Answer:
1.06 metres per second squared
Explanation:
since friction acts against foward force
20 N - 4 N = 16 N
use Newtons 2nd law F=ma Solve for a:
a= F÷m
= 16 ÷ 15
= 1.06 metres per second squared
The term for the process by which a portion of a glacier breaks off and falls into the water is called calving.
Answer:
<em>The velocity after the collision is 2.82 m/s</em>
Explanation:
<u>Law Of Conservation Of Linear Momentum
</u>
It states the total momentum of a system of bodies is conserved unless an external force is applied to it. The formula for the momentum of a body with mass m and speed v is
P=mv.
If we have a system of two bodies, then the total momentum is the sum of the individual momentums:
![P=m_1v_1+m_2v_2](https://tex.z-dn.net/?f=P%3Dm_1v_1%2Bm_2v_2)
If a collision occurs and the velocities change to v', the final momentum is:
![P'=m_1v'_1+m_2v'_2](https://tex.z-dn.net/?f=P%27%3Dm_1v%27_1%2Bm_2v%27_2)
Since the total momentum is conserved, then:
P = P'
Or, equivalently:
![m_1v_1+m_2v_2=m_1v'_1+m_2v'_2](https://tex.z-dn.net/?f=m_1v_1%2Bm_2v_2%3Dm_1v%27_1%2Bm_2v%27_2)
If both masses stick together after the collision at a common speed v', then:
![m_1v_1+m_2v_2=(m_1+m_2)v'](https://tex.z-dn.net/?f=m_1v_1%2Bm_2v_2%3D%28m_1%2Bm_2%29v%27)
The common velocity after this situation is:
![\displaystyle v'=\frac{m_1v_1+m_2v_2}{m_1+m_2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20v%27%3D%5Cfrac%7Bm_1v_1%2Bm_2v_2%7D%7Bm_1%2Bm_2%7D)
There is an m1=3.91 kg car moving at v1=5.7 m/s that collides with an m2=4 kg cart that was at rest v2=0.
After the collision, both cars stick together. Let's compute the common speed after that:
![\displaystyle v'=\frac{3.91*5.7+4*0}{3.91+4}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20v%27%3D%5Cfrac%7B3.91%2A5.7%2B4%2A0%7D%7B3.91%2B4%7D)
![\displaystyle v'=\frac{22.287}{7.91}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20v%27%3D%5Cfrac%7B22.287%7D%7B7.91%7D)
![\boxed{v' = 2.82\ m/s}](https://tex.z-dn.net/?f=%5Cboxed%7Bv%27%20%3D%202.82%5C%20m%2Fs%7D)
The velocity after the collision is 2.82 m/s
Answer:
The minimum frequency is 702.22 Hz
Explanation:
The two speakers are adjusted as attached in the figure. From the given data we know that
=3m
=4m
By Pythagoras theorem
![S_2O=\sqrt{(S_1S_2)^2+(S_1O)^2}\\S_2O=\sqrt{(3)^2+(4)^2}\\S_2O=\sqrt{9+16}\\S_2O=\sqrt{25}\\S_2O=5m](https://tex.z-dn.net/?f=S_2O%3D%5Csqrt%7B%28S_1S_2%29%5E2%2B%28S_1O%29%5E2%7D%5C%5CS_2O%3D%5Csqrt%7B%283%29%5E2%2B%284%29%5E2%7D%5C%5CS_2O%3D%5Csqrt%7B9%2B16%7D%5C%5CS_2O%3D%5Csqrt%7B25%7D%5C%5CS_2O%3D5m)
Now
The intensity at O when both speakers are on is given by
![I=4I_1 cos^2(\pi \frac{\delta}{\lambda})](https://tex.z-dn.net/?f=I%3D4I_1%20cos%5E2%28%5Cpi%20%5Cfrac%7B%5Cdelta%7D%7B%5Clambda%7D%29)
Here
- I is the intensity at O when both speakers are on which is given as 6
![W/m^2](https://tex.z-dn.net/?f=W%2Fm%5E2)
- I1 is the intensity of one speaker on which is 6
![W/m^2](https://tex.z-dn.net/?f=W%2Fm%5E2)
- δ is the Path difference which is given as
![\delta=S_2O-S_1O\\\delta=5-4\\\delta=1 m](https://tex.z-dn.net/?f=%5Cdelta%3DS_2O-S_1O%5C%5C%5Cdelta%3D5-4%5C%5C%5Cdelta%3D1%20m)
- λ is wavelength which is given as
![\lambda=\frac{v}{f}](https://tex.z-dn.net/?f=%5Clambda%3D%5Cfrac%7Bv%7D%7Bf%7D)
Here
v is the speed of sound which is 320 m/s.
