Answer:
Both the astronauts and photographer have the same displacement
Explanation:
Displacement is the minimum distance between two point. The initial point of both the astronauts and the photographer was Florida and the final point was California. So, the minimum distance for both of the astronauts and the photographer would be the distance between Florida and California would be the same.
Hence, both the astronauts and photographer will have the same displacement.
Answer:
Question 1)
a) The speed of the drums is increased from 2 ft/s to 4 ft/s in 4 s. From the below kinematic equations the acceleration of the drums can be determined.

This is the linear acceleration of the drums. Since the tape does not slip on the drums, by the rule of rolling without slipping,

where α is the angular acceleration.
In order to continue this question, the radius of the drums should be given.
Let us denote the radius of the drums as R, the angular acceleration of drum B is
α = 0.5/R.
b) The distance travelled by the drums can be found by the following kinematics formula:

One revolution is equal to the circumference of the drum. So, total number of revolutions is

Question 2)
a) In a rocket propulsion question, the acceleration of the rocket can be found by the following formula:

b) 
We use the Rydberg Equation for this which is expressed as:
<span>1/ lambda = R [ 1/(n2)^2 - 1/(n1)^2]
</span>
where lambda is the wavelength, where n represents the final and initial states. Brackett series means that the initial orbit that electron was there is 4 and R is equal to 1.0979x10^7m<span>. Thus,
</span>
1/ lambda = R [ 1/(n2)^2 - 1/(n1)^2]
1/1.0979x10^7m = 1.0979x10^7m [ 1/(n2)^2 - 1/(4)^2]
Solving for n2, we obtain n=1.
Answer: 170.67 N
Explanation:
Given
Mass of skier is 
Height of the inclination is 
Here, the potential energy of the skier is converted into kinetic energy which is consumed by the friction force by applying a constant force that does work to stop the skier.
![\Rightarrow mgh=F\cdot x\quad \quad [\text{F=constant friction force}]\\\\\Rightarrow 82.9\times 9.8\times 20=F\cdot 95.2\\\\\Rightarrow F=\dfrac{16,248.4}{95.2}\\\\\Rightarrow F=170.67\ N](https://tex.z-dn.net/?f=%5CRightarrow%20mgh%3DF%5Ccdot%20x%5Cquad%20%5Cquad%20%5B%5Ctext%7BF%3Dconstant%20friction%20force%7D%5D%5C%5C%5C%5C%5CRightarrow%2082.9%5Ctimes%209.8%5Ctimes%2020%3DF%5Ccdot%2095.2%5C%5C%5C%5C%5CRightarrow%20F%3D%5Cdfrac%7B16%2C248.4%7D%7B95.2%7D%5C%5C%5C%5C%5CRightarrow%20F%3D170.67%5C%20N)
Thus, the horizontal friction force is 170.67 N.