This question is incomplete, the complete question is;
A football quarterback throws a 0.408 kg football for a long pass. While in the motion of throwing, the quarterback moves the ball 1.909 m, starting from rest, and completes the motion in 0.439 s. Assuming the acceleration is constant, what force does the quarterback apply to the ball during the pass
;
a) F_throw = 8.083 N
b) F_throw = 9.181 N
c) F_throw = 2.284 N
d) F_throw = 16.014 N
e) None of these is correct
Answer:
the quarterback applied a force of 8.083 N to the ball during the pass
so Option a) F_throw = 8.083 N is the correct answer
Explanation:
Given that;
m = 0.408 kg
d = 1.909 m
u = 0 { from rest}
t = 0.439 s
Now using Kinetic equation
d = ut + 1/2 at²
we substitute
1.909 = (0 × 0.439) + 1/2 a(0.439)²
1.909 = 0 + 0.09636a
1.909 = 0.09636a
a = 1.909 / 0.09636
a = 19.8111 m/s²
Now force applied will be;
F = ma
we substitute
F = 0.408 × 19.8111
F = 8.0828 ≈ 8.083 N
Therefore the quarterback applied a force of 8.083 N to the ball during the pass
so Option a) F_throw = 8.083 N is the correct answer
Answer:
Explanation:
GIVEN
Force (F) = 8 N
Distance (d) = 2.5 metres
Work done = ?
WE know we have the formula
work done = F * d
Work done = 8 * 2.5
= 20 Joule
Hope it helps :)
The resultant vector is zero.
Explanation:
When you add displacement vectors using the head to tail method, you follow this procedure:
- Draw the first vector
- Draw the second vector, with its tail starting from the head of the first vector
- Draw the third vector, with its tail starting from the head of the second vector
.. and so on.
The resultant of the all vectors will be the vector connecting the tail of the 1st vector to the head of the last vector.
In this problem, the vectors make a closed polygon: this means that the head of the last vector coincides with the tail of the first vector.
Therefore, this means that the length of the resultant vector is zero.
Learn more about vector addition:
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Answer:
minimum initial velocity is 21.35 m/s
Explanation:
given data
distance S = 30 m
height h = 30 m
maximum acceleration a = 2 m/s²
to find out
minimum initial velocity that your friend could have thrown the object to enable you to catch
solution
first we get here time with the help of second equation of motion
time = 
..................1
put her value we get
time = 
time = 5.477 second
and that is time which tossed object must be take so we apply here again second equation of motion that is
-S = ut - 0.5 × gt² .......................2
-30 = u× 5.477 - 0.5 ×9.8×5.477²
solve it we get
u = 21.35 m/s
so minimum initial velocity is 21.35 m/s