f is the frequency of the sound which is to be calculated.
![16=4\times 6 \times cos^2(\pi \frac{1 \times f}{320})\\16/24= cos^2(\pi \frac{1f}{320})\\0.667= cos^2(\pi \frac{f}{320})\\cos(\pi \frac{f}{320})=\pm0.8165\\\pi \frac{f}{320}=\frac{7 \pi}{36}+2k\pi \\ \frac{f}{320}=\frac{7 }{36}+2k \\\\ {f}=320 \times (\frac{7 }{36}+2k )](https://tex.z-dn.net/?f=16%3D4%5Ctimes%206%20%5Ctimes%20cos%5E2%28%5Cpi%20%5Cfrac%7B1%20%5Ctimes%20f%7D%7B320%7D%29%5C%5C16%2F24%3D%20cos%5E2%28%5Cpi%20%5Cfrac%7B1f%7D%7B320%7D%29%5C%5C0.667%3D%20cos%5E2%28%5Cpi%20%5Cfrac%7Bf%7D%7B320%7D%29%5C%5Ccos%28%5Cpi%20%5Cfrac%7Bf%7D%7B320%7D%29%3D%5Cpm0.8165%5C%5C%5Cpi%20%5Cfrac%7Bf%7D%7B320%7D%3D%5Cfrac%7B7%20%5Cpi%7D%7B36%7D%2B2k%5Cpi%20%5C%5C%20%5Cfrac%7Bf%7D%7B320%7D%3D%5Cfrac%7B7%20%7D%7B36%7D%2B2k%20%5C%5C%5C%5C%20%7Bf%7D%3D320%20%5Ctimes%20%28%5Cfrac%7B7%20%7D%7B36%7D%2B2k%20%29)
where k=0,1,2
for minimum frequency
, k=1
![{f}=320 \times (\frac{7 }{36}+2 \times 1 )\\\\{f}=320 \times (\frac{79 }{36} )\\\\ f=702.22 Hz](https://tex.z-dn.net/?f=%7Bf%7D%3D320%20%5Ctimes%20%28%5Cfrac%7B7%20%7D%7B36%7D%2B2%20%5Ctimes%201%20%29%5C%5C%5C%5C%7Bf%7D%3D320%20%5Ctimes%20%28%5Cfrac%7B79%20%7D%7B36%7D%20%29%5C%5C%5C%5C%20f%3D702.22%20Hz)
So the minimum frequency is 702.22 Hz
Answer:
4.06 Hz
Explanation:
For simple harmonic motion, frequency is given by
where k is spring constant and m is the mass of the object.
Substituting 0.2 Kg for mass and 130 N/m for k then
![f=\frac {1}{2\pi}\times \sqrt{\frac {130}{0.2}}=4.057670803\\f\approx 4.06 Hz](https://tex.z-dn.net/?f=f%3D%5Cfrac%20%7B1%7D%7B2%5Cpi%7D%5Ctimes%20%5Csqrt%7B%5Cfrac%20%7B130%7D%7B0.2%7D%7D%3D4.057670803%5C%5Cf%5Capprox%204.06%20Hz)