Answer:
t = 23.255 s, x = 2298.98 m, v_y = - 227.90 m / s
Explanation:
After reading your extensive writing, we are going to solve the approach.
The initial speed of the plane is 250 miles / h and it is at an altitude of 2650 m; In general, planes fly horizontally for launch, therefore this is the initial horizontal speed.
As there is a mixture of units in different systems we are going to reduce everything to the SI system.
v₀ₓ = 250 miles h (1609.34 m / 1 mile) (1 h / 3600 s) = 111.76 m / s
y₀ = 2650 m
Let's set a reference system with the x-axis parallel to the ground, the y-axis is vertical. As time is a scalar it is the same for vertical and horizontal movement
Y axis
y = y₀ + v₀ t - ½ g t²
the initial vertical velocity when the cargo is dropped is zero and when it reaches the floor the height is zero
0 = y₀ + 0 - ½ g t²
t =
t = √(2 2650/ 9.8)
t = 23.255 s
Therefore, for the cargo to reach the desired point, it must be launched from a distance of
x = v₀ₓ t
x = 111.76 23.255
x = 2298.98 m
at the point and arrival the speed is
vₓ = v₀ₓ = 111.76
vertical speed is
v_y = v_{oy} - gt
v_y = 0 - gt
v_y = - 9.8 23.25 555
v_y = - 227.90 m / s
the negative sign indicates that the speed is down
in the attachment we have a diagram of the movement
To solve this problem we must rely on the equations of the simple harmonic movement that define the period as a function of length and gravity as

Where
l = Length
g = Gravity
Re-arrange to find L,

Our values are given as


Replacing,



Therefore the height would be 25.348m
Cooking takes longer because water and other liquids evaporate faster and boil at lower temperatures.At sea level water boils at 212degrees Fahrenheit but at an altitude of 7,500 feet, it boils at about 198 degrees. Foods that are prepared by boiling or simmering will cook at a lower temperature and it will take longer to cook them. Hope this helps!! ; )
Both oceanic and continental crust.
I hope this helps!
Answer:
3.75 m/s
Explanation:
From the question given above, the following data were obtained:
Acceleration (a) = 1.5 m/s²
Initial velocity (u) = 0 m/s
Time (t) = 2.5 s
Final velocity (v) =?
a = (v – u) / t
1.5 = (v – 0) / 2.5
1.5 = v / 2.5
Cross multiply
v = 1.5 × 2.5
v = 3.75 m/s
Hence, the escape velocity of the squirrel is 3.75 m/s