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3 years ago

How should the student change the circuit to give negative values for current and

Physics

1 answer:

dmitriy555 [2]3 years ago

3 0

Answer:

Flip the cell.

Explanation:

This reverses direction of energy transfer.

Alternatively, flip ammeters and voltmeters to give negative readings.

what do penguins eat for lunch?

Ice-burgers!

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Oxygen and nutrients are moved through the body by the: a. circulatory system c. lymphatic system b. digestive system d. reprodu

vagabundo [1.1K]

A. The Circulatory System.

3 0

3 years ago

Who is responsible for developing the three laws of planetary motion?

Ivenika [448]

Johannes Kepler- he did it by observing the ‘Tycho Brahe’. His 3rd law was published 10 years later to his first two laws.

3 0

3 years ago

A heat engine accepts 200,000 Btu of heat from a source at 1500 R and rejects 100,000 Btu of heat to a sink at 600 R. Calculate

diamong [38]

To solve the problem it is necessary to apply the concepts related to the conservation of energy through the heat transferred and the work done, as well as through the calculation of entropy due to heat and temperatra.

By definition we know that the change in entropy is given by

$\Delta S = \frac{Q}{T}$

Where,

Q = Heat transfer

T = Temperature

On the other hand we know that by conserving energy the work done in a system is equal to the change in heat transferred, that is

$W = Q_{source}-Q_{sink}$

According to the data given we have to,

$Q_{source} = 200000Btu$

$T_{source} = 1500R$

$Q_{sink} = 100000Btu$

$T_{sink} = 600R$

PART A) The total change in entropy, would be given by the changes that exist in the source and sink, that is

$\Delta S_{sink} = \frac{Q_{sink}}{T_{sink}}$

$\Delta S_{sink} = \frac{100000}{600}$

$\Delta S_{sink} = 166.67Btu/R$

On the other hand,

$\Delta S_{source} = \frac{Q_{source}}{T_{source}}$

$\Delta S_{source} = \frac{-200000}{1500}$

$\Delta S_{source} = -133.33Btu/R$

The total change of entropy would be,

$S = \Delta S_{source}+\Delta S_{sink}$

$S = -133.33+166.67$

$S = 33.34Btu/R$

Since $S\neq 0$ the heat engine is not reversible.

PART B)

Work done by heat engine is given by

$W=Q_{source}-Q_{sink}$

$W = 200000-100000$

$W = 100000 Btu$

Therefore the work in the system is 100000Btu

4 0

3 years ago

A skater with a mass of 72 kg is traveling east at 5.8 m/s when he collides with another skater of mass 45 kg heading 60° south

Akimi4 [234]

The final velocity is 5.87 m/s

<u>Explanation:</u>

Given-

mass, $m_{1}$ = 72 kg

speed, $v_{1}$ = 5.8 m/s

$Mass_{2}$ , $m_{2}$ = 45 kg

$speed_{2}$ , $v_{2}$ = 12 m/s

Θ = 60°

Final velocity, v = ?

Applying the conservation of momentum:

$m_{1}$ X $v_{1}$ + $m_{2}$ X $v_{2}$ = ( $m_{1}$ + $m_{2}$ ) v

72 X 5.8 + 45 X 12 X cos 60° = (72 + 45) v

v = 417.6 + 540 X $\frac{0.5}{117}$

v = 417.6 + $\frac{270}{117}$

v = 5.87 m/s

The final velocity is 5.87 m/s

8 0

3 years ago

A client experiences difficulty in performing the prone iso-abs exercise. Which of the following should be suggested as an regre

butalik [34]

Answer:

a. Quadruped arm and opposite leg raise

Explanation:

Quadruped arm and opposite leg lift

- Kneel on the floor, lean forward and place your hands down.

- Keep your knees in line with your hips and hands directly under your shoulders.

- Simultaneously raise one arm and extend the opposite leg, so that they are in line with the spine.

- Go back to the starting position.

This method is usually used as an alternative to iso-abs exercise or also known as a bridge, which allows you to exercise the abdominal and spinal area at the same time.

It is also used together with other exercises for the treatment of hyperlordosis.

8 0

3 years ago