Answer:
n = 756.25 giga electrons
Explanation:
It is given that,
If the charge on the negative plate of the capacitor, ![Q=121\ nC=121\times 10^{-9}\ C](https://tex.z-dn.net/?f=Q%3D121%5C%20nC%3D121%5Ctimes%2010%5E%7B-9%7D%5C%20C)
Let n is the number of excess electrons are on that plate. Using the quantization of charges, the total charge on the negative plate is given by :
![Q=ne](https://tex.z-dn.net/?f=Q%3Dne)
e is the charge on electron
![n=7.5625\times 10^{11}](https://tex.z-dn.net/?f=n%3D7.5625%5Ctimes%2010%5E%7B11%7D)
or
n = 756.25 giga electrons
So, there are 756.25 giga electrons are on the plate. Hence, this is the required solution.
In order to find the force (F), you would have to use the formula for it:
F=ma
where m is mass and a is acceleration.
In the problem, the mass is 2.85kg and the acceleration is 4.9m/s^2.
Therefore,
F=2.85kg(4.9m/s^2)
F=13.965kg(m/s^2)
Since N=kg(m/s^2)
F=13.965N
And because the problem requires that we use only 2 significant figures,
F=13N
Therefore, the student must exert 13N of force.
Answer:
Explanation:
Given that,
Mass of block
M = 2kg
Spring constant k = 300N/m
Velocity v = 12m/s
At t = 0, the spring is neither stretched nor compressed. Then, it amplitude is zero at t=0
xo = 0
It velocity is 12m/s at t=0
Then, it initial velocity is
Vo = 12m/s
Then, amplitude is given as
A = √[xo + (Vo²/ω²)]
Where
xo is the initial amplitude =0
Vo is the initial velocity =12m/s
ω is the angular frequency and it can be determine using
ω = √(k/m)
Where
k is spring constant = 300N/m
m is the mass of object = 2kg
Then,
ω = √300/2 = √150
ω = 12.25 rad/s²
Then,
A = √[xo + (Vo²/ω²)]
A = √[0 + (12²/12.5²)]
A = √[0 + 0.96]
A = √0.96
A = 0.98m
Generally, the internal resistance of the new battery is small, about 0.2 euros, while the old battery is large, close to 1 euro ,
Well if it was traveling for an hour then the answer is 8 miles.