Answer:
39.240 W
Explanation:
Let's start by calculating the work done by the engine. We can assume that it is the same work done by the weight of the object to bring it from 40m to the surface: as much energy it takes to bring it up, the same ammount it takes to bring it down. Said work is 
At this point we can simply apply the definition of power, that is 
, to get the power of the engine is 
Answer:
Explanation:
<h3>Given Data:</h3>
Mass = m = 68 kg
Velocity = v = 30 m/s
Time = 2 hours = 2 × 60 × 60 = 7200 s
<h3>Required:</h3>
Force = F = ?
<h3>Formula to be used:</h3>
<h3>Solution:</h3>
![\displaystyle F = \frac{(68)(30)}{7200} \\\\F = \frac{2040}{7200} \\\\F = 0.28 N\\\\\rule[225]{225}{2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20F%20%3D%20%5Cfrac%7B%2868%29%2830%29%7D%7B7200%7D%20%5C%5C%5C%5CF%20%3D%20%5Cfrac%7B2040%7D%7B7200%7D%20%5C%5C%5C%5CF%20%3D%200.28%20N%5C%5C%5C%5C%5Crule%5B225%5D%7B225%7D%7B2%7D)
Answer:
18min
Explanation:
v=d/t
t=d/v= 27/90 =0.3hrs =18min
Answer:
52 rad
Explanation:
Using
Ф = ω't +1/2αt²................... Equation 1
Where Ф = angular displacement of the object, t = time, ω' = initial angular velocity, α = angular acceleration.
Since the object states from rest, ω' = 0 rad/s.
Therefore,
Ф = 1/2αt²................ Equation 2
make α the subject of the equation
α = 2Ф/t².................. Equation 3
Given: Ф = 13 rad, t = 2.5 s
Substitute into equation 3
α = 2(13)/2.5²
α = 26/2.5
α = 4.16 rad/s².
using equation 2,
Ф = 1/2αt²
Given: t = 5 s, α = 4.16 rad/s²
Substitute into equation 2
Ф = 1/2(4.16)(5²)
Ф = 52 rad.
