Answer:
303.15K
Explanation:
the formula for converting celsius into kelvin is adding 273.15 to the number
Answer:
h = 2.64 meters
Explanation:
It is given that,
Mass of one ball, ![m_1=3\ kg](https://tex.z-dn.net/?f=m_1%3D3%5C%20kg)
Speed of the first ball,
(upward)
Mass of the other ball, ![m_2=2\ kg](https://tex.z-dn.net/?f=m_2%3D2%5C%20kg)
Speed of the other ball,
(downward)
We know that in an inelastic collision, after the collision, both objects move with one common speed. Let it is given by V. Using the conservation of momentum to find it as :
![V=\dfrac{m_1v_1+m_2v_2}{m_1+m_2}](https://tex.z-dn.net/?f=V%3D%5Cdfrac%7Bm_1v_1%2Bm_2v_2%7D%7Bm_1%2Bm_2%7D)
![V=\dfrac{3\times 20+2\times (-12)}{3+2}](https://tex.z-dn.net/?f=V%3D%5Cdfrac%7B3%5Ctimes%2020%2B2%5Ctimes%20%28-12%29%7D%7B3%2B2%7D)
V = 7.2 m/s
Let h is the height reached by the combined balls of putty rise above the collision point. Using the conservation of energy as :
![mgh=\dfrac{1}{2}mV^2](https://tex.z-dn.net/?f=mgh%3D%5Cdfrac%7B1%7D%7B2%7DmV%5E2)
![h=\dfrac{V^2}{2g}](https://tex.z-dn.net/?f=h%3D%5Cdfrac%7BV%5E2%7D%7B2g%7D)
![h=\dfrac{7.2^2}{2\times 9.8}](https://tex.z-dn.net/?f=h%3D%5Cdfrac%7B7.2%5E2%7D%7B2%5Ctimes%209.8%7D)
h = 2.64 meters
So, the height reached by the combined mass is 2.64 meters. Hence, this is the required solution.
Answer:
Because there is not as much cold as it was in France.
Explanation:
The average temperature in France during January ranges from 2.7° to 7.2° celsius which makes it the coldest month of the year. But since she comes to know that average temperature in Annville ranges 31° celsius which implies that the temperature is normal there and therefore, she packs sleeveless tops and shorts. Coats would not be required in a hot weather and hence, she does not pack it.
Answer:
4.6 kHz
Explanation:
The formula for the Doppler effect allows us to find the frequency of the reflected wave:
![f'=(\frac{v}{v-v_s})f](https://tex.z-dn.net/?f=f%27%3D%28%5Cfrac%7Bv%7D%7Bv-v_s%7D%29f)
where
f is the original frequency of the sound
v is the speed of sound
vs is the speed of the wave source
In this problem, we have
f = 41.2 kHz
v = 330 m/s
vs = 33.0 m/s
Therefore, if we substitute in the equation we find the frequency of the reflected wave:
![f'=(\frac{330 m/s}{330 m/s-33.0 m/s})(41.2 kHz)=45.8 kHz](https://tex.z-dn.net/?f=f%27%3D%28%5Cfrac%7B330%20m%2Fs%7D%7B330%20m%2Fs-33.0%20m%2Fs%7D%29%2841.2%20kHz%29%3D45.8%20kHz)
And the frequency of the beats is equal to the difference between the frequency of the reflected wave and the original frequency:
![f_B = |f'-f|=|45.8 kHz-41.2 kHz|=4.6 kHz](https://tex.z-dn.net/?f=f_B%20%3D%20%7Cf%27-f%7C%3D%7C45.8%20kHz-41.2%20kHz%7C%3D4.6%20kHz)
Answer:
2.83m
Explanation:
The information that we have is
Intensity at 2.0 m:
and ![r_{1}=2m](https://tex.z-dn.net/?f=r_%7B1%7D%3D2m)
we need an intensity level of: ![I_{2}=40dB](https://tex.z-dn.net/?f=I_%7B2%7D%3D40dB)
thus, we are looking for the distance
.
which we can find with the law for intensity and distance:
![(\frac{r_{2}}{r_{1}} )^2=\frac{I_{1}}{I_{2}}](https://tex.z-dn.net/?f=%28%5Cfrac%7Br_%7B2%7D%7D%7Br_%7B1%7D%7D%20%29%5E2%3D%5Cfrac%7BI_%7B1%7D%7D%7BI_%7B2%7D%7D)
we solve for
:
![\frac{r_{2}}{r_{1}}=\sqrt{\frac{I_{1}}{I_{2}} }\\\\r_{2}=r_{1}\sqrt{\frac{I_{1}}{I_{2}} }](https://tex.z-dn.net/?f=%5Cfrac%7Br_%7B2%7D%7D%7Br_%7B1%7D%7D%3D%5Csqrt%7B%5Cfrac%7BI_%7B1%7D%7D%7BI_%7B2%7D%7D%20%7D%5C%5C%5C%5Cr_%7B2%7D%3Dr_%7B1%7D%5Csqrt%7B%5Cfrac%7BI_%7B1%7D%7D%7BI_%7B2%7D%7D%20%7D)
and we substitute the known values:
![r_{2}=(2m)\sqrt{\frac{80dB}{40dB} }\\\\r_{2}=(2m)\sqrt{2}\\ r_{2}=2.83m](https://tex.z-dn.net/?f=r_%7B2%7D%3D%282m%29%5Csqrt%7B%5Cfrac%7B80dB%7D%7B40dB%7D%20%7D%5C%5C%5C%5Cr_%7B2%7D%3D%282m%29%5Csqrt%7B2%7D%5C%5C%20r_%7B2%7D%3D2.83m)
at a distance of 2.83m the intensity level is 40dB