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ELEN [110]
2 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]2 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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A seagull flying horizontally over the ocean at a constant speed of 2.60 m/s carries a small fish in its mouth. It accidentally
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(a) +2.60 m/s

The motion of the fish dropped by the seagul is a projectile motion, which consists of two independent motions:

- a horizontal uniform motion, at constant speed

- a vertical motion, at constant acceleration (acceleration of gravity, g=-9.8 m/s^2, downward)

In this part we are only interested in the horizontal motion. As we said the horizontal component of the fish's velocity does not change, therefore its value when the fish reaches the ocean is equal to its initial value, which is the speed at which the seagull was flying (because it was flying horizontally):

v_x = +2.60 m/s

(b) -17.2 m/s

The vertical component of the fish's velocity instead follows the equation:

v_y = u_y +gt

where

u_y = 0 is the initial vertical velocity, which is zero

g=-9.8 m/s^2 is the acceleration of gravity

t is the time

Since the fish reaches the ocean at t = 1.75 s, we can substitute this time into the formula to find the final vertical velocity:

v_y = 0+(-9.8)(1.75)=-17.2 m/s

where the negative sign indicates the direction (downward).

(c)

The horizontal component of the fish's velocity would increase

The vertical component of the fish's velocity would stay the same.

As we said from part (a) and (b):

- The horizontal component of the fish's velocity is constant during the motion and it is equal to the initial velocity of the seagull -> so if the seagull's initial speed increases, the horizontal velocity of the fish will increase too

- The vertical component of the fish's velocity does not depend on the original speed of the seagull, therefore it is not affected.

4 0
3 years ago
A ball is dropped from a high rise platform at t=0 starting from rest. After 6 seconds another ball is thrown downwards from the
AlekseyPX

Answer:

73.5 m/s

Explanation:

The position of the first ball is:

y = y₀ + v₀ t + ½ at²

y = h + (0)(18) + ½ (-9.8)(18)²

y = h − 1587.6

The position of the second ball is:

y = y₀ + v₀ t + ½ at²

y = h + (-v) (18−6) + ½ (-9.8)(18−6)²

y = h − 12v − 705.6

Setting the positions equal:

h − 1587.6 = h − 12v − 705.6

-1587.6 = -12v − 705.6

1587.6 = 12v + 705.6

882 = 12v

v = 73.5

The second ball is thrown downwards with a speed of 73.5 m/s

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