Answer:
B. w=12.68rad/s
C. α=3.52rad/s^2
Explanation:
B)
We can solve this problem by taking into account that (as in the uniformly accelerated motion)
( 1 )
where w0 is the initial angular speed, α is the angular acceleration, s is the arc length and r is the radius.
In this case s=3.7m, r=16.2cm=0.162m, t=3.6s and w0=0. Hence, by using the equations (1) we have
to calculate the angular speed w we can use
Thus, wf=12.68rad/s
C) We can use our result in B)
I hope this is useful for you
regards
The velocity of the boat after the package is thrown is 0.36 m/s.
<h3>
Final velocity of the boat</h3>
Apply the principle of conservation of linear momentum;
Pi = Pf
where;
- Pi is initial momentum
- Pf is final momentum
v(74 + 135) = 15 x 5
v(209) = 75
v = 75/209
v = 0.36 m/s
Thus, the velocity of the boat after the package is thrown is 0.36 m/s.
Learn more about velocity here: brainly.com/question/6504879
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Answer:
D = 9.9 10⁶ mi
Explanation:
In the exercise they give the expression for maximum viewing distance
D = 2 r h + h²
Ask for this distance for a height of 1100 feet
Let's calculate
D = 2 3960 1100 + 1100²
D = 8.712 10⁶ + 1.21 10⁶
D = 9.92 10⁶ mi
D = 9.9 10⁶ mi
As we know that as per Newton's II law we have
here we will have
= change in momentum
= time interval in which momentum is changed
now in order to have least injury during jumping we need to have least force on the jumper
so in order to have least force we can say that the momentum must have to change in maximum time so that amount of force must be least
So we need to increase the time in which momentum of the system is changed
