The distance an object falls, from rest, in gravity is
D = (1/2) (G) (T²)
'T' is the number seconds it falls.
In this problem,
0.92 meter = (1/2) (9.8) (T²)
Divide each side by 4.9 : 0.92 / 4.9 = T²
Take the square root
of each side: √(0.92/4.9) = T
0.433 sec = T
The horizontal speed doesn't make a bit of difference in
how long it takes to reach the floor. BUT ... if you want to
know how far from the table the pencil lands, you can find
that with the horizontal speed.
The pencil is in the air for 0.433 second.
In that time, it travels
(0.433s) x (1.4 m/s) = 0.606 meter
from the edge of the table.
Answer: slow revolution and fast rotation
Solar system has 8 planets. 4 inner rocky planets - Mercury, Venus, Earth and Mars and 4 outer gaseous planets - <u>Jupiter, Saturn, Uranus and Neptune.</u> The outer planets have few common features.
They are gaseous. There period of revolution is larger than the inner planets which means that they have slow revolution about the Sun. One day on the outer planets is smaller than the inner planets which means they have fast rotation.
<u>For example,</u> Jupiter has revolves around sun in 11.86 Earth years and rotates about axis in 9.8 Earth hours. Uranus revolves around sun in 84 Earth years and rotates on its axis 17.9 Earth hours.
Answer:
measuring the zero intensity point, we can deduce the movement of the screen.
The distance from the center of the pattern to the first zero is proportional to the distance to the screen,
Explanation:
The expression for the diffraction phenomenon is
a sin θ = m λ
for the case of destructive interference. In general the detection screen is quite far from the grid, let's use trigonometry to find the angles
tan θ = y / L
in these experiments the angles are small
tan θ = sin θ / cos θ = sin θ
sunt θ = y / L
we substitute
a 
= m λ
y = m L λ / a
therefore, by carefully measuring the zero intensity point, we can deduce the movement of the screen.
The distance from the center of the pattern to the first zero is proportional to the distance to the screen, so you can know where the displacement occurs, it should be clarified that these displacements are very small so the measurement system must be capable To measure quantities on the order of hundredths of a millimeter, a micrometer screw could be used.
