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ELEN [110]
3 years ago
15

How much work was done by a hot air balloon to lift up a 100 Newton to a height of 300 meters?

Physics
1 answer:
Monica [59]3 years ago
5 0

So, the work was done by that hot air-balloon is <u>30,000 J or 30 kJ</u>.

<h3>Introduction</h3>

Hi ! In this question, I will help you. <u>Work is the amount of force exerted to cause an object to move a certain distance from its starting point</u>. In physics, the amount of work will be proportional to the increase in force and increase in displacement. Amount of work can be calculated by this equation :

\boxed{\sf{\bold{W = F \times s}}}

With the following condition :

  • W = work (J)
  • F = force (N)
  • s = shift or displacement (m)

Now, the s (displacement) can be written as ∆h (altitude change) because the object move to vertical line. The formula can also be changed to:

\boxed{\sf{\bold{W = F \times \Delta h}}}

With the following condition :

  • W = work (J)
  • F = force (N)
  • \sf{\Delta h} = change of altitude (m)

If an object has mass, then the object will also be affected by gravity. Always remember that F = m × g. So that :

\sf{W = F \times \Delta h}

\boxed{\sf{\bold{W = m \times g \times \Delta h}}}

With the following condition :

  • W = work (J)
  • m = mass of the object (kg)
  • g = acceleration of the gravity (m/s²)
  • \sf{\Delta h} = change of altitude (m)

<h3>Problem Solving</h3>

We know that :

  • F = force = 100 N
  • \sf{\Delta h} = change of altitude 300 m

What was asked :

  • W = work = ... J

Step by step :

\sf{W = F \times \Delta h}

\sf{W = 100 \times 300}

\boxed{\sf{W = 30,000 \: J = 30 \: kJ}}

<h3>Conclusion</h3>

So, the work was done by that hot air-balloon is 30,000 J or 30 kJ.

<h3>See More :</h3>
  • Work that he had done to lift object brainly.com/question/26341717
  • Converting work to potential energy brainly.com/question/26487284
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What is the mass of a truck in grams of it produces a force of 1500N while accelerating at a rate of 6 m/s²?​
aleksley [76]

Answer:

250,000

Explanation:

<h2> </h2>

<h2>formula = ( F=ma </h2>

  • F=1500N
  • a=6m/s^2
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  • m=?
  • 1500/6 = m
  • m=250 kg
  • 1kg =1000gm so 250kg =250,000gm
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A) 3 x 10 ^ 8
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What are used to transfer electrical energy from a power plant all the way to a home? generators turbines power transformers pow
defon

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What is the sprinters output at 2.0 s, 4.0 s and 6.0 s?
kari74 [83]

A 59 kg sprinter, starting from rest, runs 47 m in 7.0 s at constant acceleration.?  

What is the sprinter's power output at 2.0 s, 4.0 s, and 6.0 s?  

Instantaneous Power is the force times velocity  

P = Fv  

Because the acceleration is constant, the force will be constant as well  

F = ma  

P = mav  

for constant acceleration, the velocity at each time is found using  

v = at  

P = ma(at) = ma²t  

find the acceleration using kinematic equation  

s = ½at²  

a = 2s/t²  

a = 2(47) / 7.0²  

a = 1.918 m/s²  

P(2.0) = 59(1.918²)2.0 = 434.25 W = 0.43 kW  

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P(6.0) = 59(1.918²)6.0 = 1302.76 W = 1.3 kW  

I hope this helped.  


5 0
3 years ago
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