Answer:
Sometimes atoms gain or lose electrons. The atom then loses or gains a "negative" charge. These atoms are then called ions. Positive Ion - Occurs when an atom loses an electron negative charge it has more protons than electrons.
Explanation:
:)
Answer:
a) 10.51 J
b) 3.48 m/s
Explanation:
Given data :
mass of train ( M ) = 2.2 kg
Given initial velocity ( u ) = 1.6 m/s
<u>a) calculating work done by the force over the journey of the train</u>
F = mx + b ------ ( 1 )
m = slope = ( Δ f / Δ x ) = 2.8 / -7.5 = - 0.373 N/m
x = distance travelled on the x axis by the train = 7.5 m
F = force experienced by the train = 2.8 N
x = 0
∴ b = 2.8
hence equation 1 can be written as
F = ( -0.373) x + 2.8 ----- ( 2 )
hence to determine the work done by the force
W =
Note: the limits are actually 7.5 and 0
∴ W ( work done ) = -10.49 + 21 = 10.51 J
<u>b) calculate the speed of the train at the end of its journey</u>
we will apply the work energy theorem
W = 1/2 m*v^2 - 1/2 m*u^2
∴ V^2 = 2 / M ( W + 1/2 M*u^2 ) ( input values into equation )
V^2 = 12.11
hence V = 3.48 m/s
It can help by stoping a bike because the brakes are not always enogh
Answer:
a.) τ = 2.85 s b.) Q = 3.19 * 10^-5 C c.) t = 1.691 s
Explanation:
So we are told that it is a RC circuit. We are told
= 12.0 V, R = 1.07 MΩ and C = 2.66 µF.
a.) The time constant for RC circuit, τ = RC. Substituting our known values we get:
τ = RC where R = (1.07 * 10 ^ 6)Ω and C = (2.66 * 10 ^ -6) F
τ = (1.07 * 10 ^ 6)Ω * (2.66 * 10 ^ -6) F = 2.8462 s ≈ 2.85 s
τ = 2.85 s
b.) The relationship between capacitance, potential, charge is given:
![Q = CV[1-e^{-t/RC} ]](https://tex.z-dn.net/?f=Q%20%3D%20CV%5B1-e%5E%7B-t%2FRC%7D%20%5D)
The capacitor is fully charge when t approaches infinity, therefore:
![Q = \lim_{t \to \infty} a_n CV[1-e^{-t/RC} ]](https://tex.z-dn.net/?f=Q%20%3D%20%20%5Clim_%7Bt%20%5Cto%20%5Cinfty%7D%20a_n%20CV%5B1-e%5E%7B-t%2FRC%7D%20%5D)
When t approaches infinity, the term e becomes very small (e^-∞ = 0), therefore we can simplify the equation and plug in our values
![Q = (2.66*10^{-6}) F * (12.0)V *[1 - 0] = 3.192 * 10^{-5}](https://tex.z-dn.net/?f=Q%20%3D%20%282.66%2A10%5E%7B-6%7D%29%20F%20%2A%20%2812.0%29V%20%2A%5B1%20-%200%5D%20%3D%203.192%20%2A%2010%5E%7B-5%7D)
Q = 3.19 * 10^-5 C
c.) Using the same equation as before, we can substitute Q in and solve for Q:
![(14.3 * 10 ^ 6) C = (2.66*10^{-6})F *(12.0)V*[1-e^{-t/(2.85s)}]\\0.552 = e^{-t/(2.85s)}\\t = -1 * 2.85 * ln(0.552) \\t = 1.69120678 s](https://tex.z-dn.net/?f=%2814.3%20%2A%2010%20%5E%206%29%20C%20%3D%20%282.66%2A10%5E%7B-6%7D%29F%20%2A%2812.0%29V%2A%5B1-e%5E%7B-t%2F%282.85s%29%7D%5D%5C%5C0.552%20%3D%20e%5E%7B-t%2F%282.85s%29%7D%5C%5Ct%20%3D%20-1%20%2A%202.85%20%2A%20ln%280.552%29%20%5C%5Ct%20%3D%201.69120678%20s)
t = 1.691 s
Hope this helps! I'm not sure what the units you want, so convert to the desired units.