Answer:
T=Lnsin
Please check the attached
Explanation:
The torque can simply be calculated by multiplying the length of the rod by the perpendicular force n as shown in the attached figure.
Note that sin90=1
T=Lsin(nsin90)
T=Lsinxn
T=Lnsin
Place a beaker covering its free surface with a paper over a coin as shown in the figure- 1. The coin can be seen through the si
1 answer:
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Answer:
(a). The average speed is 51.83 m/s.
(b). The average velocity over one revolution is zero.
Explanation:
Given that,
Angular velocity = 110 rev/m
Radius = 4.50 m
(a). We need to calculate the average speed
Using formula of average speed
(b). The average velocity over one revolution is zero because the net displacement is zero in one revolution.
Hence, (a). The average speed is 51.83 m/s.
(b). The average velocity over one revolution is zero.
Answer:
b) a = -k / m x , c) d²x / dt² = - A w² cos (wt+Ф) , d) and e) T = 2π √m / k
h) a = - A w² cos (wt+Ф)
Explanation:
a) see free body diagram in the attachment
b) We write Newton's second law
Fe = m a
-k x = ma
a = -k / m x
c) the acceleration is
a = d²x / dt²
If x = A cos wt
v = dx / dt = -A w sin (wt +Ф)
a = d²x / dt² = - A w² cos (wt+Ф)
d) we substitute in Newton's second law
d²x / dt² = -k / m x
We call
w² = k / m
e) substitute to find w
-A w² cos (wt+Ф) = -k / m A cos (wt+Ф)
w² = k / m
Angular velocity and frequency are related
w = 2π f
f = 1 / T
We substitute
T = 2π / w
T = 2π √m / k
g) v= - A w sin (wt+Ф)
h) acceleration is
a = - A w² cos (wt+Ф)
