Answer:
2.58 m/s
Explanation:
We can use the following equation of motion to find out the constant acceleration of the block:

where v = 3.8 m/s is the final velocity after it has traveled 7.4 m,
is the initial velocity of the block when it starts from rest, , and
is the distance traveled.



We can use the same motion equation to calculate block speed at the end of 3.4m track


Answer:
Thrust developed = 212.3373 kN
Explanation:
Assuming the ship is stationary
<u>Determine the Thrust developed</u>
power supplied to the propeller ( Punit ) = 1900 KW
Duct distance ( diameter ; D ) = 2.6 m
first step : <em>calculate the area of the duct </em>
A = π/4 * D^2
= π/4 * ( 2.6)^2 = 5.3092 m^2
<em>next : calculate the velocity of propeller</em>
Punit = (A*v*β ) / 2 * V^2 ( assuming β = 999 kg/m^3 ) also given V1 = 0
∴V^3 = Punit * 2 / A*β
= ( 1900 * 10^3 * 2 ) / ( 5.3092 * 999 )
hence V2 = 8.9480 m/s
<em>Finally determine the thrust developed </em>
F = Punit / V2
= (1900 * 10^3) / ( 8.9480)
= 212.3373 kN
The unmodulated carrier wave is going into the box, and when it comes out, its AMPLITUDE has been modulated.
Answer:
175 m
Explanation:
In a velocity vs time graph, displacement is the area under the curve.
We can calculate this as area of a trapezoid:
A = ½ (10 m/s + 60 m/s) (5 s)
A = 175 m
Or, we can split the area into a rectangle and a triangle.
A = (10 m/s) (5 s) + ½ (60 m/s − 10 m/s) (5 s)
A = 175 m
Explanation:
Given that,
Initial speed of the airfield, u = 0
Final speed, v = 27.8 m/s
Acceleration of the airfield, 
Length of the runway, d = 150 m
Let v' is the speed of the airplane to reach the required speed for takeoff. Finding v' using third equation of motion as :

This speed is not enough as the airfield must reach a speed before takeoff of at least 27.8 m/s. Now, the required length of the runways is :

So, the minimum length of the runways is 193.21 meters.