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3 years ago

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Light enters from air into diamond which has refractive index of 2.42 calculate the speed of light in diamond the speed of light

in air is 3*10^8 m/s

1 answer:

ELEN [110]3 years ago

8 0

The speed of light in diamond is $1.24\cdot 10^8 m/s$

Explanation:

The index of refraction of a material is equal to the ratio between the speed of light in vacuum and the speed of light in the material:

$n=\frac{c}{v}$

where

n is the index of refraction

c is the speed of light in vacuum

v is the speed of light in the material

In this problem, we have:

n = 2.42 is the refractive index of diamond

$c=3.0\cdot 10^8 m/s$ is the speed of light in vacuum

Solving for v, we find the speed of light in diamond:

$v=\frac{c}{n}=\frac{3\cdot 10^8}{2.42}=1.24\cdot 10^8 m/s$

Learn more about refraction:

brainly.com/question/3183125

brainly.com/question/12370040

#LearnwithBrainly

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10. Someone takes 11 minutes to walk up a hill 120m high. His weight is 550N.

belka [17]

Answer:

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2 years ago

Un puente de acero de 100 m de largo a 8° C aumenta su temperatura a 24°C ¿Cuánto medirá su longitud? Valor del coeficiente de d

BartSMP [9]

La longitud <em>final</em> del puente de acero es 100.018 metros.

Asumamos que la dilatación <em>térmica</em> experimentada por el puente de acero es <em>pequeña</em>, de modo que podemos emplear la siguiente aproximación <em>lineal</em> para determinar la longitud <em>final</em> del puente de acero ( $L$ ), en metros:

$L = L_{o}\cdot [1+\alpha\cdot (T_{f}-T_{o})]$ (1)

Donde:

$L_{o}$ - Longitud inicial del puente, en metros.
$\alpha$ - Coeficiente de dilatación, sin unidad.
$T_{o}$ - Temperatura inicial, en grados Celsius.
$T_{f}$ - Temperatura final, en grados Celsius.

Si tenemos que $L_{o} = 100\,m$ , $\alpha = 11.5\times 10^{-6}$ , $T_{o} = 8\,^{\circ}C$ y $T_{f} = 24\,^{\circ}C$ , entonces la longitud final del puente de acero es:

$L = (100\,m)\cdot [1+(11.5\times 10^{-6})\cdot (24\,^{\circ}C - 8\,^{\circ}C)]$

$L = 100.018\,m$

La longitud <em>final</em> del puente de acero es 100.018 metros.

Para aprender más sobre dilatación térmica, invitamos cordialmente a ver esta pregunta verificada: brainly.com/question/24953416

5 0

2 years ago

A puck of mass 0.110 kg slides across ice in the positive x-direction with a kinetic friction coefficient between the ice and pu

lara [203]

Answer:

a) Ffr = -0.18 N

b) a= -1.64 m/s2

c) t = 9.2 s

d) x = 68.7 m.

e) W= -12.4 J

f) Pavg = -1.35 W

g) Pinst = -0.72 W

Explanation:

a)

While the puck slides across ice, the only force acting in the horizontal direction, is the force of kinetic friction.
This force is the horizontal component of the contact force, and opposes to the relative movement between the puck and the ice surface, causing it to slow down until it finally comes to a complete stop.
So, this force can be written as follows, indicating with the (-) that opposes to the movement of the object.

$F_{frk} = -\mu_{k} * F_{n} (1)$

where μk is the kinetic friction coefficient, and Fn is the normal force.

