We need to consider for this exercise the concept Drag Force and Torque. The equation of Drag force is
![F_D = c_D A \frac{\rho V^2}{2}](https://tex.z-dn.net/?f=F_D%20%3D%20c_D%20A%20%5Cfrac%7B%5Crho%20V%5E2%7D%7B2%7D)
Where,
F_D = Drag Force
= Drag coefficient
A = Area
= Density
V = Velocity
Our values are given by,
(That is proper of a cone-shape)
![A = 9m^2](https://tex.z-dn.net/?f=A%20%3D%209m%5E2)
![\rho = 1.2Kg/m^3](https://tex.z-dn.net/?f=%5Crho%20%3D%201.2Kg%2Fm%5E3)
![V = 6.5m/s](https://tex.z-dn.net/?f=V%20%3D%206.5m%2Fs)
Part A ) Replacing our values,
![F_D = 0.5*9*\frac{1.2*6.5^2}{2}](https://tex.z-dn.net/?f=F_D%20%3D%200.5%2A9%2A%5Cfrac%7B1.2%2A6.5%5E2%7D%7B2%7D)
![F_D = 114.075N](https://tex.z-dn.net/?f=F_D%20%3D%20114.075N)
Part B ) To find the torque we apply the equation as follow,
![\tau = F*d](https://tex.z-dn.net/?f=%5Ctau%20%3D%20F%2Ad)
![\tau = (114.075N)(7)](https://tex.z-dn.net/?f=%5Ctau%20%3D%20%28114.075N%29%287%29)
![\tau = 798.525N.m](https://tex.z-dn.net/?f=%5Ctau%20%3D%20798.525N.m)
Work of the force = 10 N
Time required for the work = 50 sec
Height = 7 m
We are given with the value of work and time in the question.
Substitute the values in the formula of power and then you'll get the power required.
We know that,
w = Work
p = Power
t = Time
By the formula,
Given that,
Work (w) = 7 m = 70 Joules
Time (t) = 50 sec
Substituting their values,
p = 70/50
p = 1.4 watts
Therefore, the power required is 1.4 watts.
Hope it helps!
W=gm
where g - gravitation
m - mass
w - weight
as gravitation equals to zero, multiplying by 0 gives W=0
It is not possible to tell whether and object is heavy or light
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}}}](https://tex.z-dn.net/?f=%20%5Cboxed%7B%5Csf%7B%5Cbold%7BW%20%3D%20F%20%5Ctimes%20s%7D%7D%7D%20)
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}}}](https://tex.z-dn.net/?f=%20%5Cboxed%7B%5Csf%7B%5Cbold%7BW%20%3D%20F%20%5Ctimes%20%5CDelta%20h%7D%7D%7D%20)
With the following condition :
- W = work (J)
- F = force (N)
= 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}](https://tex.z-dn.net/?f=%20%5Csf%7BW%20%3D%20F%20%5Ctimes%20%5CDelta%20h%7D%20)
![\boxed{\sf{\bold{W = m \times g \times \Delta h}}}](https://tex.z-dn.net/?f=%20%5Cboxed%7B%5Csf%7B%5Cbold%7BW%20%3D%20m%20%5Ctimes%20g%20%5Ctimes%20%5CDelta%20h%7D%7D%7D%20)
With the following condition :
- W = work (J)
- m = mass of the object (kg)
- g = acceleration of the gravity (m/s²)
= change of altitude (m)
<h3>Problem Solving</h3>
We know that :
- F = force = 100 N
= change of altitude 300 m
What was asked :
Step by step :
![\sf{W = F \times \Delta h}](https://tex.z-dn.net/?f=%20%5Csf%7BW%20%3D%20F%20%5Ctimes%20%5CDelta%20h%7D%20)
![\sf{W = 100 \times 300}](https://tex.z-dn.net/?f=%20%5Csf%7BW%20%3D%20100%20%5Ctimes%20300%7D%20)
![\boxed{\sf{W = 30,000 \: J = 30 \: kJ}}](https://tex.z-dn.net/?f=%20%5Cboxed%7B%5Csf%7BW%20%3D%2030%2C000%20%5C%3A%20J%20%3D%2030%20%5C%3A%20kJ%7D%7D%20)
<h3>Conclusion</h3>
So, the work was done by that hot air-balloon is 30,000 J or 30 kJ.
<h3>See More :</h3>
Answer:
last is the answer.
increase the voltage in order to send energy faster.