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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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Explanation: In order to solve this problem we have to use the Coulomb force given by:

F=k*q^2/ d^2 where he consider the same charge for each point

so, we have

q^2= F*d^2/k= 1.137* (10 C*9*10m)^2/9*10^9 N*m^2/C^2)=3.2 * 10^-3 C.

5 0
3 years ago
two boys started runing stright to ward each other from two pionts 100m aparts.one run at speed of 4m/s and other at 6m/s.how fa
kodGreya [7K]

Answer:

One boys rate is "r" and the other's is (r+10)

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3 years ago
How much total heat transfer is necessary to lower the temperature of 0.175 kg of steam from 125.5 °C to −19.5 °C, including the
GaryK [48]

Answer:

Explanation:

The heat required to change the temperature of  steam from 125.5  °C to 100 °C is:

Q_1 = ms_{steam} (125.5^0C - 100^0C) \\ \\ Q_1 = 0.175 \ kg ( 1520 \ J/kg.K ) (25.5^0 \ C) \\ \\ Q_1 = 6783 \ J

The heat required to change the steam at 100°C to water at 100°C is;

Q_2 = mL_v \\ \\ Q_2 = (0.175 \ kg) (2.25*10^6 \ J/kg ) \\ \\ Q_2 = 393750 \ J

The heat required to change the temperature from 100°C to 0°C is

Q_3 = ms_{water} (100^) \ C) \\ \\ Q_3 = (0.175\ kg)(4186 \ J/kg.K) (100 ^0c ) \\ \\ Q_3 = 73255 \ J

The heat required to change the water at 0°C to ice at 0°C  is:

Q_4 = mL_f \\ \\ Q_4 = (0.175 \ kg)(3.34*10^5 \ J/kg) \\\\ Q_4 = 58450 \ J

The heat required to change the temperature of ice from 0°Cto -19.5°C is:

Q_5 = ms _{ice} (100^0 C) \\ \\ Q_5 = (0.175 \ kg)(2090 \ J/kg.K)(19.5^0C)  \\ \\ Q_5 = 7132.125 \ J

The total heat required to change the steam into ice is:

Q = Q_1 + Q_2 + Q_3 + Q_4 +Q_5 \\ \\Q = (6788+393750+73255+58450+7132.125)J \\ \\ Q = 539325.125 \ J \\ \\ Q = 5.39*10^5 \ J

b)

The time taken to convert steam from 125 °C to 100°C is:

t_1 = \frac{Q_1}{P} = \frac{6738 \ J}{835 \ W}  = 8.12 \ s

The time taken to convert steam at  100°C to water at  100°C is:

t_2 = \frac{Q_2}{P} =\frac{393750}{834} =471.56 \ s

The time taken to convert water to 100° C to 0° C is:

t_3 = \frac{Q_3}{P} =\frac{73255}{834} = 87.73 \ s

The time taken to convert water at 0° to ice at 0° C is :

t_4 = \frac{Q_4}{P} =\frac{58450}{834} = 70.08  \ s

The time taken to convert ice from 0° C to -19.5° C is:

t_5 = \frac{Q_5}{P} =\frac{7132.125}{834} = 8.55  \ s

5 0
3 years ago
Is it possible for a baseball to have as large a momentum as a much more massive bowling ball
kirza4 [7]
Yes it is possible. Momentum is calculated by the mass of the object times its velocity.
For example, say a bowling ball weighs 3.0kg and is travelling at a speed of 3.0m/s. Its momentum would be 3.0×3.0=9.0 kg·m/s.
Now say we have a baseball weighing 0.20kg and it is travelling at a speed of 47.0m/s. Its momentum would be 0.20×47.0=9.4 kg·m/s, which is more than that of the bowling ball.
8 0
3 years ago
a car that is moving in a straight line has an acceleration of 8.0 m/s/s for 4.0 seconds. which one of the following is true of
Nuetrik [128]

Answer:

32 ms

Explanation:

v=32ms.

Explanation:

I will assume that you mean that the acceleration is 8.0 ms2, as 8.0ms is a value for velocity, not acceleration.

Here, we use the formula v=u+at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is time. Let's substitute values, then:

v=u+at,

v=0+8.0⋅4.0

v=32,

v=32ms.

Hope it Helps! :D .

7 0
3 years ago
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