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natima [27]
3 years ago
9

A weightlifter expends 294 w of power lifting a weight to a height of 2 m in 10 seconds. what mass has he lifted?

Physics
1 answer:
Masja [62]3 years ago
3 0
The power of the weightlifter is equal to the ratio between the work done by it and the time taken:
P= \frac{W}{t}
since we know the power and the time, we can find the work done:
W=Pt=(294 W)(10 s)=2940 J

For the work-energy theorem, this work done by the lifter is equal to the gravitational potential energy gained by the mass, which is given by:
U=mg \Delta h
where m is the mass and \Delta h is the increase in height. Rearranging the formula and using W=U, we can find the mass:
m= \frac{U}{g\Delta h}= \frac{2940 J}{(9.81 m/s^2)(2 m)}=149.8 kg
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Please help
Allisa [31]

<u>We are given:</u>

Mass of the rocket = 10 kg

Weight of the Rocket = 100 N

Upward thrust applied by the rocket = 400 N

<u>Net upward force on the rocket:</u>

We are given that gravity pulls the rocket with a force of 100 N

Also, the rocket applied a force of 400N against gravity

Net upward force = Upward thrust - Force applied by gravity

Net upward force = 400 - 100

Net upward force = 300 N

<u>Upward Acceleration of the Rocket:</u>

From newton's second law:

F = ma

<em>replacing the variables</em>

300 = 10 * a

a = 30 m/s²

5 0
3 years ago
Check that equation 19.1 is dimensionally correct. the factor cw (specific heat of water is sometimes omitted because its numeri
Sladkaya [172]
Omitting the 1 will not change the value of the number, but will change the units at the end of the problem
8 0
3 years ago
Determine the force of gravity or weight of each of the following objects or systems. Use
Inessa05 [86]

Answer:

1: 19,6 N | 2: 686 N | 3: 705,6 N

Explanation:

Just apply the formula:

1: 2 . 9,8 = 19.6 N

2: (60+10) . 9,8 = 686 N

3: [30 + (6 . 7)] . 9,8

[30 + 42] . 9,8

72 . 9,8 = 705.6 N

Just sum all the weights in the system and then multiply by gravity (9,8 in this case)

4 0
3 years ago
What types of elements are useful for dating materials?
Semenov [28]
Well im not sure if this is the correct dating materials but here are some examples of Fundamentals of radiometric dating<span>Radioactive decay.
Accuracy of radiometric dating.
Closure temperature.
The age equation.
Uranium–lead dating method.
Samarium–neodymium dating method.
Potassium–argon dating method.
<span>Rubidium–strontium dating method.</span></span>
3 0
3 years ago
Read 2 more answers
A proton is released from rest at the origin in a uniform electric field in the positive x direction with magnitude 850 N/C. Wha
Gwar [14]

A proton is released from rest at the origin in a uniform electric field in the positive x direction with magnitude 850 N/C. The change in the electric potential energy of the proton-field system when the proton travels to x = 2.50m is -3.40 × 10⁻¹⁶ J (Option B)

<h3 /><h3>How is the change in electric potential energy of the proton-field system calculated?</h3>

  • Work done on the proton =Negative of the change in the electric potential energy of the proton field
  • In the given case, W = -qΔV
  • -W = qΔV
  • = qEcosθ
  • Therefore, work done on the proton = -e(8.50×10^2 N/C)(2.5m)(1)
  • = -3.40×10^-^1^6 J
  • Any change in the potential energy indicates the work done by the proton.
  • Therefore the positive sign shows that the potential energy increases when the proton does the work.
  • The negative sign shows that the potential energy decreases when the proton does the work.

To learn more about electric potential energy, refer

brainly.com/question/14306881

#SPJ4

3 0
1 year ago
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