Answer:
while we often confuse mass with weight, 47 kg is 47 x 9.8 = 460.6 Newtons. 9.8 is acceleration of gravity in m/sec/sec
Answer:
The charge carried by the droplet is ![1.330\times10^{-19}\ C](https://tex.z-dn.net/?f=1.330%5Ctimes10%5E%7B-19%7D%5C%20C)
Explanation:
Given that,
Distance =8.4 cm
Time = 0.250 s
Suppose tiny droplets of oil acquire a small negative charge while dropping through a vacuum in an experiment. An electric field of magnitude
points straight down and if the mass of the droplet is ![2.93\times10^{-15} kg](https://tex.z-dn.net/?f=2.93%5Ctimes10%5E%7B-15%7D%20kg)
We need to calculate the acceleration
Using equation of motion
![s=ut+\dfrac{1}{2}at^2](https://tex.z-dn.net/?f=s%3Dut%2B%5Cdfrac%7B1%7D%7B2%7Dat%5E2)
Put the value into the formula
![8.4\times10^{-2}=0+\dfrac{1}{2}\times a\times(0.250)^2](https://tex.z-dn.net/?f=8.4%5Ctimes10%5E%7B-2%7D%3D0%2B%5Cdfrac%7B1%7D%7B2%7D%5Ctimes%20a%5Ctimes%280.250%29%5E2)
![a=\dfrac{8.4\times10^{-2}\times2}{(0.250)^2}](https://tex.z-dn.net/?f=a%3D%5Cdfrac%7B8.4%5Ctimes10%5E%7B-2%7D%5Ctimes2%7D%7B%280.250%29%5E2%7D)
![a=2.688\ m/s^2](https://tex.z-dn.net/?f=a%3D2.688%5C%20m%2Fs%5E2)
We need to calculate the charge carried by the droplet
Using formula of electric filed
![E=\dfrac{F}{q}](https://tex.z-dn.net/?f=E%3D%5Cdfrac%7BF%7D%7Bq%7D)
![q=\dfrac{ma}{E}](https://tex.z-dn.net/?f=q%3D%5Cdfrac%7Bma%7D%7BE%7D)
Put the value into the formula
![q=\dfrac{2.93\times10^{-15}\times2.688}{5.92\times10^4}](https://tex.z-dn.net/?f=q%3D%5Cdfrac%7B2.93%5Ctimes10%5E%7B-15%7D%5Ctimes2.688%7D%7B5.92%5Ctimes10%5E4%7D)
![q=1.330\times10^{-19}\ C](https://tex.z-dn.net/?f=q%3D1.330%5Ctimes10%5E%7B-19%7D%5C%20C)
Hence, The charge carried by the droplet is ![1.330\times10^{-19}\ C](https://tex.z-dn.net/?f=1.330%5Ctimes10%5E%7B-19%7D%5C%20C)
Answer: I dont have a question but you have a fo.rt.ni.te pfp and Im guessing you play fort so add me my epic is qt-orange im NA_EAST
Explanation:
The transfer of heat through flowing material is convection
Answer:
![v_{f}=17.47 m/s](https://tex.z-dn.net/?f=v_%7Bf%7D%3D17.47%20m%2Fs)
Explanation:
Let's use the conservation of momentum to solve it.
(1)
- The total initial momentum will be:
![m_{1}v_{1i}+m_{2}v_{2i}](https://tex.z-dn.net/?f=m_%7B1%7Dv_%7B1i%7D%2Bm_%7B2%7Dv_%7B2i%7D)
- The total final momentum will be:
, but as they stick together after the collision, v1f = v2f = vf.
So we can rewrite (1), using the above information:
![m_{1}v_{1i}+m_{2}v_{2i}=m_{1}v_{f}+m_{2}v_{f}](https://tex.z-dn.net/?f=m_%7B1%7Dv_%7B1i%7D%2Bm_%7B2%7Dv_%7B2i%7D%3Dm_%7B1%7Dv_%7Bf%7D%2Bm_%7B2%7Dv_%7Bf%7D)
![m_{1}v_{1i}+m_{2}v_{2i}=v_{f}(m_{1}+m_{2})](https://tex.z-dn.net/?f=m_%7B1%7Dv_%7B1i%7D%2Bm_%7B2%7Dv_%7B2i%7D%3Dv_%7Bf%7D%28m_%7B1%7D%2Bm_%7B2%7D%29)
![v_{f}=\frac{m_{1}v_{1i}+m_{2}v_{2i}}{m_{1}+m_{2}}](https://tex.z-dn.net/?f=v_%7Bf%7D%3D%5Cfrac%7Bm_%7B1%7Dv_%7B1i%7D%2Bm_%7B2%7Dv_%7B2i%7D%7D%7Bm_%7B1%7D%2Bm_%7B2%7D%7D)
![v_{f}=\frac{1100\cdot 14+2500\cdot 19}{1100+2500}](https://tex.z-dn.net/?f=v_%7Bf%7D%3D%5Cfrac%7B1100%5Ccdot%2014%2B2500%5Ccdot%2019%7D%7B1100%2B2500%7D)
Finally, the magnitude of the velocity of the wreckage of the two cars immediately after the collision is:
![v_{f}=17.47 m/s](https://tex.z-dn.net/?f=v_%7Bf%7D%3D17.47%20m%2Fs)
I hope it helps you!