A long. 1.0 kg rope hangs from a support that breaks, causing the rope to fall, if the pull exceeds 43 N. A student team has bui
1 answer:
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Answer:
106.7 N
Explanation:
We can solve the problem by using the impulse theorem, which states that the product between the average force applied and the duration of the collision is equal to the change in momentum of the object:
where
F is the average force
is the duration of the collision
m is the mass of the ball
v is the final velocity
u is the initial velocity
In this problem:
m = 0.200 kg
u = 20.0 m/s
v = -12.0 m/s
Solving for F,
And since we are interested in the magnitude only,
F = 106.7 N
Answer:
Option (2)
Explanation:
From the figure attached,
Horizontal component,
= 7.22 m
Vertical component,
= 9.58 m
Similarly, Horizontal component of vector C,
= C[Cos(60)]
= 6[Cos(60)]
=
= 3 m
= 5.20 m
Resultant Horizontal component of the vectors A + C,
m
= 4.38 m
Now magnitude of the resultant will be,
From ΔOBC,
=
=
= 6.1 m
Direction of the resultant will be towards vector A.
tan(∠COB) =
=
=
m∠COB =
= 46°
Therefore, magnitude of the resultant vector will be 6.1 m and direction will be 46°.
Option (2) will be the answer.
