The calculation of the centripetal acceleration of an object following a circular path is based on the equation,
a = v² / r
where a is the acceleration, v is the velocity, and r is the radius.
Substituting the known values from the given above,
4.4 m/s² = (15 m/s)² / r
The value of r from the equation is 51.14 m.
Answer: 51.14 m
Answer:
To create a second harmonic the rope must vibrate at the frequency of 3 Hz
Explanation:
First we find the fundamental frequency of the rope. The fundamental frequency is the frequency of the rope when it vibrates in only 1 loop. Therefore,
f₁ = v/2L
where,
v = speed of wave = 36 m/s
L = Length of rope = 12 m
f₁ = fundamental frequency
Therefore,
f₁ = (36 m/s)/2(12 m)
f₁ = 1.5 Hz
Now the frequency of nth harmonic is given in general, as:
fn = nf₁
where,
fn = frequency of nth harmonic
n = No. of Harmonic = 2
f₁ = fundamental frequency = 1.5 Hz
Therefore,
f₂ = (2)(1.5 Hz)
<u>f₂ = 3 Hz</u>
Answer:
2.48 m/s
Explanation:
We can use the kinematic equation,
s = ut +½at²
Where
s = displacement
u = initial velocity
t = time taken
a = acceleration
Using the equation in vertical direction,
321 = 0×t +½×g×t², u = 0 because initial vertical velocity is 0
We get t = 8.01 s
Using the equation in the horizontal direction,
52 = u×8.01 +½×0×(8.01)²,. a = 0 because no unbalanced force act on object in that direction
So u = 2.48 m/s
Here try this. The pic is the answer