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My answers
Answer: minimum speed of launch must be 7.45m/s
Explanation:
Given the following:
Height or distance (s) = 2.83m
The final velocity(Vf) at maximum height = 0
Upward motion, acceleration due to gravity(g) us negative = -9.8m/s^2
From the 3rd equation of motion:
V^2 = u^2 - 2gs
Where V = final velocity
u = initial velocity
Therefore, u = Vi
u = √Vf^2 - 2gs
u = √0^2 - 2(-9.8)(2.83)
u = √0 + 55.468
u = √55.468
u = 7.4476 m/s
u = 7.45m/s
Answer:
The angle formed of the rope with the surface = 40°
Force applied = 125Newtons
The displacement covered by the box =25metres
W= FDcos theta
[125×40×cos(40°) ] Joules
= [ (3125×0.76604444311)]Joules
= 2393.88888472 joules(ans)
Hope it helps
<span>a thin fibrous cartilage between the surfaces of some joints, e.g., the knee.</span>
Integrating the velocity equation, we will see that the position equation is:

<h3>How to get the position equation of the particle?</h3>
Let the velocity of the particle is:

To get the position equation we just need to integrate the above equation:


Then:


Replacing that in our integral we get:


Where C is a constant of integration.
Now we remember that 
Then we have:

To find the value of C, we use the fact that f(0) = 0.

C = -1 / 3
Then the position function is:

Integrating the velocity equation, we will see that the position equation is:

To learn more about motion equations, refer to:
brainly.com/question/19365526
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