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bija089 [108]
3 years ago
6

Since sinusoidal waves are cyclical, a particular phase difference between two waves is identical to that phase difference plus

a cycle. For example, if two waves have a phase difference of π4, the interference effects would be the same as if the two waves had a phase difference of π4+2π. The complete criterion for constructive interference between two waves is therefore written as follows: phase difference=0+2πnfor any integer n Write the full criterion for destructive interference between two waves
Physics
1 answer:
r-ruslan [8.4K]3 years ago
4 0

Answer:

nπ + π/2 for any integer n

Explanation:

Since destructive interference occurs every odd multiple of half wavelength, that is π/2, 3π/2, 5π/2 where the interference is half wavelength and in general, (n + 1/2)π where n is an integer. So, nπ + π/2 for any integer n

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Consider a pool shaped like the bottom half of a sphere, that is being filled. The radius of the pool is 10 ft. If the height is
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Explanation:

Below is an attachment containing the solution.

4 0
3 years ago
A resistor and a capacitor are connected in series across an ideal battery having a constant voltage across its terminals. Long
tino4ka555 [31]

Answer:

A) equal to the battery's terminal voltage.

Explanation:

When the capacitor is fully charged after long hours of charging , its  potential becomes equal to the emf of the battery and its polarity is opposite to that of battery . Hence net emf becomes equal . The capacitor itself becomes a battery which is connected in the circuit with opposite polarity . This results in the net emf and  current becoming zero . There is no charging current when the capacitor is fully charged .

4 0
3 years ago
A 45 kg wagon is being pulled with a rope that makes an angle of 38 degrees with the horizontal. The applied force is 410 N and
mel-nik [20]

Answer:

7.35m/s²

Explanation:

From the question we are not told what to find but we can as well find the acceleration of the wagon.

According to newton second law of motion

\sum F_x = ma_x\\Fm - Ff = ma_x\\Fm - \mu R = ma_x\\Fm- \mu mg = ma_x\\

Fm is the moving force = 410N

\mu is the coefficient of friction = 0.18

m is the mass = 45kg

g is the acceleration dur to gravity = 9.8m/s²

a is the acceleration of the wagon

Substitute the given data into the equation ang get ax

Fm- \mu mg = ma_x\\410 - (0.18)(9.8)(45) = 45a_x\\410 - 79.38 = 45a_x\\330.62 = 45a_x\\a_x = 330.62/45\\a_x = 7.35m/s^2\\

Hence the acceleration of the wagon is 7.35m/s²

8 0
3 years ago
Suppose that the process were repeated, except that in step 3 a neutral acrylic rod instead of a finger is used to touch the ele
Marta_Voda [28]

Answer:

b) True Only if the finger is isolated from ground

c) True. The total charge does not change since the system is isolated

Explanation:

When the electroscope is touched with an acrylic rod, some charges are transferred from the electroscope to the rod, until the charge in both is equal.

In the case it know when the electroscope is touched with a finger, two things can happen.

- The body is isolated from the ground, the efective charge is redistributed between the two bodies. Case similar to insulating rod

- The body is connected to ground, the charge is transferred to the finger and from here to the ground until the total charge is transferred and the Earth and the final charge of the electroscope is zero.

Let's review the final statements

a) False, when part of the load is touched, it passes to the rod, so when it separates it does not return to the initial load

b) True Only if the finger is isolated from ground

c) True. The total load does not change since the system is isolated

d) False. The value of the load changes =, but its sign does not

8 0
3 years ago
A 1.4-kg cart is attached to a horizontal spring for which the spring constant is 50 N/m . The system is set in motion when the
marin [14]

Answer:

A)

0.395 m

B)

2.4 m/s

Explanation:

A)

m = mass of the cart = 1.4 kg

k = spring constant of the spring = 50 Nm⁻¹

x = initial position of spring from equilibrium position = 0.21 m

v_{i} = initial speed of the cart = 2.0 ms⁻¹

A = amplitude of the oscillation = ?

Using conservation of energy

Final spring energy = initial kinetic energy + initial spring energy

(0.5) kA^{2} = (0.5) m v_{i}^{2} + (0.5) k x_{i}^{2} \\kA^{2} = m v_{i}^{2} + k x_{i}^{2} \\(50) A^{2} = (1.4) (2.0)^{2} + (50) (0.21)^{2} \\A = 0.395 m

B)

m = mass of the cart = 1.4 kg

k = spring constant of the spring = 50 Nm⁻¹

A = amplitude of the oscillation = 0.395 m

v_{o} = maximum speed at the equilibrium position

Using conservation of energy

Kinetic energy at equilibrium position = maximum spring potential energy at extreme stretch of the spring

(0.5) m v_{o}^{2} = (0.5) kA^{2}\\m v_{o}^{2} = kA^{2}\\(1.4) v_{o}^{2} = (50) (0.395)^{2}\\v_{o} = 2.4 ms^{-1}

5 0
3 years ago
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