three charged particals are located at the corners of an equil triangle shown in the figure showing let (q 2.20 Uc) and L 0.650
Answer:
Vprom = 0.00347[km/min]
Explanation:
We can calculate each of the average speeds and then perform the overall average between the two speeds.
V1 = 6/54
V1 = 0.111[km/min]
V2 = 1/16
V2 = 0.0625[km/min]
![V_{prom} = \frac{V_{1} + V_{2}}{2} \\V_{prom} = \frac{0.1111 + 0.0625}{2}\\V_{prom} = 0.00347 [km/min]](https://tex.z-dn.net/?f=V_%7Bprom%7D%20%3D%20%5Cfrac%7BV_%7B1%7D%20%2B%20V_%7B2%7D%7D%7B2%7D%20%20%5C%5CV_%7Bprom%7D%20%3D%20%5Cfrac%7B0.1111%20%2B%200.0625%7D%7B2%7D%5C%5CV_%7Bprom%7D%20%3D%200.00347%20%5Bkm%2Fmin%5D)
What's the weight of the pear ?
Weight = (mass) x (gravity) = (1 kg) x (9.8 m/s²) = 9.8 Newtons.
OK. We know there's a force of 9.8 Newtons acting downwards on the pear.
Is the pear accelerating ? No ! It's just laying there on the table.
If it's not accelerating, then we know that the net force on it must be zero.
So there must be ANOTHER force acting UPWARDS on it, to exactly
cancel out the downward force of its weight. THAT's the "normal" force ...
the upward force that the table exerts on the pear. It must also be 9.8N,
but UPwards, so that if you add it to the weight, the sum is zero.
