Answer:
The displacement in t = 0,
y (0) = - 0.18 m
Explanation:
Given f = 40 Hz , A = 0.25m , μ = 0.02 kg / m, T = 20.48 N
v = √ T / μ
v = √20.48 N / 0.02 kg /m = 32 m/s
λ = v / f
λ = 32 m/s / 40 Hz = 0.8
K = 2 π / λ
K = 2π / 0.8 = 7.854
φ = X * 360 / λ
φ = 0.5 * 360 / 0.8 = 225 °
Using the model of y' displacement
y (t) = A* sin ( w * t - φ )
When t = 0
y (0) = 0.25 m *sin ( w*(0) - 225 )
y (0) = 0.25 * -0.707
y (0) = - 0.18 m
Answer:
Explanation:
Case 1:
mass = m
initial velocity = vo
final velocity = 0
height = y
Use third equation of motion
v² = u² - 2as
0 = vo² - 2 g y
y = vo² / 2g ... (1)
Case 2:
mass = 2m
initial velocity = 2vo
final velocity = 0
height = y '
Use third equation of motion
v² = u² - 2as
0 = 4vo² - 2 g y'
y ' = 4vo² / 2g
y' = 4 y
Thus, the second rock reaches the 4 times the distance traveled by the first rock.
Answer:
when the direction of the Current changes.
Explanation:
Electromagnet refers to an iron ore wrapped around with a coil of wire, in presence of electric current. As it acts like a magnet, when current is passed through it.
The north & south poles of magnetic fields produced by such magnet, change with direction of current passed through it.
Answer:
Work, in physics, measure of energy transfer that occurs when an object is moved over a distance by an external force
Explanation:
at least part of which is applied in the direction of the displacement. ... To express this concept mathematically, the work W is equal to the force f times the distance d, or W = fd.
Answer:
The new potential energy decreases by the factor of 2 to the old potential energy.
Explanation:
Capacitance of a parallel plate capacitor is given by the relation :
C = (ε₀A)/d
Here ε₀ is vacuum permittivity, A is area of the capacitor plate and d is the distance between them.
Potential energy of the capacitor, U = ![\frac{1}{2}CV^{2}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7DCV%5E%7B2%7D)
Here V is the potential difference between the plates.
According to the problem, the distance between the plates get double but area remains same. So,
d₁ = 2d
Here d₁ is new distance between the plates.
Hence, new capacitance is :
C₁ = (ε₀A)/d₁ = (ε₀A)/2d = C/2
The capacitor have same potential difference that is V. Hence, the new potential energy is :
U₁ =
= ![\frac{1}{2}\frac{C}{2} V^{2}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7BC%7D%7B2%7D%20V%5E%7B2%7D)
U₁ = U/2
![\frac{U_{1} }{U} = \frac{1}{2}](https://tex.z-dn.net/?f=%5Cfrac%7BU_%7B1%7D%20%7D%7BU%7D%20%3D%20%5Cfrac%7B1%7D%7B2%7D)