Answer:
0.786 Hz, 1.572 Hz, 2.358 Hz, 3.144 Hz
Explanation:
The fundamental frequency of a standing wave on a string is given by
where
L is the length of the string
T is the tension in the string
is the mass per unit length
For the string in the problem,
L = 30.0 m
T = 20.0 N
Substituting into the equation, we find the fundamental frequency:
The next frequencies (harmonics) are given by
with n being an integer number and f being the fundamental frequency.
So we get:
Answer:
Option D is the correct answer.
Explanation:
Since value of angular acceleration is constant, the body has only centripetal acceleration.
Centripetal acceleration
We have radius = 7.112 cm = 0.07112 m
Frequency, f = 1975 rpm = 32.92 rps
Angular frequency, ω = 2πf = 2 x π x 32.92 = 206.82 rad/s
Substituting in centripetal acceleration equation,
Option D is the correct answer.
Answer:
160J
Explanation:
Given force = 8N and total distance = 20 meters
Workdone = force x distance
= 8 x 20
= 160J
Therefore, workdone by Riley in pulling the hoover is 160J
Answer:
The velocity of a falling object
Explanation:
The positive X axis is towards right and positive Y axis is towards up, so North direction is positive
A vector with less than 1 magnitude is not negative, because its magnitude may be in between 0 and 1 which is positive vector.
Any vector whose magnitude is greater than 1 is never be a negative vector.
The velocity of a falling object is towards bottom, that is towards negative Y axis. So that vector is negative.
A. Using the third equation of motion:
v2 = u2 + 2as
from the question;
the jet was initially at rest
hence u = 0
a = 1.75m/s2
s = 1500m
v2 = 02 + 2(1.75)(1500)
v2 = 5250
v = √5250
v = 72.46m/s
hence it moves with a velocity of 72.46m/s.
b. s = ut + 1/2at2
1500 = 0(t) + 1/2(1.75)t2
1500 × 2 = 2× 1/2(1.75)t2
3000 = 1.75t2
1714.29 = t2
41.4 = t
hence the time taken for the plane to down the runway is 41.4s.
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