After all that you have learned in this unit, construct a pamphlet (brochure) in Microsoft Publisher helping new freshman to enc
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Answer:
the magnitude of first force = 3 × 5= 15 N
ANd, the magnitude of second force = 5 × 5 = 25 N
Explanation:
The computation of the magnitude of the each force is shown below:
Provided that
Ratio of forces = 3: 5
Let us assume the common factor is x
Now
first force = 3x
And, the second force = 5x
Resultant force = 35 N
The Angle between the forces = 60 degrees
Based on the above information
Resultant force i.e. F = √ F_1^2 +F_2^2 + 2 F_1F_2cos
35 = √[(3x)²+ (5x)²+ 2 (3x)(5x) cos 60°]
35 =√ 9x² + 25x² + 15x² (cos 60° = 0.5)
35 = √49 x²
x = 5
So, the magnitude of first force = 3 × 5= 15 N
ANd, the magnitude of second force = 5 × 5 = 25 N
Answer:
2.47 m
Explanation:
Let's calculate first the time it takes for the ball to cover the horizontal distance that separates the starting point from the crossbar of d = 52 m.
The horizontal velocity of the ball is constant:
and the time taken to cover the horizontal distance d is
So this is the time the ball takes to reach the horizontal position of the crossbar.
The vertical position of the ball at time t is given by
where
is the initial vertical velocity
g = 9.8 m/s^2 is the acceleration of gravity
And substituting t = 2.56 s, we find the vertical position of the ball when it is above the crossbar:
The height of the crossbar is h = 3.05 m, so the ball passes
above the crossbar.