Answer / Explanation:
Referencing the diagram of a resonance in a vertical tube as illustrated in the attached diagram below,
we can say that:
Vertical wave is the result of superposition of dual and identical waves.
We should also note that the only difference is that they travel in opposite directions to each other.
So, let assume a tuning fork is vibrated with frequency ν
at the open end of the tube of varying length which can be changed during vibration with the help of water level change.
Also, let us assume the minimum length of the tube be L, when the resonance is heard.
Also, we assume that 1/4 of the wavelength λ
of the standing wave is equal to length of the minimum length L of the tube.
Hence, we have:
λ = 4
L
.
So, if speed of sound in air is V
, then frequency of the tuning fork can be expressed as:
U = v / λ = v / 4L
Noting that we have been given the speed of sound in air as V = 343
m
/
s
.
where:
U = v / λ = v / 4L
therefore,
343
m
/
s / 4 x 0.5525 m
= 155.20 Hz
Noting that 1 wave length in centimeter is 29979245800 Hz
Therefore, I wavelength in centimeter = 29979245800 Hz / 155.20 Hz
= 193 cm
Answer:
86.51° North of West or 273.49°
Explanation:
Let V' = velocity of boat relative to the earth, v = velocity of boat relative to water and V = velocity of water.
Now, by vector addition V' = V + v'.
Since v' = 6.10 m/s in the north direction, v' = (6.10 m/s)j and V = 100 m/s in the east direction, V = (100 m/s)i. So that
V' = V + v'
V' = (100 m/s)i + (6.10 m/s)j
So, we find the direction,Ф the boat must steer to from the components of V'.
So tanФ = 6.10 m/s ÷ 100 m/s
tanФ = 0.061
Ф = tan⁻¹(0.061) = 3.49°
So, the angle from the north is thus 90° - 3.49° = 86.51° North of West or 270° + 3.49° = 273.49°
Answer:
The distance that the spring compresses is:
![v\sqrt{\frac{m}{k}}](https://tex.z-dn.net/?f=v%5Csqrt%7B%5Cfrac%7Bm%7D%7Bk%7D%7D)
Explanation:
<u>Kinetic and Elastic Potential Energy</u>
The kinetic energy of an object of mass m traveling at a speed v is:
![\displaystyle K=\frac{1}{2}mv^2](https://tex.z-dn.net/?f=%5Cdisplaystyle%20K%3D%5Cfrac%7B1%7D%7B2%7Dmv%5E2)
The elastic potential energy of a spring of constant k that compresses a distance x is:
![\displaystyle E=\frac{1}{2}kx^2](https://tex.z-dn.net/?f=%5Cdisplaystyle%20E%3D%5Cfrac%7B1%7D%7B2%7Dkx%5E2)
The block of mass m is moving at a speed v when compresses a spring of constant k. The kinetic energy will eventually transform into elastic energy, but before that, both energies will be equal. It happens when:
![\displaystyle \frac{1}{2}mv^2=\frac{1}{2}kx^2](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7B1%7D%7B2%7Dmv%5E2%3D%5Cfrac%7B1%7D%7B2%7Dkx%5E2)
Simplifying:
![\displaystyle mv^2=kx^2](https://tex.z-dn.net/?f=%5Cdisplaystyle%20mv%5E2%3Dkx%5E2)
Dividing by k:
![\displaystyle x^2=\frac{mv^2}{k}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20x%5E2%3D%5Cfrac%7Bmv%5E2%7D%7Bk%7D)
Taking square root:
![\displaystyle x=\sqrt{\frac{mv^2}{k}}=v\sqrt{\frac{m}{k}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20x%3D%5Csqrt%7B%5Cfrac%7Bmv%5E2%7D%7Bk%7D%7D%3Dv%5Csqrt%7B%5Cfrac%7Bm%7D%7Bk%7D%7D)
The distance that the spring compresses is ![\mathbf{v\sqrt{\frac{m}{k}}}](https://tex.z-dn.net/?f=%5Cmathbf%7Bv%5Csqrt%7B%5Cfrac%7Bm%7D%7Bk%7D%7D%7D)
Answer: 17.03 rad/s^2
Explanation: One of the equation of motion that defines a circular motion with constant acceleration is given below as
ω = ω0 + αt
Where ω = final angular velocity = 9.3 rev/s
ω0 = initial angular velocity = 7.1 rev/s
α = angular acceleration = ?
t = time taken = 6.0s
By substituting the parameters, we have that
9.3 = 7.1 +α(6)
9.3 - 7.1 = 6α
16.3 = 6α
α = 16.3/6
α = 2.71 rev/s^2
We can also give the answer in rad/s^2 ( another unit of angular acceleration) by multipying by 2π
Hence, α = 2.71 × 2π , where π = 3.142
α = 17.03 rad/s^2
Answer is B, it can be positive or negative, as long as it has a charge.