Answer:
The airplane should release the parcel m before reaching the island
Explanation:
The height of the plane is , and its speed is v=150 m/s
When an object moves horizontally in free air (no friction), the equation for the y measured with respect to ground is
[1]
And the distance X is
x = V.t [2]
Being t the time elapsed since the release of the parcel
If we isolate t from the equation [1] and replace it in equation [2] we get
Using the given values:
x = m
Answer:
R = 0.0503 m
Explanation:
This is a projectile launching exercise, to find the range we can use the equation
R = v₀² sin 2θ / g
How we know the maximum height
² =
² - 2 g y
= 0
= √ 2 g y
= √ 2 9.8 / 15
= 1.14 m / s
Let's use trigonometry to find the speed
sin θ = / vo
vo = / sin θ
vo = 1.14 / sin 60
vo = 1.32 m / s
We calculate the range with the first equation
R = 1.32² sin(2 60) / 30
R = 0.0503 m
Answer:
the wavelength λ of the light when it is traveling in air = 560 nm
the smallest thickness t of the air film = 140 nm
Explanation:
From the question; the path difference is Δx = 2t (since the condition of the phase difference in the maxima and minima gets interchanged)
Now for constructive interference;
Δx=
replacing ;
Δx = 2t ; we have:
2t =
Given that thickness t = 700 nm
Then
2× 700 = --- equation (1)
For thickness t = 980 nm that is next to constructive interference
2× 980 = ----- equation (2)
Equating the difference of equation (2) and equation (1); we have:'
λ = (2 × 980) - ( 2× 700 )
λ = 1960 - 1400
λ = 560 nm
Thus; the wavelength λ of the light when it is traveling in air = 560 nm
b)
For the smallest thickness
∴
Thus, the smallest thickness t of the air film = 140 nm
A. Formula: F=ma or F/m=a
10,000N/1,267kg≈7.9m/
B. Formula: a= and s=d/t
speed= 394.6/15
s=26.3m/s
a=
a=1.75m/
C. 7.9-1.75=difference of 6.15m/
D. The force that most likely caused this difference is friction forces