Answer:
y = 80.2 mille
Explanation:
The minimum size of an object that can be seen is determined by the diffraction phenomenon, if we use the Rayleigh criterion that establishes that two objects can be distinguished without the maximum diffraction of a body coincides with the minimum of the other body, therefore so much for the pupil of the eye that it is a circular opening
θ = 1.22 λ/ d
in a normal eye the diameter of the pupils of d = 2 mm = 0.002 m, suppose the wavelength of maximum sensitivity of the eye λ = 550 nm = 550 10⁻⁹ m
θ = 1.22 550 10⁻⁹ / 0.002
θ = 3.355 10⁻⁴ rad
Let's use trigonometry to find the distance supported by this angle, the distance from the moon to the Earth is L = 238900 mille = 2.38900 10⁵ mi
tan θ = y / L
y = L tan θ
y = 2,389 10⁵ tan 3,355 10⁻⁴
y = 8.02 10¹ mi
y = 80.2 mille
This is the smallest size of an object seen directly by the eye
Answer:
Explanation:
How long does the football need to rise?
4.70/3 = 2.35 s
What height will the football reach?
h = ½(9.81)2.35² = 27.1 m
With what speed does the punter need to kick the football?
vy = g•t = 9.81(2.35) = 23.1 m/s
vx = d/t = 56.0/4.70 = 11.9 m/s
v = √(vx²+vy²) = 26.0 m/s
At what angle (θ), with the horizontal, does the punter need to kick the football?
θ = arctan(vy/vx) = 62.7°
Answer:
F = 19.1 N
Explanation:
To find the force exerted by the string on the block you use the following formula:
(1)
k: spring constant = 95.5 N/m
x: displacement of the block from its equilibrium position = 0.200 m
you replace the values of k and x in the equation (1):
Hence, the force exterted on the block is 19.1 N
Inertia is when a object in motion will stay in motion or in a standing still state unless acted upon by a unbalancing force.
Friction is when a object slows down because it is rubbing against another object.
If a object is sliding across a surface, theoretically, it would not stop but because it is on a flat surface it would experience friction, this will disperse some of the kinetic energy that it has thus slowing the object down eventually, after some time, to a stop.
Hope this helps! :)
Answer:
V(average)=6.37 V
Explanation:
Given Data
Peak Voltage=10V
Frequency=10 kHZ
To Find
Average Voltage
Solution
For this first we need to find Voltage peak to peak
So
Voltage (peak to peak)= 2× voltage peak
Voltage (peak to peak)= 2×10
Voltage (peak to peak)= 20 V
Now from Voltage (peak to peak) formula we can find the Average Voltage
So
Voltage (peak to peak)=π×V(average)
V(average)=Voltage (peak to peak)/π
V(average)=20/3.14
V(average)=6.37 V
