Answer:
![f=272.15Hz](https://tex.z-dn.net/?f=f%3D272.15Hz)
Explanation:
Given data
Length of tube L=0.632 m
Speed of sound v=344 m/s
To find
Fundamental frequency f
Solution
The fundamental frequency of the tube can be given as:
![f=\frac{v}{2L}\\ f=\frac{344m/s}{2(0.632m)}\\ f=272.15Hz](https://tex.z-dn.net/?f=f%3D%5Cfrac%7Bv%7D%7B2L%7D%5C%5C%20f%3D%5Cfrac%7B344m%2Fs%7D%7B2%280.632m%29%7D%5C%5C%20f%3D272.15Hz)
Answer:
Yes
Explanation:
There are two types of interference possible when two waves meet at the same point:
- Constructive interference: this occurs when the two waves meet in phase, i.e. the crest (or the compression, in case of a longitudinale wave) meets with the crest (compression) of the other wave. In such a case, the amplitude of the resultant wave is twice that of the original wave.
- Destructive interferece: this occurs when the two waves meet in anti-phase, i.e. the crest (or the compression, in case of a longitudinal wave) meets with the trough (rarefaction) of the other wave. In this case, the amplitude of the resultant wave is zero, since the amplitudes of the two waves cancel out.
In this problem, we have a situation where the compression of one wave meets with the compression of the second wave, so we have constructive interference.
The vertical component of the velocity after the given time is -9.8 m/s while the horizontal component of the velocity is 16 m/s.
The given parameters;
- initial horizontal velocity, vₓ = 16 m/s
- initial vertical velocity,
![v_y =0](https://tex.z-dn.net/?f=v_y%20%3D0)
- time interval 1 seconds
The components of the velocity can be horizontal or vertical velocity.
The vertical component of the velocity is affected by acceleration due to gravity while the horizontal component of the velocity is not affected by gravity.
The vertical component of the velocity is calculated as;
![v_y = v_0_y -gt\\\\v_y = 0 - (1\times 9.8)\\\\v_y = -9.8 \ m/s](https://tex.z-dn.net/?f=v_y%20%3D%20v_0_y%20-gt%5C%5C%5C%5Cv_y%20%3D%200%20-%20%281%5Ctimes%209.8%29%5C%5C%5C%5Cv_y%20%3D%20-9.8%20%5C%20m%2Fs)
The horizontal component of the velocity is constant since it is not affected by gravity.
The horizontal component of the velocity = 16 m/s
Thus, the vertical component of the velocity after the given time is -9.8 m/s while the horizontal component of the velocity is 16 m/s.
Learn more here:brainly.com/question/20349275
Answer:
mass of box 1 = 2.20 kg
mass of box 2 = 5.93 kg
Explanation:
Let the mass of box 1 and box 2 is respectively
and ![m_2](https://tex.z-dn.net/?f=m_2)
so we will have
Force applied on box 1 then acceleration
![a = \frac{F}{m_1 + m_2}](https://tex.z-dn.net/?f=a%20%3D%20%5Cfrac%7BF%7D%7Bm_1%20%2B%20m_2%7D)
![1.50 = \frac{12.2}{m_1 + m_2}](https://tex.z-dn.net/?f=1.50%20%3D%20%5Cfrac%7B12.2%7D%7Bm_1%20%2B%20m_2%7D)
![m_1 + m_2 = 8.13](https://tex.z-dn.net/?f=m_1%20%2B%20m_2%20%3D%208.13)
Now we know that contact force between them in above case is given as
![F_n = m_2a](https://tex.z-dn.net/?f=F_n%20%3D%20m_2a)
![8.90 = m_2(1.5)](https://tex.z-dn.net/?f=8.90%20%3D%20m_2%281.5%29)
![m_2 = 5.93 kg](https://tex.z-dn.net/?f=m_2%20%3D%205.93%20kg)
now we have
![m_1 = 2.20 kg](https://tex.z-dn.net/?f=m_1%20%3D%202.20%20kg)
First, let's put 22 km/h in m/s:
![22 \frac{km}{h} \times \frac{1000m}{1km} \times \frac{1h}{3600s}=6.11 \frac{m}{s}](https://tex.z-dn.net/?f=22%20%5Cfrac%7Bkm%7D%7Bh%7D%20%5Ctimes%20%20%5Cfrac%7B1000m%7D%7B1km%7D%20%20%5Ctimes%20%20%5Cfrac%7B1h%7D%7B3600s%7D%3D6.11%20%5Cfrac%7Bm%7D%7Bs%7D%20%20)
Now the radial force required to keep an object of mass m, moving in circular motion around a radius R, is given by
![F_{rad}=m \frac{v^2}{R}](https://tex.z-dn.net/?f=F_%7Brad%7D%3Dm%20%5Cfrac%7Bv%5E2%7D%7BR%7D%20)
The force of friction is given by the normal force (here, just the weight, mg) times the static coefficient of friction:
![F_{fric}= mg \mu_{s}](https://tex.z-dn.net/?f=F_%7Bfric%7D%3D%20mg%20%5Cmu_%7Bs%7D)
Notice we don't use the kinetic coefficient even though the bike is moving. This is because when the tires meet the road they are momentarily stationary with the road surface. Otherwise the bike is skidding.
Now set these equal, since friction is the only thing providing the ability to accelerate (turn) without skidding off the road in a line tangent to the curve: