Answer:
H = 0.673
Explanation:
given,
side of cubical crate = 0.74
weight of the crate = 600 N
magnitude of force = 330 N
the Horizontal distance of its Center of mass
= 0.74/2
= 0.37
Let the required Height be H
By Balancing the Torques, we get
H x 330 N = 0.37 x 600
330 H = 222
H = 0.673
hence, the height above the floor where force is acting is equal to 0.673 m
Answers:
a) -171.402 m/s
b) 17.49 s
c) 1700.99 m
Explanation:
We can solve this problem with the following equations:
(1)
(2)
(3)
Where:
is the bomb's final jeight
is the bomb'e initial height
is the bomb's initial vertical velocity, since the airplane was moving horizontally
is the time
is the acceleration due gravity
is the bomb's range
is the bomb's initial horizontal velocity
is the bomb's fina velocity
Knowing this, let's begin with the answers:
<h3>b) Time</h3>
With the conditions given above, equation (1) is now written as:
(4)
Isolating :
(5)
(6)
(7)
<h3>a) Final velocity</h3>
Since , equation (3) is written as:
(8)
(9)
(10) The negative sign ony indicates the direction is downwards
<h3>c) Range</h3>
Substituting (7) in (2):
(11)
(12)
The mass affects the kinetic energy because the more the mass the more energy is given to the object and the speed<span> affects by making it go faster and longer, so whenever speed goes up so does energy.</span>
Answer:
6.58m
Explanation:
The kinetic energy = Workdone on the roller
Workdone = Force * distance
Given
KE = Workdone = 362J
Force = 55N
Required
Distance
Substitute into the formula;
Workdone = Force * distance
362 = 55d
d = 362/55
d = 6.58m
Hence the student must push at a distance of 6.58m
