Its average speed, pretending that it traveled at a constant speed, is
v = s / t
= 600 m
5 x 60 s
= 2 m/s
but to be a velocity it needs a direction as well as a speed.
( Sorry. Can’t find a division line to put between the 600 m and the 5 x 60 s )
Answer:
The frequency of the signal is 2 GHz
Explanation:
Given;
period of the clock signal, T = 500 ps = 500 x 10⁻¹² s
the frequency of the signal is given by;
F = 2 GHz
Therefore, the frequency of the signal is 2 x 10⁹ Hz or 2 GHz
Answer:
D) θ₂= 36. 6º
Explanation:
In this diffraction experiment it is described by the equation
sin θ = m λ
The first dark strip occurs for m = 1 and since the angle is generally small we can approximate sine to the value of the angle
θ₁ = λ/ a
This equation is valid for linear slits, in the case of a circular slit the problem must be solved in polar coordinates, so the equation changes slightly
θ₂ = 1.22 λ / a
In the proposed exercise we start with a linear slit of width a, where tes1 = 30º and end with a circular slit of the same diameter
θ₂ = 1.22 λ / a
Let's clear (Lam/a) of equalizing the two equations
θ₁ = θ₂/ 1.22
θ₂ = 1.22 θ₁
θ₂ = 1.22 30
θ₂= 36. 6º
When reviewing the correct results is D
Answer:
6.93m
Explanation:
Using the method of SOH CAH TOA
The base of the tree is 12m which is the adjacent side of the triangle to be generated, the height of the tree will be facing the angle of elevation and it is the opposite side.
Using the formula tan(theta) = Opposite/Adjacent
Given theta = 30°
Adjacent is 12m which is the distance the tree is making with the shadow on the ground
Opposite side will be the height of the tree.
Tan(theta) = height/12
Tan30° = height/12
Height = 12tan30°
Height = 6.93m
Therefore the height of the tree is 6.93m
