The smallest difference in voltage that can be resolved is referred to as the resolution. The resolution can be calculated with the following formula:
resolution=voltage range / digital range
The voltage range in our case is from -500mV to 500mV, which gives 1000mV.
The digital range on the other hand is 2^(number of bits).
It depends on what type of bit board we are using. If the ADC we are using is a 16 bit board, then 2^16=<span>65536.
So, the resolution is:
resolution=1000mV/</span><span>65536=0.015 mV</span>
Ok so I’m pretty sure the answer would be 2 because the mass of the rock would have the same mass on Earth as it has on the moon. Also the Density of a solid object remains constant meaning it doesn’t change. But the weight would change because the Earths gravitational pull is more than that of the moon. I hope this helped!
According to above question ~
Let's find the charge (q) by using formula ~
Hence, 12 coulombs of charge flow past any point in the wire in 3 seconds
There are two general types of collisions, inelastic and elastic.
Inelastic collisions occur when two objects collide but neither of them bounce away from each other.
Collisions in which the objects do not touch each other are elastic. (Ex: Rutherford Scattering)
Given Information:
Pendulum 1 mass = m₁ = 0.2 kg
Pendulum 2 mass = m₂ = 0.6 kg
Pendulum 1 length = L₁ = 5 m
Pendulum 2 length = L₂ = 1 m
Required Information:
Affect of mass on the frequency of the pendulum = ?
Answer:
The mass of the ball will not affect the frequency of the pendulum.
Explanation:
The relation between period and frequency of pendulum is given by
f = 1/T
The period of pendulum is given by
T = 2π√(L/g)
Where g is the acceleration due to gravity and L is the length of the string
As you can see the period (and frequency too) of pendulum is independent of the mass of the pendulum. Therefore, the mass of the ball will not affect the frequency of the pendulum.
Bonus:
Pendulum 1:
T₁ = 2π√(L₁/g)
T₁ = 2π√(5/9.8)
T₁ = 4.49 s
f₁ = 1/T₁
f₁ = 1/4.49
f₁ = 0.22 Hz
Pendulum 2:
T₂ = 2π√(L₂/g)
T₂ = 2π√(1/9.8)
T₂ = 2.0 s
f₂ = 1/T₂
f₂ = 1/2.0
f₂ = 0.5 Hz
So we can conclude that the higher length of the string increases the period of the pendulum and decreases the frequency of the pendulum.
