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3 years ago

How is energy and work related.

Physics

2 answers:

mel-nik [20]3 years ago

7 0

<h3>The capacity of doing work.</h3>

every form of energy is work. Work done on the body is stored in the form of energy. To do more work, more energy is required.

Explanation:

Anna007 [38]3 years ago

5 0

Answer:

Varies

Explanation:

They both relate to the process of doing something.

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A spring on a horizontal surface can be stretched and held 0.5 m from its equilibrium position with a force of 60 N. a. How much

Lostsunrise [7]

Answer:

a)1815Joules b) 185Joules

Explanation:

Hooke's law states that the extension of a material is directly proportional to the applied force provided that the elastic limit is not exceeded. Mathematically;

F = ke where;

F is the applied force

k is the elastic constant

e is the extension of the material

From the formula, k = F/e

F1/e1 = F2/e2

If a force of 60N causes an extension of 0.5m of the string from its equilibrium position, the elastic constant of the spring will be ;

k = 60/0.5

k = 120N/m

a) To get the work done in stretching the spring 5.5m from its position,

Work done by the spring = 1/2ke²

Given k = 120N/m, e = 5.5m

Work done = 1/2×120×5.5²

Work done = 60× 5.5²

Work done = 1815Joules

b) work done in compressing the spring 1.5m from its equilibrium position will be gotten using the same formula;

Work done = 1/2ke²

Work done =1/2× 120×1.5²

Works done = 60×1.5²

Work done = 135Joules

8 0

3 years ago

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3 years ago

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Tasya [4]

uhm i think i don’t understand what it says but i can answer if it was English ! i also tried to translate it but it didn’t help !

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3 years ago

A particular heat engine has a mechanical power output of 4.00 kW and an efficiency of 26.0%. The engine expels 8.55 103 J of ex

Ivahew [28]

To develop the problem we will start by finding the energy taken by each cycle through the efficiency of the motor and the exhausted energy. Later the work will be found for the conservation of energy in which this is equivalent to the difference between the two calculated energy values. Finally the estimated time will be calculated with the work and the power given,

$\text{Efficiency of the heat engine} = \eta = 26\% = 0.26$

$\text{Energy taken in by the heat engine during each cycle} = Q_h$

$\text{Energy exhausted by the heat engine in each cycle} = Q_c = 8.55*10^3 J$

$\eta = 1 - \frac{Q_{c}}{Q_{h}}$

$0.26 = 1 - \frac{8.55\ast 10^{3}}{Q_{h}}$

$\frac{8.55* 10^{3}}{Q_{h}} = 0.74$

$Q_h = \frac{8.55*10^3}{0.74}$

$Q_h = 11.554*10^3J$

PART A)

Work done by the heat engine in each cycle = W

$W = Q_h-Q_c$

$W = 11.554*10^3J-8.55*10^3J$

$W = 3004J$

According to the value given we have that,

$P = 4.0kW$

$P = 4000W$

Power is defined as the variation of energy as a function of time therefore,

$P = \frac{W}{t}$

$4000W = \frac{3004J}{t}$

$t = \frac{3004}{4000}$

$t = 0.75s$

Therefore the interval for each cycle is 0.75s

5 0

3 years ago

During a long jump, an Olympic champion's center of mass rose about 1.2 m from the launch point to the top of the arc. 1) What m

Tanzania [10]

Answer: Minimum speed needed by the Olympic champion at launch if he was traveling at 6.8 m/s at the top of the arc is 11.65 m/s.

Explanation:

Velocity is only in horizontal direction at the top most point which is similar to the velocity in the horizontal direction at the time of launch.

Now, according to the law of conservation of energy the formula used is as follows.

$mgh = \frac{1}{2} mv^{2}_{y}\\v_{y} = \sqrt{2gh}\\= \sqrt{2 \times 9.8 m/s^{2} \times 1.2}\\= 4.85 m/s$

As speed at which the person is travelling was 6.8 m/s. Hence, the initial velocity will be calculated as follows.

$v = \sqrt{v^{2}_{x} + v^{2}_{y}}\\= \sqrt{(6.8)^{2} + (4.85 m/s)^{2}}\\= 11.65 m/s$

Thus, we can conclude that minimum speed needed by the Olympic champion at launch if he was traveling at 6.8 m/s at the top of the arc is 11.65 m/s.

6 0

3 years ago

How is energy and work related.​

How is energy and work related.