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4 years ago

What is dispersion? A physics definition please.

1 answer:

6 0

In optics, dispersion is the phenomenon in which the phase velocity of a wave depends on its frequency. Media having this common property may be termed dispersive media. Sometimes the term chromatic dispersion is used for specificity.

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Which has the greater acceleration

Alik [6]

Answer: Object B

The velocity of object A is depicted in the graph as a straight line (constant speed therefore no acceleration).
The graph indicates that the velocity of object B increases (the object is accelerating).

5 0

3 years ago

Find p1, the gauge pressure at the bottom of tube 1. (Gauge pressure is the pressure in excess of outside atmospheric pressure.)

Taya2010 [7]

Answer:

(a). The gauge pressure at the bottom of tube 1 is $P_{1}=\rho g h_{1}$

(b). The speed of the fluid in the left end of the main pipe $\sqrt{\dfrac{2g(h_{2}-h_{1})}{(1-(\gamma)^2)}}$

Explanation:

Given that,

Gauge pressure at bottom = p₁

Suppose, an arrangement with a horizontal pipe carrying fluid of density p . The fluid rises to heights h1 and h2 in the two open-ended tubes (see figure). The cross-sectional area of the pipe is A1 at the position of tube 1, and A2 at the position of tube 2.

Find the speed of the fluid in the left end of the main pipe.

(a). We need to calculate the gauge pressure at the bottom of tube 1

Using bernoulli equation

$P_{1}=\rho g h_{1}$

(b). We need to calculate the speed of the fluid in the left end of the main pipe

Using bernoulli equation

Pressure for first pipe,

$P_{1}=\rho gh_{1}$ .....(I)

Pressure for second pipe,

$P_{2}=\rho gh_{2}$ .....(II)

From equation (I) and (II)

$P_{2}-P_{1}=\dfrac{1}{2}\rho(v_{1}^2-v_{2}^2)$

Put the value of P₁ and P₂

$\rho g h_{2}-\rho g h_{1}=\dfrac{1}{2}\rho(v_{1}^2-v_{2}^2)$

$gh_{2}-gh_{1}=\dfrac{1}{2}(v_{1}^2-v_{2}^2)$

$2g(h_{2}-h_{1})=v_{1}^2-v_{2}^2$ ....(III)

We know that,

The continuity equation

$v_{1}A_{1}=v_{2}A_{2}$

$v_{2}=v_{1}(\dfrac{A_{1}}{A_{2}})$

Put the value of v₂ in equation (III)

$2g(h_{2}-h_{1})=v_{1}^2-(v_{1}(\dfrac{A_{1}}{A_{2}}))^2$

$2g(h_{2}-h_{1})=v_{1}^2(1-(\dfrac{A_{1}}{A_{2}}))^2$

Here, $\dfrac{A_{1}}{A_{2}}=\gamma$

So, $2g(h_{2}-h_{1})=v_{1}^2(1-(\gamma)^2)$

$v_{1}=\sqrt{\dfrac{2g(h_{2}-h_{1})}{(1-(\gamma)^2)}}$

Hence, (a). The gauge pressure at the bottom of tube 1 is $P_{1}=\rho g h_{1}$

(b). The speed of the fluid in the left end of the main pipe $\sqrt{\dfrac{2g(h_{2}-h_{1})}{(1-(\gamma)^2)}}$

8 0

3 years ago

A batter hits a foul ball. The 0.140-kg baseball that was approaching him at 40.0 m/s leaves the bat at 30.0 m/s in a direction

Korvikt [17]

Answer:

The magnitude of the impulse delivered to the baseball is 7.0 Ns

Explanation:

Given;

mass of the foul ball, m = 0.14 kg

initial velocity, u = 40 m/s

final velocity, v = 30 m/s in perpendicular direction

Impulse is given as change in momentum;

initial momentum in horizontal direction, Pi = mu

Pi = 0.14 x 40 = 5.6 Ns

final momentum in perpendicular direction, Pf = mv

Pf = 0.14 x 30

Pf = 4.2 Ns

The resultant impulse is given by;

J² = 5.6² + 4.2²

J² = 49

J = √49

J = 7.0 Ns

Therefore, the magnitude of the impulse delivered to the baseball is 7.0 Ns

5 0

3 years ago

A person stands on a scale in an elevator. As the elevator starts, the scale has a constant reading of 592 N. As the elevator la

gladu [14]

Answer:

Explanation:

According to newton's secomd law, ∑F = ma

∑F is the summation of the force acting on the body

m is the mass of the body

a is the acceleration

Given the normal force when the elevator starts N1 = 592N

Normal force after the elevator stopped N2 = 400N

When the elevator starts, its moves upward, the sum of force ∑F = Normal (N)force on the elevator - weight of the person( Fg)

When moving up;

N1 - Fg = ma

N1 = ma + Fg ...(1)

Stopping motion of the elevator occurs after the elevator has accelerates down. The sum of forces in this case will give;

N2 - Fg = -ma

N2 = -ma+Fg ...(2)

Adding equation 1 and 2 we will have;

N1+N2 = 2Fg

592N + 400N = 2Fg

992N 2Fg

Fg = 992/2

Fg = 496N

The weight of the person is 496N

<em>\b) To get the person mass, we will use the relationship Fg = mg</em>

g = 9.81m/s

496 = 9.81m

mass m = 496/9.81

mass = 50.56kg

c) To get the magnitude of acceleration of the elevator, we will subtract equation 1 from 2 to have;

N1-N2 = 2ma

592-400 = 2(50.56)a

192 = 101.12a

a = 192/101.12

a = 1.90m/s²

3 0

4 years ago

Box A has a mass of 35.0kg and Box B has a mass of 20.0kg. What is the tension on

lions [1.4K]

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3 0

3 years ago