Answer:
41.3 m/s^2 option (e)
Explanation:
force, F = 6.81 N
mass, m = 165 g = 0.165 kg
Let a be the acceleration of the puck.
Use newtons' second law
Force = mass x acceleration
6.81 = 0.165 x a
a = 41.27 m/s^2
a = 41.3 m/s^2
Thus, the acceleration of the puck is 41.3 m/s^2.
0.29 m/s (wave velocity = wavelength (lamda)/period (T) in metres)
35 / 1.2 = 29.16
29.16 ÷ 100 = 0.29
Wave velocity in string:
The properties of the medium affect the wave's velocity in a string. For instance, if a thin guitar string is vibrated while a thick rope is not, the guitar string's waves will move more quickly. As a result, the linear densities of the two strings affect the string's velocity. Linear density is defined as the mass per unit length.
Instead of the sinusoidal wave, a single symmetrical pulse is taken into consideration in order to comprehend how the linear mass density and tension will affect the wave's speed on the string.
Learn more about density here:
brainly.com/question/15164682
#SPJ4
Answer/Explanation:
In chemistry, concentration is the abundance of a constituent divided by the total volume of a mixture. Several types of mathematical description can be distinguished: mass concentration, molar concentration, number concentration, and volume concentration.
A concentration refers to a subject or study within a specific major. As part of the degree coursework, concentration classes count toward your major requirements as well.
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:
![Q=3.9825\times 10^{-9} C](https://tex.z-dn.net/?f=Q%3D3.9825%5Ctimes%2010%5E%7B-9%7D%20C)
Explanation:
We are given that a parallel- plate capacitor is charged to a potential difference V and then disconnected from the voltage source.
1 m =100 cm
Surface area =S=![\frac{360}{10000}=0.036 m^2](https://tex.z-dn.net/?f=%5Cfrac%7B360%7D%7B10000%7D%3D0.036%20m%5E2)
![\Delta d=0.8 cm=0.008 m](https://tex.z-dn.net/?f=%5CDelta%20d%3D0.8%20cm%3D0.008%20m)
![\Delta V=100 V](https://tex.z-dn.net/?f=%5CDelta%20V%3D100%20V)
We have to find the charge Q on the positive plates of the capacitor.
V=Initial voltage between plates
d=Initial distance between plates
Initial Capacitance of capacitor
![C=\frac{\epsilon_0 S}{d}](https://tex.z-dn.net/?f=C%3D%5Cfrac%7B%5Cepsilon_0%20S%7D%7Bd%7D)
Capacitance of capacitor after moving plates
![C_1=\frac{\epsilon_0 S}{(d+\Delta d)}](https://tex.z-dn.net/?f=C_1%3D%5Cfrac%7B%5Cepsilon_0%20S%7D%7B%28d%2B%5CDelta%20d%29%7D)
![V=\frac{Q}{C}](https://tex.z-dn.net/?f=V%3D%5Cfrac%7BQ%7D%7BC%7D)
Potential difference between plates after moving
![V=\frac{Q}{C_1}](https://tex.z-dn.net/?f=V%3D%5Cfrac%7BQ%7D%7BC_1%7D)
![V+\Delta V=\frac{Q}{C_1}](https://tex.z-dn.net/?f=V%2B%5CDelta%20V%3D%5Cfrac%7BQ%7D%7BC_1%7D)
![\frac{Qd}{\epsilon_0S}+100=\frac{Q(d+\Delta d)}{\epsilon_0S}](https://tex.z-dn.net/?f=%5Cfrac%7BQd%7D%7B%5Cepsilon_0S%7D%2B100%3D%5Cfrac%7BQ%28d%2B%5CDelta%20d%29%7D%7B%5Cepsilon_0S%7D)
![\frac{Q(d+\Delta d)}{\epsilon_0 S}-\frac{Qd}{\epsilon_0S}=100](https://tex.z-dn.net/?f=%5Cfrac%7BQ%28d%2B%5CDelta%20d%29%7D%7B%5Cepsilon_0%20S%7D-%5Cfrac%7BQd%7D%7B%5Cepsilon_0S%7D%3D100)
![\frac{Q\Delta d}{\epsilon_0 S}=100](https://tex.z-dn.net/?f=%5Cfrac%7BQ%5CDelta%20d%7D%7B%5Cepsilon_0%20S%7D%3D100)
![\epsilon_0=8.85\times 10^{-12}](https://tex.z-dn.net/?f=%5Cepsilon_0%3D8.85%5Ctimes%2010%5E%7B-12%7D)
![Q=\frac{100\times 8.85\times 10^{-12}\times 0.036}{0.008}](https://tex.z-dn.net/?f=Q%3D%5Cfrac%7B100%5Ctimes%208.85%5Ctimes%2010%5E%7B-12%7D%5Ctimes%200.036%7D%7B0.008%7D)
![Q=3.9825\times 10^{-9} C](https://tex.z-dn.net/?f=Q%3D3.9825%5Ctimes%2010%5E%7B-9%7D%20C)
Hence, the charge on positive plate of capacitor=![Q=3.9825\times 10^{-9} C](https://tex.z-dn.net/?f=Q%3D3.9825%5Ctimes%2010%5E%7B-9%7D%20C)