Answer:
Spring constant, k = 24.1 N/m
Explanation:
Given that,
Weight of the object, W = 2.45 N
Time period of oscillation of simple harmonic motion, T = 0.64 s
To find,
Spring constant of the spring.
Solution,
In case of simple harmonic motion, the time period of oscillation is given by :

m is the mass of object


m = 0.25 kg


k = 24.09 N/m
or
k = 24.11 N/m
So, the spring constant of the spring is 24.1 N/m.
Is halved. A 6Ω resistor connected to a voltage source which voltage is decreased from 12V to 6V the current passing through the resistor is halved.
The key to solve this problem is applying Ohm's Law V = R I, clearing I from the equation, we obtain I = V/R. Then, the current is directly proportional to the voltage and inversely proportional to the resistance.
V = 12V and R = 6Ω
I = 12V/6Ω = 2A
V = 6V and R = 6Ω
V = 6V/6Ω = 1A
As we can see the current is halved if the voltage descreased from 12V to 6V
The solution for this problem would be:(10 - 500x) / (5 - x)
so start by doing:
x(5*50*2) - xV + 5V = 0.02(5*50*2)
500x - xV + 5V = 10
V(5 - x) = 10 - 500x
V = (10 - 500x) / (5 - x)
(V stands for the volume, but leaves us with the expression for x)
Answer:
Plot ln K vs 1/T
(a) -0.5004; (b) 0.002 539 K⁻¹; (c) -197.1 K⁻¹; (d) 1.64 kJ/mol
Explanation:
This is an example of the Arrhenius equation:

Thus, if we plot ln k vs 1/T, we should get a straight line with slope = -Eₐ/R and a y-intercept = lnA
Data:

Calculations:
(a) Rise
Δy = y₂ - y₁ = -0.9545 - (-0.4541) = -0.9545 + 0.4541 = -0.5004
(b) Run
Δx = x₂ - x₁ = 0.004 444 - 0.001 905 = 0.002 539 K⁻¹
(c) Slope
Δy/Δx = -0.5004/0.002 539 K⁻¹ = -197.1 K⁻¹
(d) Activation energy
Slope = -Eₐ/R
Eₐ = -R × slope = -8.314 J·K⁻¹mol⁻¹ × (-197.1 K⁻¹) = 1638 J/mol = 1.64 kJ/mol
Answer:
5.22 x 10^5 V
Explanation:
guessed on castle learning and got it right