Answer:
acceleration m/s²
Explanation:
The units on the left side of the equation are units of length. The units on the right side of the equation must be units of length also. Acceleration times time squared gives length.
I’m not sure if this will help but I found: https://prezi.com/l0fa6du3b9kp/going-off-the-grid-assignment/?fallback=1 and
We need to consider for this exercise the concept Drag Force and Torque. The equation of Drag force is

Where,
F_D = Drag Force
= Drag coefficient
A = Area
= Density
V = Velocity
Our values are given by,
(That is proper of a cone-shape)



Part A ) Replacing our values,


Part B ) To find the torque we apply the equation as follow,



Answer:
v₃ = 9.62[m/s]
Explanation:
To solve this type of problem we must use the principle of conservation of linear momentum, which tells us that the momentum is equal to the product of mass by velocity.
We must analyze the moment when the astronaut launches the toolkit, the before and after. In order to return to the ship, the astronaut must launch the toolkit in the opposite direction to the movement.
Let's take the leftward movement as negative, which is when the astronaut moves away from the ship, and rightward as positive, which is when he approaches the ship.
In this way, we can construct the following equation.

where:
m₁ = mass of the astronaut = 157 [kg]
m₂ = mass of the toolkit = 5 [kg]
v₁ = velocity combined of the astronaut and the toolkit before throwing the toolkit = 0.2 [m/s]
v₂ = velocity for returning back to the ship after throwing the toolkit [m/s]
v₃ = velocity at which the toolkit should be thrown [m/s]
Now replacing:
![-(157+5)*0.2=(157*0.1)-(5*v_{3})\\(5*v_{3})= 15.7+32.4\\v_{3}=9.62[m/s]](https://tex.z-dn.net/?f=-%28157%2B5%29%2A0.2%3D%28157%2A0.1%29-%285%2Av_%7B3%7D%29%5C%5C%285%2Av_%7B3%7D%29%3D%2015.7%2B32.4%5C%5Cv_%7B3%7D%3D9.62%5Bm%2Fs%5D)
Answer:
Exposure time limitation, shielding and distance.
Explanation:
- Limitation of exposure time, since the dose received is directly proportional to the exposure time, so that, at a shorter time, lower dose. For this reason, planning is suggested, to reduce time.
-
Use of shields. This allows a reduction in the dose received by the technician when filtered by the barrier (screen). There are two types of shields or screens, the primary barriers (attenuate the radiation of the primary beam) and the secondary barriers (avoid diffuse radiation).
-
Distance to the radioactive source. The dose received is inversely proportional to the square of the distance to the radioactive source. Therefore, if the distance is doubled, the dose received will decrease by a quarter. Reason for this, it is advisable to use devices or remote controls whenever possible.