Answer:
mechanical energy per unit mass is 887.4 J/kg
power generated is 443.7 MW
Explanation:
given data
average velocity = 3 m/s
rate = 500 m³/s
height h = 90 m
to find out
total mechanical energy and power generation potential
solution
we know that mechanical energy is sum of potential energy and kinetic energy
so
E = 
×m×v² + m×g×h .............1
and energy per mass unit is
E/m = ×v² + g×h
put here value
E/m = ×3² + 9.81×90
E/m = 887.4 J/kg
so mechanical energy per unit mass is 887.4 J/kg
and
power generated is express as
power generated = energy per unit mass ×rate×density
power generated = 887.4× 500× 1000
power generated = 443700000
so power generated is 443.7 MW
Answer:
a) y₂ = 49.1 m
, t = 1.02 s
, b) y = 49.1 m
, t= 1.02 s
Explanation:
a) We will solve this problem with the missile launch kinematic equations, to find the maximum height, at this point the vertical speed is zero
² = 
² - 2 g (y –yo)
The origin of the coordinate system is on the floor and the ball is thrown from a height
y-yo = ![v_{oy}² /2 g y- 0 = 10.0²/2 9.8 y - 0 = 5.10 m The height from the ground is the height that rises from the reference system plus the depth of the ground from the reference system y₂ = 5.1 + 44 y₂ = 49.1 m Let's use the other equation to find the time [tex]v_{y}](https://tex.z-dn.net/?f=v_%7Boy%7D%C2%B2%20%2F2%20g%0A%3C%2Fp%3E%3Cp%3E%20%20%20%20%20%20%20%20%20%20%20%20y-%200%20%3D%2010.0%C2%B2%2F2%209.8%0A%3C%2Fp%3E%3Cp%3E%20%20%20%20%20%20%20%20%20%20%20%20y%20-%200%20%3D%205.10%20m%0A%3C%2Fp%3E%3Cp%3E%20%20%20%20%20%20%20%20%20%20%20%20%0A%3C%2Fp%3E%3Cp%3EThe%20height%20from%20the%20ground%20is%20the%20height%20that%20rises%20from%20the%20reference%20system%20plus%20the%20depth%20of%20the%20ground%20from%20the%20reference%20system%0A%3C%2Fp%3E%3Cp%3E%20%20%20%20%20%20%20%20%20%20%20%20%20y%E2%82%82%20%3D%205.1%20%2B%2044%0A%3C%2Fp%3E%3Cp%3E%20%20%20%20%20%20%20%20%20%20%20%20%20y%E2%82%82%20%3D%2049.1%20m%0A%3C%2Fp%3E%3Cp%3E%3C%2Fp%3E%3Cp%3ELet%27s%20use%20the%20other%20equation%20to%20find%20the%20time%0A%3C%2Fp%3E%3Cp%3E%20%20%20%20%20%20%20%20%20%20%20%20%20%20%5Btex%5Dv_%7By%7D)
= - g t
t = / g
t = 10 / 9.8
t = 1.02 s
b) the maximum height
y- 44.0 = ² / 2 g
y - 44.0 = 5.1
y = 5.1 +44.0
y = 49.1 m
The time is the same because it does not depend on the initial height
t = 1.02 s
The x -component of the object's acceleration is 2 m/s².
<h3>What's the resultant force along x- direction?</h3>
- Forces along x axis direction are as follows
- 4N along +x axis, so it's taken as +4 N
- 2N along -x axis , so it's taken as -2N.
- Resultant force along x direction = 4N - 2N = 2 N which is along + ve x direction.
<h3>What's the acceleration along x axis direction?</h3>
- As per Newton's second law, Force = mass × acceleration of the object
- Force along x axis= mass × acceleration along x axis= 2N
- Acceleration = 2/ mass = 2/1 = 2 m/s²
Thus, we can conclude that the acceleration along x axis is 2 m/s².
Disclaimer: The question was given incomplete on the portal. Here is the complete question.
Question: The forces in (Figure 1) are acting on a 1.0 kg object. What is ax, the x-component of the object's acceleration?
Learn more about the acceleration here:
brainly.com/question/460763
#SPJ1
B) The amount of work done
Well you’d have a force due to gravity, the normal force which will be perpendicular to the sources (meaning you’ll have components to this vector), and you’d have the force of friction opposing the motion of the box. I’m also assuming there’s no air resistance. In this case you’d have three vector forces.