Since the puck is not accelerated in the vertical direction, and there are only two forces acting on it vertically (the normal force Fn, upward, and the weight Fg, downward), we conclude that both must be equal and opposite each other:

$F_{n} = F_{g} = m*g (2)$

We can replace (2) in (1), and substituting μk by its value, to find the value of the kinetic friction force, as follows:

$F_{frk} = -\mu_{k} * F_{n} = -0.167*9.8m/s2*0.11kg = -0.18 N (3)$

b)

According Newton's 2nd Law, the net force acting on the object is equal to its mass times the acceleration.
In this case, this net force is the friction force which we have already found in a).
Since mass is an scalar, the acceleration must have the same direction as the force, i.e., points to the left.
We can write the expression for a as follows:

$a= \frac{F_{frk}}{m} = \frac{-0.18N}{0.11kg} = -1.64 m/s2 (4)$

c)

Applying the definition of acceleration, choosing t₀ =0, and that the puck comes to rest, so vf=0, we can write the following equation:

$a = \frac{-v_{o} }{t} (5)$

Replacing by the values of v₀ = 15 m/s, and a = -1.64 m/s2, we can solve for t, as follows:

$t =\frac{-15m/s}{-1.64m/s2} = 9.2 s (6)$

d)

From (1), (2), and (3) we can conclude that the friction force is constant, which it means that the acceleration is constant too.
So, we can use the following kinematic equation in order to find the displacement before coming to rest:

$v_{f} ^{2} - v_{o} ^{2} = 2*a*\Delta x (7)$

Since the puck comes to a stop, vf =0.
Replacing in (7) the values of v₀ = 15 m/s, and a= -1.64 m/s2, we can solve for the displacement Δx, as follows:

$\Delta x = \frac{-v_{o}^{2}}{2*a} =\frac{-(15.0m/s)^{2}}{2*(-1.64m/s2} = 68.7 m (8)$

e)

The total work done by the friction force on the object , can be obtained in several ways.
One of them is just applying the work-energy theorem, that says that the net work done on the object is equal to the change in the kinetic energy of the same object.
Since the final kinetic energy is zero (the object stops), the total work done by friction (which is the only force that does work, because the weight and the normal force are perpendicular to the displacement) can be written as follows:

$W_{frk} = \Delta K = K_{f} -K_{o} = 0 -\frac{1}{2}*m*v_{o}^{2} =-0.5*0.11*(15.0m/s)^{2} = -12.4 J (9)$

f)

By definition, the average power is the rate of change of the energy delivered to an object (in J) with respect to time.
$P_{Avg} = \frac{\Delta E}{\Delta t} (10)$
If we choose t₀=0, replacing (9) as ΔE, and (6) as Δt, and we can write the following equation:

$P_{Avg} = \frac{\Delta E}{\Delta t} = \frac{-12.4J}{9.2s} = -1.35 W (11)$

g)

The instantaneous power can be deducted from (10) as W= F*Δx, so we can write P= F*(Δx/Δt) = F*v (dot product)
Since F is constant, the instantaneous power when v=4.0 m/s, can be written as follows:

$P_{inst} =- 0.18 N * 4.0m/s = -0.72 W (12)$

7 0

3 years ago

The center-seeking change in velocity of an object moving in a circle is:

NeTakaya

The center-seeking change in velocity of an object moving in a circle is the centripetal acceleration.

So, by Newton's laws, we know that an object moving with a given velocity will remain in constant motion with a constant velocity until we apply an acceleration.

So we define acceleration as the rate of change of the velocity, also remember that velocity is a vector (has magnitude and direction), so, if there is a change the direction of the velocity, we have an acceleration that causes that.

In circular motion, the velocity vector is always perpendicular to the radius of the circle, and it can only be possible if the velocity direction is changing constantly. This will happen because of something called centripetal acceleration.

This acceleration points radially inwards (to the center of the circle) so is also perpendicular to the velocity of the moving object, and this is what causes the constant change in the direction of the velocity of the moving object.

Just to give an example, if you have a string with a mass on one end, and with your hand, you rotate the mass (from the string), the tension of the string would be the centripetal acceleration.

If you want to learn more about circular motion, you can read:

brainly.com/question/2285236

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2 years ago

Read 2 more answers

Bill and Bob are carrying loads of trash to the dump. They are walking down the sidewalk, side by side but Bob begins to lag beh

anzhelika [568]

Answer:

I think it is C) Newton's 2nd Law. Bob is pulling the heavier load. He needs a greater force to move as fast as Bill.

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3 years ago