⁻⁵) × 1000 × ΔT

ΔV₂ = 0.09 ΔT

New volume of the container at that temperature = V₀ + ΔV₂ = 1000 + 0.09 ΔT

At the temperature where overflow occurs, the two volumes are initially first the same.

980 + 0.392ΔT = 1000 + 0.09 ΔT

0.392ΔT - 0.09ΔT = 1000 - 980

0.302ΔT = 20

ΔT = (20/0.302) = 66.23°C

T - 20° = 66.23°

T = 66.23 + 20 = 86.23°C

The temperature beyond which the substance overflows the container is 86.23°C.

Hope this Helps!!!

8 0

3 years ago