The space shuttle, in circular orbit around the Earth, collides with a small asteroid which ends up in the shuttle's storage bay
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Answer:
beat frequency = 13.87 Hz
Explanation:
given data
lengths l = 2.00 m
linear mass density μ = 0.0065 kg/m
String A is under a tension T1 = 120.00 N
String B is under a tension T2 = 130.00 N
n = 10 mode
to find out
beat frequency
solution
we know here that length L is
L = n × ........1
so λ =
and velocity is express as
V = .................2
so
frequency for string A = f1 =
f1 =
f1 =
and
f2 =
so
beat frequency is = f2 - f1
put here value
beat frequency = -
beat frequency = 13.87 Hz
This motion of all can be divided in to two, first the upward motion and second downward motion.
In case of projectiles time taken for upward motion = time taken for downward motion.
So we have time taken for upward motion = 8.14/2 = 4.07 seconds
We have equation of motion, v = u+at, where v is the final velocity , u is the initial velocity, a is the acceleration and t is the time taken.
In case of upward motion
v = 0 m/s
t = 4.07 seconds
a = -9.8
Substituting
So initial velocity of ball = 39.886 m/s.