Answer:
F = 2 I A / c
Explanation:
The radiation pressure on a reflective surface is
P = 2 S / c
Where S is the Poynting Vector and c the speed of light
Furthermore pressure is defined as the ratio of force to area
P = F / A
Let's replace
F / A = 2 S / c
F = 2 S A / c
The poynting vector is the power per unit area that is equal to the intensity
S = I
F = 2 I A / c
Answer:
correct answer is C
Explanation:
In this exercise, you are asked to complete the sentences so that the sentence makes sense.
1) in projectile launching, the only force that acts is gravity in the vertical direction, so the time of going up is EQUAL to the time of going down
correct answer C
2) when a body falls freely, the acceleration is the ratio of gravity, therefore if it starts from rest, its height is
y = v₀ t - ½ gt²
v₀ = 0
y = -1/2 g t²
so the position is not proportional to the square of the time
correct answer is C
Answer:
The strength of gravity decreases.
An example of that would be if you were in space; you float around because there's no gravity.
Answer:
a = 0,1[m/s^2]
Explanation:
First we need to indentify the initial data.
And using this kinematic equation we have:
![v = 4[m/s]\\v_{0}= 2 [m/s] \\t = 20[s]\\\\v= v_{0}+a*t\\a=\frac{v-v_{0}}{t} \\a= \frac{4-2}{20} \\a=0.1[m/s^{2}]](https://tex.z-dn.net/?f=v%20%3D%204%5Bm%2Fs%5D%5C%5Cv_%7B0%7D%3D%202%20%5Bm%2Fs%5D%20%5C%5Ct%20%3D%2020%5Bs%5D%5C%5C%5C%5Cv%3D%20v_%7B0%7D%2Ba%2At%5C%5Ca%3D%5Cfrac%7Bv-v_%7B0%7D%7D%7Bt%7D%20%5C%5Ca%3D%20%5Cfrac%7B4-2%7D%7B20%7D%20%5C%5Ca%3D0.1%5Bm%2Fs%5E%7B2%7D%5D)
Answer:
1.) answer B
2.) answer D
3.) answer A
Explanation:
In all of these problems, it is essential to draw pictures in order to understand which trigonometric function to use according to the angle that the vector in question forms with the component requested. For all of them try to picture a right angle triangle with the vector as the hypotenuse, and the components as the triangle's shorter sides. Please refer to the three pictures attached as image for this answer a,d notice that the vector quantity known for all cases is represented in red, and the component to find is represented in green.
Problem 1) : the vector velocity makes an angle of 24 degrees with the edge of the table. So picture that vector as the hypotenuse of a right angle triangle for which you know the value: 1.8 m/s
So in this case, where you know the angle, the hypotenuse, and need to find the adjacent side to the angle, you use the cosine function as follows:
requested component 
which we round to 1.6 to match answer C).
For problem 2.) wee need to find the component opposite to the given angle in the triangle for which we also know the hypotenuse. So we use the sine function as follows:
requested component 
which we round to 135.9 m to match answer D).
For problem 3.) we need to find the horizontal component to the acceleration which corresponds to the adjacent side to the known angle, so we use the cosine function as follows:
requested component 
which we round tp 7.7 to match answer A).