B
Scientific laws are facts that describe observed reoccurring natural events.
Answer:
Length of the string, l = 0.486 meters
Explanation:
It is given that,
Mass of the string, ![m=2.4\times 10^{-3}\ kg](https://tex.z-dn.net/?f=m%3D2.4%5Ctimes%2010%5E%7B-3%7D%5C%20kg)
Tension in the string, T = 120 N
Frequency of transverse wave, f = 260 Hz
Wavelength of the wave, ![\lambda=0.6\ m](https://tex.z-dn.net/?f=%5Clambda%3D0.6%5C%20m)
The speed of a transverse wave (v) is given by :
........(1)
<em>Where,</em>
![\mu=\dfrac{m}{l}](https://tex.z-dn.net/?f=%5Cmu%3D%5Cdfrac%7Bm%7D%7Bl%7D)
Also, speed of a wave,
.........(2)
From equation (1) and (2) :
![l=\dfrac{f^2\lambda^2m}{T}](https://tex.z-dn.net/?f=l%3D%5Cdfrac%7Bf%5E2%5Clambda%5E2m%7D%7BT%7D)
![l=\dfrac{(260\ Hz)^2\times (0.6\ m)^2\times 2.4\times 10^{-3}\ kg}{120\ N}](https://tex.z-dn.net/?f=l%3D%5Cdfrac%7B%28260%5C%20Hz%29%5E2%5Ctimes%20%280.6%5C%20m%29%5E2%5Ctimes%202.4%5Ctimes%2010%5E%7B-3%7D%5C%20kg%7D%7B120%5C%20N%7D)
l = 0.486 m
So, the length of the string is 0.486 meters. Hence, this is the required solution.
Answer:
d = 9.69 cm
Explanation:
given,
mass of the block = 1.2 Kg
spring force constant(k) = 730 N/m
spring is compressed = d = ?
rough patch width = 5 cm
μ_k = 0.44
work done by friction = energy lost
![\mu_k mg \times x = \dfrac{1}{2}kd^2-\dfrac{1}{2}mv^2](https://tex.z-dn.net/?f=%5Cmu_k%20mg%20%5Ctimes%20x%20%3D%20%5Cdfrac%7B1%7D%7B2%7Dkd%5E2-%5Cdfrac%7B1%7D%7B2%7Dmv%5E2)
![0.44\times 1.2 \times 9.8 \times 0.05 = \dfrac{1}{2}\times 730 \times d^2-\dfrac{1}{2}\times 1.2 \times 2.3^2](https://tex.z-dn.net/?f=0.44%5Ctimes%201.2%20%5Ctimes%209.8%20%5Ctimes%200.05%20%3D%20%5Cdfrac%7B1%7D%7B2%7D%5Ctimes%20730%20%5Ctimes%20d%5E2-%5Cdfrac%7B1%7D%7B2%7D%5Ctimes%201.2%20%5Ctimes%202.3%5E2)
![3.432 = 365 d^2](https://tex.z-dn.net/?f=3.432%20%3D%20365%20d%5E2)
![d = \dfrac{3.432}{365}](https://tex.z-dn.net/?f=d%20%3D%20%5Cdfrac%7B3.432%7D%7B365%7D)
d = 0.0969 m
d = 9.69 cm
Answer:
3.63 s
Explanation:
We can solve the problem by using the equivalent SUVAT equations for the angular motion.
To find the angular acceleration, we can use the following equation:
![\omega_f^2 - \omega_i ^2 =2 \alpha \theta](https://tex.z-dn.net/?f=%5Comega_f%5E2%20-%20%5Comega_i%20%5E2%20%3D2%20%5Calpha%20%5Ctheta)
where
is the final angular speed
is the initial angular speed
is the angular distance covered
is the angular acceleration
Re-arranging the formula, we can find
:
![\alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}=\frac{(3.14\cdot 10^4 rad/s)^2-(1.10\cdot 10^4 rad/s)^2}{2(2.00\cdot 10^4 rad)}=2.16\cdot 10^4 rad/s^2](https://tex.z-dn.net/?f=%5Calpha%3D%5Cfrac%7B%5Comega_f%5E2-%5Comega_i%5E2%7D%7B2%5Ctheta%7D%3D%5Cfrac%7B%283.14%5Ccdot%2010%5E4%20rad%2Fs%29%5E2-%281.10%5Ccdot%2010%5E4%20rad%2Fs%29%5E2%7D%7B2%282.00%5Ccdot%2010%5E4%20rad%29%7D%3D2.16%5Ccdot%2010%5E4%20rad%2Fs%5E2)
Now we want to know the time the bit takes starting from rest to reach a speed of
. So, we can use the following equation:
![\alpha = \frac{\omega_f-\omega_i}{t}](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%5Cfrac%7B%5Comega_f-%5Comega_i%7D%7Bt%7D)
where:
is the angular acceleration
is the final speed
is the initial speed
t is the time
Re-arranging the equation, we can find the time:
![t=\frac{\omega_f-\omega_i}{\alpha}=\frac{7.85\cdot 10^4 rad/s-0}{2.16\cdot 10^4 rad/s^2}=3.63 s](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B%5Comega_f-%5Comega_i%7D%7B%5Calpha%7D%3D%5Cfrac%7B7.85%5Ccdot%2010%5E4%20rad%2Fs-0%7D%7B2.16%5Ccdot%2010%5E4%20rad%2Fs%5E2%7D%3D3.63%20s)
ANSWER is c
HOPR THIS WILL BE HELPFUL