Answer:
(a) The spring constant is 59.23 N/m
(b) The total energy involved in the motion is 0.06 J
Explanation:
Given;
mass, m = 240 g = 0.24 kg
frequency, f = 2.5 Hz
amplitude of the oscillation, A = 4.5 cm = 0.045 m
The angular speed is calculated as;
ω = 2πf
ω = 2 x π x 2.5
ω = 15.71 rad/s
(a) The spring constant is calculated as;
![\omega = \sqrt{\frac{k}{m} } \\\\\omega ^2 = \frac{k}{m} \\\\k = m\omega ^2\\\\where;\\\\k \ is \ the \ spring \ constant\\\\k = (0.24) \times (15.71)^2\\\\k = 59.23 \ N/m](https://tex.z-dn.net/?f=%5Comega%20%3D%20%5Csqrt%7B%5Cfrac%7Bk%7D%7Bm%7D%20%7D%20%5C%5C%5C%5C%5Comega%20%5E2%20%3D%20%5Cfrac%7Bk%7D%7Bm%7D%20%5C%5C%5C%5Ck%20%3D%20m%5Comega%20%5E2%5C%5C%5C%5Cwhere%3B%5C%5C%5C%5Ck%20%5C%20is%20%5C%20the%20%5C%20spring%20%5C%20constant%5C%5C%5C%5Ck%20%3D%20%280.24%29%20%5Ctimes%20%2815.71%29%5E2%5C%5C%5C%5Ck%20%3D%2059.23%20%5C%20N%2Fm)
(b) The total energy involved in the motion;
E = ¹/₂kA²
E = (0.5) x (59.23) x (0.045)²
E = 0.06 J
A convergent boundary is shown.
As per Newton's II law we know that
![F = ma](https://tex.z-dn.net/?f=%20F%20%3D%20ma)
here
F = applied unbalanced force
m = mass of object
a = acceleration of object
now it is given that force F = 50 N North applied on block on earth due to which block will accelerate by 4 m/s^2
so here from above equation
![50 = m* 4](https://tex.z-dn.net/?f=50%20%3D%20m%2A%204)
![m = \frac{50}{4} = 12.5 kg](https://tex.z-dn.net/?f=m%20%3D%20%5Cfrac%7B50%7D%7B4%7D%20%3D%2012.5%20kg)
Now we took another situation where block is placed on surface of moon and again force F = 25 N is applied on the block
So we will again use Newton's II law
![F = ma](https://tex.z-dn.net/?f=F%20%3D%20ma)
![25 = 12.5 * a](https://tex.z-dn.net/?f=25%20%3D%2012.5%20%2A%20a)
![a = \frac{25}{12.5}](https://tex.z-dn.net/?f=%20a%20%3D%20%5Cfrac%7B25%7D%7B12.5%7D)
![a = 2 m/s^2](https://tex.z-dn.net/?f=a%20%3D%202%20m%2Fs%5E2)
so block will accelerate on moon by acceleration 2 m/s^2
Answer: C = Q/4πR
Explanation:
Volume(V) of a sphere = 4πr^3
Charge within a small volume 'dV' is given by:
dq = ρ(r)dV
ρ(r) = C/r^2
Volume(V) of a sphere = 4/3(πr^3)
dV/dr = (4/3)×3πr^2
dV = 4πr^2dr
Therefore,
dq = ρ(r)dV ; dq =ρ(r)4πr^2dr
dq = C/r^2[4πr^2dr]
dq = 4Cπdr
FOR TOTAL CHANGE 'Q', we integrate dq
∫dq = ∫4Cπdr at r = R and r = 0
∫4Cπdr = 4Cπr
Q = 4Cπ(R - 0)
Q = 4CπR - 0
Q = 4CπR
C = Q/4πR
The value of C in terms of Q and R is [Q/4πR]
The further the planet is from the sun the smaller the year is