Ok, I think this is right but I am not sure:
Q = ϵ
0AE
A= π π
r^2
=(8.85x10^-12 C^2/Nm^2)
( π π (0.02m)^2)
(3x10^6 N/C) =3.3x10^-8 C = 33nC N = Q/e = (3.3x10^-8 C)/(1.60x10^-19 C/electron) = 2.1x10^11 electrons
<span>The best way to cool soft and thick foods when using the refrigerator is by having them to be placed and poured on a pan or another way is by having them to be placed in one container in which they are in a water bath, to be heated of.</span>
Answer:
Use the ammeter to measure the current that flows through each wire, because a larger current that flows through the wire corresponds to a smaller resistivity
Explanation:
Since they are connected to a constant voltage power source, the potential difference does not change. The potential difference is proportional to the product of the current and the resistance and, the resistance opposes the flow of electric current. It is clear to see that a large current that flows through the current means there is a lesser resistance to the flow of current at constant potential difference across the circuit.
Answer:
a) 12.8212 N
b) 12.642 N
Explanation:
Mass of bucket = m = 0.54 kg
Rate of filling with sand = 56.0 g/ sec = 0.056 kg/s
Speed of sand = 3.2 m/s
g= 9.8 m/sec2
<u>Condition (a);</u>
Mass of sand = Ms = 0.75 kg
So total mass becomes = bucket mass + sand mass = 0.54 +0.75=1.29 kg
== > total weight = 1.29 × 9.8 = 12.642 N
Now impact of sand = rate of filling × velocity = 0.056 × 3.2 = 0.1792 kg. m /sec2=0.1792 N
Scale reading is sum of impact of sand and weight force ;
i-e
scale reading = 12.642 N+0.1792 N = 12.8212 N
<u>Codition (b);</u>
bucket mass + sand mass = 0.54 +0.75=1.29 kg
==>weight = mg = 1.29 × 9.8 = 12.642 N (readily calculated above as well)
Answer:
a) d = 6.0 m
Explanation:
Since car is accelerating at uniform rate then here we can say that the distance moved by the car with uniform acceleration is given as
![d = \frac{(v_f + v_i)}{2} \times t](https://tex.z-dn.net/?f=d%20%3D%20%5Cfrac%7B%28v_f%20%2B%20v_i%29%7D%7B2%7D%20%5Ctimes%20t)
here we know that
![v_f = 10 m/s](https://tex.z-dn.net/?f=v_f%20%3D%2010%20m%2Fs)
![v_i = 0](https://tex.z-dn.net/?f=v_i%20%3D%200)
![t = 1.2 s](https://tex.z-dn.net/?f=t%20%3D%201.2%20s)
now we will have
![d = \frac{(10 + 0)}{2}\times 1.2](https://tex.z-dn.net/?f=d%20%3D%20%5Cfrac%7B%2810%20%2B%200%29%7D%7B2%7D%5Ctimes%201.2)
![d = 5 \times 1.2](https://tex.z-dn.net/?f=d%20%3D%205%20%5Ctimes%201.2)
![d = 6.0 m](https://tex.z-dn.net/?f=d%20%3D%206.0%20m)