Answer:
Proof in explanataion
Explanation:
The basic dimensions are as follows:
MASS = M
LENGTH = L
TIME = T
i)
Given equation is:
where,
H = height (meters)
u = speed (m/s)
g = acceleration due to gravity (m/s²)
Sin Ф = constant (no unit)
So there dimensions will be:
H = [L]
u = [LT⁻¹]
g = [LT⁻²]
Sin Ф = no dimension
Therefore,
![[L] = \frac{[LT^{-1}]^2}{[LT^{-2}]}\\\\\ [L] = [L^{(2-1)}T^{(-2+2)}]](https://tex.z-dn.net/?f=%5BL%5D%20%3D%20%5Cfrac%7B%5BLT%5E%7B-1%7D%5D%5E2%7D%7B%5BLT%5E%7B-2%7D%5D%7D%5C%5C%5C%5C%5C%20%5BL%5D%20%3D%20%5BL%5E%7B%282-1%29%7DT%5E%7B%28-2%2B2%29%7D%5D)
<u>[L] = [L]</u>
Hence, the equation is proven to be homogenous.
ii)
where,
F = Force = Newton = kg.m/s² = [MLT⁻²]
G = Gravitational Constant = N.m²/kg² = (kg.m/s²)m²/kg² = m³/kg.s²
G = [M⁻¹L³T⁻²]
m₁ = m₂ = mass = kg = [M]
r = distance = m = [L]
Therefore,
![[MLT^{-2}] = \frac{[M^{-1}L^{3}T^{-2}][M][M]}{[L]^2}\\\\\ [MLT^{-2}] = [M^{(-1+1+1)}L^{(3-2)}T^{-2}]\\\\](https://tex.z-dn.net/?f=%5BMLT%5E%7B-2%7D%5D%20%3D%20%5Cfrac%7B%5BM%5E%7B-1%7DL%5E%7B3%7DT%5E%7B-2%7D%5D%5BM%5D%5BM%5D%7D%7B%5BL%5D%5E2%7D%5C%5C%5C%5C%5C%20%5BMLT%5E%7B-2%7D%5D%20%3D%20%5BM%5E%7B%28-1%2B1%2B1%29%7DL%5E%7B%283-2%29%7DT%5E%7B-2%7D%5D%5C%5C%5C%5C)
<u>[MLT⁻²] = [MLT⁻²]</u>
Hence, the equation is proven to be homogenous.
Answer:
5.77 Volt
Explanation:
N = 1000, Diameter = 42 cm = 0.42 m, t = 12 ms = 12 x 10^-3 s
Change in magnetic field, B = 5 x 10^-5 T
The peak value of emf is given by
e = N x dФ / dt
e = N x A x dB/dt
e = (1000 x 3.14 x 0.21 x 0.21 x 5 x 10^-5) / (1.2 x 10^-3)
e = 5.77 Volt
Answer:
Explanation:
pV = nrT
n = PV/RT
n = (100*10^3)(.01)/(300*0.082057)
n = 40.62 moles
a. 0.5 T
- The amplitude A of a simple harmonic motion is the maximum displacement of the system with respect to the equilibrium position
- The period T is the time the system takes to complete one oscillation
During a full time period T, the mass on the spring oscillates back and forth, returning to its original position. This means that the total distance covered by the mass during a period T is 4 times the amplitude (4A), because the amplitude is just half the distance between the maximum and the minimum position, and during a time period the mass goes from the maximum to the minimum, and then back to the maximum.
So, the time t that the mass takes to move through a distance of 2 A can be found by using the proportion
and solving for t we find
b. 1.25T
Now we want to know the time t that the mass takes to move through a total distance of 5 A. SInce we know that
- the mass takes a time of 1 T to cover a distance of 4A
we can set the following proportion:
And by solving for t, we find
