Answer:
176.22 nm
Explanation:
= Index of refraction of oil = 1.64
= Wavelength of the light = 578 nm
= minimum thickness of the oil
For destructive interference, we have
For minimum thickness, m = 1
Answer:
500 hours
Explanation:
The sum of total hours over its lifetime will be given by
Where T is total time, r is rate in decimal and To is the original charge hours. Substituting the original charge hours with 5 hours and rate as 0.99 then the time will be
Therefore, the time is equivalent to 500 hours
Answer:
x=2.4t+4.9t^2
Explanation:
This equation is one of the kinematic equations to solve for distance. The original equation is as follows:
X=Xo+Vt+1/2at^2
We know that the ball starts at rest meaning that its initial velocity and position is zero.
X=0+Vt+1/2at^2
Since it is going down the ramp, you can use the acceleration of gravity constant. (9.81 m/s^2) and simplify that with the 1/2.
X=Vt+4.9t^2
Note: Since the positive direction in this problem is down, you are adding the 4.9t^2, but if a question says that the downward direction is negative, you would subtract those values.
Now, substitute in your velocity value.
X=2.4t+4.9t^2
Answer:
An Energy held by an object because of its position relative by other objects
Explanation:
Answer:
While falling, the magnitude of the acceleration of the egg is 9.82 m/s²
While stopping, the magnitude of the deceleration of the egg is 79.3 m/s²
Explanation:
Hi there!
The equation of velocity of the falling egg is the following:
v = v0 + a · t
Where:
v = velocity at time t.
v0 = initial velocity.
a = acceleration.
t = time
Let´s calculate the acceleration of the egg while falling. Notice that the result should be the acceleration of gravity, ≅ 9.8 m/s².
v = v0 + a · t
11.1 m/s = 0 m/s + a · 1.13 s (since the egg is dropped, the initial velocity is zero). Solving for "a":
11.1 m/s / 1.13 s = a
a = 9.82 m/s²
While falling, the magnitude of the acceleration of the egg is 9.82 m/s²
Now, using the same equation, let´s find the acceleration of the egg while stopping. We know that at t = 0.140 s after touching the ground, the velocity of the egg is zero. We also know that the velocity of the egg before hiiting the ground is 11.1 m/s, then, v0 = 11.1 m/s:
v = v0 + a · t
0 = 11.1 m/s + a · 0.140 s
-11.1 m/s / 0.140 s = a
a = -79.3 m/s²
While stopping, the magnitude of the deceleration of the egg is 79.3 m/s²