The magnitude of the force required to stop the weight in 0.333 seconds is 67.6 N.
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Magnitude of required force to stop the weight</h3>
The magnitude of the force required to stop the weight in 0.333 seconds is calculated by applying Newton's second law of motion as shown below;
F = ma
F = m(v/t)
F = (mv)/t
F = (5 x 4.5)/0.333
F = 67.6 N
Thus, the magnitude of the force required to stop the weight in 0.333 seconds is 67.6 N.
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Answer:
The average velocity is
and
respectively.
Explanation:
Let's start writing the vertical position equation :

Where distance is measured in meters and time in seconds.
The average velocity is equal to the position variation divided by the time variation.
= Δx / Δt = 
For the first time interval :
t1 = 5 s → t2 = 8 s
The time variation is :

For the position variation we use the vertical position equation :

Δx = x2 - x1 = 1049 m - 251 m = 798 m
The average velocity for this interval is

For the second time interval :
t1 = 4 s → t2 = 9 s


Δx = x2 - x1 = 1495 m - 125 m = 1370 m
And the time variation is t2 - t1 = 9 s - 4 s = 5 s
The average velocity for this interval is :

Finally for the third time interval :
t1 = 1 s → t2 = 7 s
The time variation is t2 - t1 = 7 s - 1 s = 6 s
Then


The position variation is x2 - x1 = 701 m - (-1 m) = 702 m
The average velocity is

Answer:
The magnitude of vector B is 43 units and it points in the negative y-direction.
Explanation:
Resultant of vectors = vector sum of all the vectors
Vector A = 29j
Vector B = ?
Resultant of vector A and B = R = -14j
R = A + B
-14j = 29j + B
B = -14j - 29j = - 43j
Hence, the magnitude of vector B is 43 units and it points in the negative y-direction.
Answer:
270°
Explanation:
A full rotation is 360° so half (2/4) of it is 180° meaning a quarter (1/4) is 90°
So you can do 90+90+90=270
or 180+90=270