I don't know for sure but i think that it is because the force him pushing her pushed him back as well
Hope i helped
Solving this using the time, we know that range = horizontal velocity x time of flight
since
there are no horizontal forces acting on the ball, there are no
horizontal accelerations and the initial horizontal velocity of 36 cos
28 will be constant throughout. If we use the correct time of flight given the launch parameters, we have
range = 36 cos 28 x 3.44 s = 109.3 m
Acc to
Publisso
Peer review is subjecting the author's scholarly work and research to the scrutiny of other experts in the same field to check its validity and evaluate its suitability for publication.
A peer review helps the publisher decide whether a work should be accepted.
Answer:
.a = 849.05 m / s²
Explanation
The centripetal acceleration is
a = v² / r
Linear and angular velocity are related
v = w r
Angular velocity and frequency are related by
w = 2π f
Let's replace
a = w² r
a = 4π² f² r
Let's reduce to the SI system
f = 2.30 rev / s (2π rad / 1 rev) = 14.45 rad / s
.r = 10.3 cm = 0.103 m
Let's calculate
a = 4π² 14.45² 0.103
.a = 849.05 m / s²
Answer:
The correct choice is A. Yes, the more exercise the better.
