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sammy [17]
3 years ago
12

What is work - energy theorem ??​

Physics
2 answers:
velikii [3]3 years ago
7 0

Answer:

The work-energy theorem explains the idea that the net work - the total work done by all the forces combined - done on an object is equal to the change in the kinetic energy of the object

Explanation:

The work-energy theorem states that the work done by the net force acting on a body is equal to the change produced in the kinetic energy of the body. Let F. be the variable force. We have, W=xf∫xiF.

Elden [556K]3 years ago
3 0

The work-energy theorem explains the idea that the net work - the total work done by all the forces combined - done on an object is equal to the change in the kinetic energy of the object. After the net force is removed (no more work is being done) the object's total energy is altered as a result of the work that was done.

This idea is expressed in the following equation:

is the total work done

is the change in kinetic energy

is the final kinetic energy

is the initial kinetic energy

mark me as brainliest ❤️

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What is tyndall effect​
ivann1987 [24]

Answer:

light scattering by particles in a colloid or in a very fine suspension

4 0
3 years ago
Read 2 more answers
A fire woman dropped a person onto the safety net. Right
dolphi86 [110]

Answer:

m = 28.7[kg]

Explanation:

To solve this problem we must use the definition of kinetic energy, which can be calculated by means of the following equation.

E_{k}=\frac{1}{2}*m*v^{2}\\

where:

Ek = kinetic energy = 1800 [J]

m = mass [kg]

v = 11.2 [m/s]

1800=\frac{1}{2}*m*(11.2)^{2}\\m = 28.7[kg]

7 0
3 years ago
The air in a car tire la compressed when the car rolls over a rock. If the air
stealth61 [152]

Answer:

the signs of heat and work are; -Q and -W

Explanation:

The first law of thermodynamics is given by; ΔU = Q − W

where;

ΔU is the change in internal energy of a system,

Q is the net heat transfer (the sum of all heat transfer into and out of the system)

W is the net work done (the sum of all work done on or by the system).

Now, The system in this case is the tire and since the air gets warmer, heat must have left the system. Therefore Q is negative (-Q).

Since work is done by the system, W remains negative.

Thus, the signs of heat and work are; -Q and - W

8 0
3 years ago
What are two ways you can increase power by climbing the stairs?
Nina [5.8K]

Answer:

change the

  • time
  • distance
  • force.

Explanation:

P = W/time

W = F*d

You have control over how fast you go up the stairs.

You also have control over how far up the stairs you go.

Therefore the answer is

  • time
  • distance

If you don't like distance as an answer, you can carry something up the stairs -- anything that increases F will do.

8 0
3 years ago
Suppose that a person gets hit by a bus moving at 30 mi/h with a 58,000 lbs of force in the direction of motion. If the mass of
alexandr402 [8]

The impulse of a force is due to the change in the motion of an object

A. The persons speed after impact is approximately 59.38 mi/h

B. The expected speed is <u>29.89 mi/h</u> which is less than the findings

Reason:

Known parameters are;

The speed of the bus, v = 30 mi/h

The force with which the person was hit, F = 58,000 lbs

Mass of the bus, M = 40,000 lbs

Mass of the person, m = 150 lbs

Duration of the impact, Δt = 0.007 seconds

A. The speed of the person at the end of the impact, <em>v</em>, is given as follows;

The impulse of the force = F × Δt = m × Δv

For the person, we get;

58,000 lbf ≈ 1866094.816 lb·ft./s²

58,000 lbf × 0.007 s = 150 lbs × Δv

1,866,094.816 lb·ft./s²

\Delta v = \dfrac{1,866,094.816\ lbs \times 0.007 \, s}{150 \, lbs} \approx  87.084  \ ft./s

Δv = v₂ - v₁

The initial speed of the person at the instant, can be as v₁ = 0

The final speed, v₂ = Δv - v₁

∴ v₂ ≈  87.084 ft./s - 0 = 87.084 ft./s

≈ <u>87.084 ft./s</u>

<u />v_2 \approx \dfrac{87.084 \ ft./s}{y} \times\dfrac{1 \ mi}{5280 \ ft.} \times \dfrac{3,600 \ s}{1 \, hour} \approx 59.38 \ mi/h<u />

The speed of the person at the end of the impact, v₂ ≈ <u>59.38 mi/h</u>

B. Where the momentum is conserved, we have;

m₁·v₁ + m₂v₂ = (m₁ + m₂)·v

v = \dfrac{m_1 \cdot v_1 + m_2 \cdot v_2}{m_2 + m_1}

v = \dfrac{40,000 \times 30  + 150 \times 0}{40,000 + 150} \approx 29.89

The expected speed of the person at the end of the impact is 29.89 mi/h, and therefore, <u>the findings does not agree with the expectation</u>

Learn more here:

brainly.com/question/18326789

3 0
3 years ago
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