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zheka24 [161]
3 years ago
11

Phase difference formula

Physics
1 answer:
Anestetic [448]3 years ago
3 0

Answer:

Δx is the path difference between the two waves.

...

Phase Difference And Path Difference Equation.

Formula Unit

Phase Difference \Delta \phi=\frac{2\pi\Delta x}{\lambda } Radian or degree

Path Difference \Delta x=\frac{\lambda }{2\pi }\Delta \phi meter

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solniwko [45]
Blue is pulling 100n harder
7 0
3 years ago
if v=5.00 meters/second and makes an angle of 60 degrees with the negative direction of the y-axis, calculate the possible value
lora16 [44]

The answer is both 5sin(60)=4.33 and -5sin(60)=-4.33

5 0
3 years ago
Read 2 more answers
Which has greater kinetic energy, a car traveling at 30 km/HR or a car of half the mass traveling at 60 km/HR? a. The 60 km/HR c
artcher [175]

Answer:

a.The 60 km/HR car

Explanation:

Kinetic Energy: This can be defined as the energy of a body due to motion. The S.I unit of kinetic energy is Joules (J).

It can be expressed mathematically as

Ek = 1/2mv²......................... Equation 1

Where Ek = kinetic energy, m = mass, v = velocity.

(i) A car travelling at 30 km/hr, with a mass of m,

Ek = 1/2(m)(30)²

Ek = 450m J.

(ii) A car travelling at 60 km/hr, with a mass of m/2

Ek = 1/2(m/2)(60)²

Ek = 900m J.

Thus , the car travelling at 60 km/hr at half mass has a greater kinetic energy to the car traveling at 30 km/hr at full mass.

The right option is a.The 60 km/HR car

6 0
4 years ago
1. What is work done in holding a 15kg suitcase while waiting for a bus for 15 minutes?
Anettt [7]
The man is holding the suitcase at the same height above the surface of earth. So the gravitation potential energy remains the same. 

<span>work done is force * displacement = weight * 0 = 0</span>
7 0
3 years ago
From Gauss's law, the electric field set up by a uniform line of charge is given by the following expression where is a unit vec
Evgesh-ka [11]

Answer:

\Delta V=\lambda *ln(r_{2}/r_{1}) /\ (2\pi*\epsilon_{o})

Explanation:

Using the Gauss Law, we obtain the electric Field for a uniform large line of charge:

2\pi r L*E=\lambda *L/\epsilon_{o}

E=\lambda /\(2 \pi* r *\epsilon_{o})

We calculate the potential difference from the electric field:

\Delta V=-\int\limits^{r_{1}}_{r_{2}} E \, dr =-\int\limits^{r_{1}}_{r_{2}} \lambda dr/ (2\pi*r*\epsilon_{o})=\lambda *ln(r_{2}/r_{1}) /\ (2\pi*\epsilon_{o})

5 0
4 years ago
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