Math is the process of using the given information, along with
all the general stuff that you know, to find the missing information.
With no given information, we have no way to even guess at
an answer.
Answer:
t = 1
Explanation:
you dont know the value of the variable
Speed would such a block have if pushed horizontally 106 m along a frictionless track by such a laser is 0.127 m / s
First, it is necessary to find the radiation pressure on the surface. You will find it using the following formula:
P = P / (πr ^ 2) c
where P is the pressure and c is the speed of light in vacuum
P = 27 * 10 ^ 6 / π (0.2 / 2) ^ 2 * (3 * 10 ^ 8)
= 286.62× = 2866N / m ^ 2.
Then you must calculate the force (F) and the acceleration (a). This is done through the formulas:
F = P * (πr ^ 2)
F = 2866 * π * (0.2 / 2) ^ 2 = 0.089N
As, a = F / m
a = 0.089 / 104 = 0.00085m / s ^ 2
You can now calculate the speed.
V = √2ad
V = √2 *0.00085 * 106
V = 0.127 m / s
The complete question is: You've recently read about a chemical laser that generates a 20.0-cm-diameter, 27.0 MW laser beam. One day, after physics class, you start to wonder if you could use the radiation pressure from this laser beam to launch small payloads into orbit. To see if this might be feasible, you do a quick calculation of the acceleration of a 20.0-cm-diameter, 104 kg, perfectly absorbing block. What speed would such a block have if pushed horizontally 106 m along a frictionless track by such a laser? Express your answer with the appropriate units.
Answer: The origin might be from MARS
Explanation: Meteorites origin are either from Asteroid, Moon or Mars. Their origin is usually classified according to their age after using radioisotope dating. Meteorites that originate from Mars are between the age of 4.5 billion to 200 million years. Since the meteorite in question is only a billion years old, the origin would be from Mars.
Answer:
I_total = L² (m + M / 3)
Explanation:
The moment of inertia is defined by
I = ∫ r² dm
It is appreciated that it is a scalar quantity, for which it is additive, in this case the system is formed by two bodies and the moment of inertia must be the sum of each moment of inertia with respect to the same axis of rotation.
The moment of inertia of a bar with respect to an axis that passes through ends is
I_bar = 1/3 M L²
The moment of inertia of a particle is
I_part = m x²
We have to assume the point where the particle sticks to the bar, suppose it sticks to the end
x = L
Total moment of inertia is the sum of these two moments of inertia
I_total = I_bar + I_particule
I_total = 1/3 M L² + m L²
I_total = L² (m + M / 3)