Answer:
<em>The speed of the passengers is 5.24 m/s</em>
Explanation:
<u>Uniform Circular Motion
</u>
It occurs when an object in a circular path travels equal angles in equal times.
The angular speed can be calculated in two different ways:

Where:
v = tangential speed
r = radius of the circle described by the rotating object
Also:

Where:
f = frequency
Since the frequency is calculated when the number of revolutions n and the time t are known:

The Ferris wheel has a diameter of 100 m and makes n=1 rotation in t=60 seconds, thus the frequency is:

The angular speed is:

Now we calculate the tangential speed, solving this formula for v:


The radius is half the diameter, r=100/2=50 m:

Calculating:
v = 5.24 m/s
The speed of the passengers is 5.24 m/s
The density is 81.4 g/m3. Before you start plugging numbers into the density formula (D=M/V), you should convert 104 kg to grams, which ends up being 104,000 grams. Then you can plug in the 104,000 grams and 1,278 m3 into the formula. When you divide the mass by the volume, you get a really long decimal, which you can round to 81.4 g/m3, or whatever place your teacher wants you to round to.
Answer:t=0.3253 s
Explanation:
Given
speed of balloon is 
speed of camera 
Initial separation between camera and balloon is 
Suppose after t sec of throw camera reach balloon then,
distance travel by balloon is


and distance travel by camera to reach balloon is


Now






There are two times when camera reaches the same level as balloon and the smaller time is associated with with the first one .
(b)When passenger catches the camera time is 
velocity is given by



and position of camera is same as of balloon so
Position is 

Answer:
The answer is 13 however make sure if they ask for a certain measurement like meter answer it by saying 13 meters.
Explanation:
This basically turns into basic algebra if you know the formula for work. The formula for work is W=F*d
Here are the variables that you know 650J=50N*d so you need d.
All you do is divide 650J by 50N and you get a total of 13 (meters since I don't know what they want you to put it in).
Answer:
Potential energy is 
Explanation:
The potential energy depends on the mass, the acceleration of gravity g and the height at which the object or person is.
Potential energy 
In this case we would need to know the exact mass of the hiker in order to calculate the potential energy.
But we know the values of g and h


So, the potential energy

m is the mass of the hiker, wich is not in the description of the problem.