Answer: apparent weighlessness.
Explanation:
1) Balance of forces on a person falling:
i) To answer this question we will deal with the assumption of non-drag force (abscence of air).
ii) When a person is dropped, and there is not air resistance, the only force acting on the person's body is the Earth's gravitational attraction (downward), which is the responsible for the gravitational acceleration (around 9.8 m/s²).
iii) Under that sceneraio, there is not normal force acting on the person (the normal force is the force that the floor or a chair exerts on a body to balance the gravitational force when the body is on it).
2) This is, the person does not feel a pressure upward, which is he/she does not feel the weight: freefalling is a situation of apparent weigthlessness.
3) True weightlessness is when the object is in a place where there exists not grativational acceleration: for example a point between two planes where the grativational forces are equal in magnitude but opposing in direction and so they cancel each other.
Therefore, you conclude that, assuming no air resistance, a person in this ride experiencing apparent weightlessness.
Answer:
Final speed of the train is 7.5 m/s
Explanation:
It is given that,
Uniform acceleration of the train is, a = 1.5 m/s²
It starts from rest and travels for 5.0 s. We have to find the final velocity of the train. By using first equation of motion as :
Here, train starts from rest so, u = 0
v = 7.5 m/s
So, the final velocity of the train is 7.5 m/s. Hence, this is the required solution.
Answer:
7,14545 mph and 3,1936 m/s
Explanation:
The average speed is calculated by dividing the displacement over time, then it is 26,2 miles/(3 2/3 hours), here 3 (2/3) hours is a mixed number, that represents 11/3 hours or 3,66 hours. Then the average speed is 7,14545 mph, now to turn this into meters per second, we notice as mentioned that 1 mile =1609 meters and 1 hour=3600 seconds. Then 7,14545 miles/hour* (1 hour/3600 seconds) * (1609 meters/1 mile)=3,1936 m/s
since both components, length and time, are measurable
<span>since Rate = length ÷ time </span>
<span>∴ rate is also measurable and ∴ quantitative.
</span>
Answer:
1 * 10^-7 [J]
Explanation:
To solve this problem we must use dimensional analysis.
1 ergos [erg] is equal to 1 * 10^-7 Joules [J]
![1[erg]*\frac{1*10^{-7} }{1}*[\frac{J}{erg} ] \\= 1*10^{-7}[J]](https://tex.z-dn.net/?f=1%5Berg%5D%2A%5Cfrac%7B1%2A10%5E%7B-7%7D%20%7D%7B1%7D%2A%5B%5Cfrac%7BJ%7D%7Berg%7D%20%5D%20%5C%5C%3D%201%2A10%5E%7B-7%7D%5BJ%5D)
