Answer:
0
Explanation:
The displacement is zero since it goes in a full circle and ends up where it started.
Answer:
41.8° to the vertical ( west of North) , 44.6 s
Explanation:
Using Pythagoras theorem, taking the direction it must go line as the hypotenuse
vh² = vs² + vd² vs is the speed of the stream, vh is the speed in the direction it must go to the vertical and vd is the speed going directly north
3 ² = 2² + vd²
9 - 4 = vd²
√5 = vd
2.24 m/s = vd which is the resultant speed
the direction it will go = tan θ = 2 / 2.24
θ = tan⁻¹(2 / 2.24) = 41.8° to the vertical ( west of North)
b)t = distance / v = 100 / 2.24 = 44.6 s
Answer:
(3) The period of the satellite is independent of its mass, an increase in the mass of the satellite will not affect its period around the Earth.
(4) he gravitational force between the Sun and Neptune is 6.75 x 10²⁰ N
Explanation:
(3) The period of a satellite is given as;
![T = 2\pi \sqrt{\frac{r^3}{GM} }](https://tex.z-dn.net/?f=T%20%3D%202%5Cpi%20%5Csqrt%7B%5Cfrac%7Br%5E3%7D%7BGM%7D%20%7D)
where;
T is the period of the satellite
M is mass of Earth
r is the radius of the orbit
Thus, the period of the satellite is independent of its mass, an increase in the mass of the satellite will not affect its period around the Earth.
(4)
Given;
mass of the ball, m₁ = 1.99 x 10⁴⁰ kg
mass of Neptune, m₂ = 1.03 x 10²⁶ kg
mass of Sun, m₃ = 1.99 x 10³⁰ kg
distance between the Sun and Neptune, r = 4.5 x 10¹² m
The gravitational force between the Sun and Neptune is calculated as;
![F_g = \frac{Gm_2m_3}{r^2} \\\\F_g = \frac{6.67\times 10^{-11} \times 1.03 \times 10^{26}\times 1.99\times 10^{30}}{(4.5\times 10^{12})^2} \\\\F_g = 6.751 \times 10^{20} \ N](https://tex.z-dn.net/?f=F_g%20%3D%20%5Cfrac%7BGm_2m_3%7D%7Br%5E2%7D%20%5C%5C%5C%5CF_g%20%3D%20%5Cfrac%7B6.67%5Ctimes%2010%5E%7B-11%7D%20%5Ctimes%201.03%20%5Ctimes%2010%5E%7B26%7D%5Ctimes%201.99%5Ctimes%2010%5E%7B30%7D%7D%7B%284.5%5Ctimes%2010%5E%7B12%7D%29%5E2%7D%20%5C%5C%5C%5CF_g%20%3D%206.751%20%5Ctimes%2010%5E%7B20%7D%20%5C%20N)
There are a few ways to do this- unfortunately different fields are better at it than others! Medical research is generally pretty good, some other fields likewise very good, some not as much.
Basically, though, what they do is use standadisation- they agree on the terminology, units of data, statistical measures, and so forth, that will be used in that scientific field. As much as possible, every scientist in the field uses those standards so everyone working in the field should recognise it.
For instance, in clinical trials, there is very good agreement worldwide on what the different metrics we use are- e.g. in cancer research, we usually want to know the 5-year survival rate (meaning the percentage of patients still alive 5 years after diagnosis). So anyone with the right training should be able to pick up a clinical trial report and understand what the results are and what the report is saying.