Answer:
(a) The speed of the target proton after the collision is:
, and (b) the speed of the projectile proton after the collision is:
.
Explanation:
We need to apply at the system the conservation of the linear momentum on both directions x and y, and we get for the x axle:
, and y axle:
. Now replacing the value given as:
,
for the projectile proton and according to the problem
are perpendicular so
, and assuming that
, we get for x axle:
and y axle:
, then solving for
, we get:
and replacing at the first equation we get:
, now solving for
, we can find the speed of the projectile proton after the collision as:
and
, that is the speed of the target proton after the collision.
The time period resulting in oscillations will be 1.986 seconds.
<h3>What is the period of oscillation?</h3>
The period is the amount of time it takes for a particle to perform one full oscillation. T is the symbol for it. Taking the reciprocal of the frequency yields the frequency of the oscillation.
The time period of the oscillation is;

Hence the time period resulting oscillations will be 1.986 seconds.
To learn more about the time period of oscillation refer to the link;
brainly.com/question/20070798
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Answer:
the elements towards the bottom left corner
Answer:
1/3 the distance from the fulcrum
Explanation:
On a balanced seesaw, the torques around the fulcrum calculated on one side and on another side must be equal. This means that:

where
W1 is the weight of the boy
d1 is its distance from the fulcrum
W2 is the weight of his partner
d2 is the distance of the partner from the fulcrum
In this problem, we know that the boy is three times as heavy as his partner, so

If we substitute this into the equation, we find:

and by simplifying:

which means that the boy sits at 1/3 the distance from the fulcrum.
Answer:
0.108 rad/s².
Explanation:
Given that
Initial angular velocity ,ωi = 0 rad/s
Final angular velocity ωf= 0.5 rev/s
We know that
1 rev/s = 6.28 rad/s
ωf= 3.14 rad/s
t= 28.9 s
We know that (if acceleration is constant)
ωf=ωi + α t
α=Angular acceleration
3.14 = 0 + α x 28.9
Therefore the acceleration will be 0.108 rad/s².
Therefore the answer will be 0.108 rad/s².