Answer:
The angle between the red and blue light is 1.7°.
Explanation:
Given that,
Wavelength of red = 656 nm
Wavelength of blue = 486 nm
Angle = 37°
Suppose we need to find the angle between the red and blue light as it leaves the prism
We need to calculate the angle for red wavelength
Using Snell's law,
Put the value into the formula
We need to calculate the angle for blue wavelength
Using Snell's law,
Put the value into the formula
We need to calculate the angle between the red and blue light
Using formula of angle
Put the value into the formula
Hence, The angle between the red and blue light is 1.7°.
The energy conservation and trigonometry we can find the results for the questions about the movement of the acrobat are;
a) The maximum speed is v = 4.89 m / s
b) The maximum height is h = 1.22 m
The energy conservation is one of the most fundamental principles of physics, stable that if there are no friction forces the mechanistic energy remains constant. Mechanical energy is the sum of the kinetic energy plus the potential energies.
Em = K + U
Let's write the energy in two points.
Starting point. Highest part of the oscillation
Em₀ = U = m g h
Final point. Lower part of the movement
= K = ½ m v²
Energy is conserved.
Emo =
m g h = ½ m v²
v² = 2 gh
Let's use trigonometry to find the height, see attached.
h = L - L cos θ
h = L (1- cos θ)
They indicate that the initial angle is tea = 48º and the length is L = 3.7 m, let's calculate.
h = 3.7 (1- cos 48)
h = 1.22 m
this is the maximum height of the movement.
Let's calculate the velocity.
v = 4.89 m / s
In conclusion using the conservation of energy and trigonometry we can find the results for the questions about the movement of the acrobat are;
a) The maximum speed is v = 4.89 m / s
b) The maximum height is h = 1.22 m
Learn more here: brainly.com/question/13010190
Answer:
we assume that it starts with a velocity of 10m/s. At 2m height above ground level, its velocity decreases at 3m above ground level
for its way down the velocity at 3m on its way down is more than its velocity at 2m on its way down.
Explanation:
A student throws a small rock straight upwards. The rock rises to its highest point and then falls back down. How does the speed of the rock at 2m on the way down compare with its speed at 2m on the way up?
It decreases in speed on its way down and increases in speed on its way down.
it decreases in speed on its way up because the the vertical motion is against the earths gravitational pull on an object to the earth's center
.It increases in speed on his way down because its under the influence of gravity
from newton's equation of motion we can check by
using V^2=u^2+2as
we assume that it starts with a velocity of 10m/s. At 2m height above ground level, its velocity decreases at 3m above ground level
for its way down the velocity at 3m on its way down is more than its velocity at 2m on its way down.
Answer:
Explanation:
This does not violate Newton's 1st law because the net force would still be 0 in order to produce uniform motion (aka constant velocity). The other forces acting on the vehicles is air resistance which is non-zero. So we need car internal force to counter balance this force, which require extra gas for the car.
