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3 years ago

What can you say about the speed of light in diamond?

1 answer:

7 0

So n=c/v, n= index, c=speed of light and v= speed of light in diamond. 2.42=c/v so v=c/2.42, c≈3x108 m/sec so v=1.24x108 m/sec.
Hope this helps.

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Speed is the rate of change in

vaieri [72.5K]

Speed is the rate of change of distance.

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3 years ago

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An amusement park ride consists of airplane-shaped cars attached to steel rods. Each rod has a length of 15m and a cross-section

Oliga [24]

Explanation:

Given that,

Length of each rod, L = 15 m

Area of cross section of the rod, $A=8\ cm^2=0.0008\ m^2$

Young's modulus for steel, $Y=2\times 10^{11}\ Pa$

Total weight of two people, F = 1900 N

We need to find the stretching in the rod when the ride is at rest. The formula of Young's modulus of a material is given by :

$Y=\dfrac{FL}{A\Delta L}$

Here, $\Delta L$ is the stretching in the rod

$\Delta L=\dfrac{FL}{YA}\\\\\Delta L=\dfrac{1900\times 15}{2\times 10^{11}\times 0.0008}\\\\\Delta L=1.78\times 10^{-4}\ m$

So, the rod will stretched to $1.78\times 10^{-4}\ m$

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3 years ago

If you stood on a planet having a mass four times that of earth mass and a radius two times of earth radius , you would weigh?

trasher [3.6K]

Answer: Weight will remain same as on the Earth

Explanation :

From Universal law of gravitation :

$F=G\dfrac{M_em}{R^2}$ ......(1)

$M_e$ is the mass of earth

m is the mass of the person

R is the radius of the earth

We know that F= mg.....(2)

So, from equation (1) and (2)

$mg=\dfrac{GM_em}{R^2}$

$g=\dfrac{GM_e}{R^2}$ .....(3)

It is given that, If we stood on a planet having a mass four times that of earth mass and a radius two times of earth radius then from equation (3)

$g=\dfrac{G\times 4M_e}{4R^2}$

The value of g is will not change. So, our weight would not change. It would remain the same as on the Earth.

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3 years ago

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Trata de que bobo 1 2 3

kkurt [141]

Answer:

what?

Explanation:

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3 years ago

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Suppose an automobile engine can produce 215 N⋅m of torque, and assume this car is suspended so that the wheels can turn freely.

natulia [17]

Answer:

α = 68.8 rad/s²

Explanation:

Given: m = 13.5 kg r= 0.155 m radius.

Inertia is given by I = ½mr²

I = ½*13.5kg*(0.155m)²

I = 0.162 kg*m²

Given: m = 2.05kg ring inside radius ri= 0.18 m; outside radius ro= 0.32 m

I = ½m*(r₁² + r₂²)

I = ½*2.05kg*(0.18² + 0.32²)

I = 0.138 kg*m²

Tread: m = 9.5 kg r= 0.31 m

I = mr² (hoop/ring or cylindrical shell formula)

I = 9.5kg*0.31m²

I = 0.9129kg*m²

Drive shaft: m= 31 kg solid cylinder r = 3.1 cm = 0.031m.

I = ½mr²

I = 31kg*0.031m²

I = 0.0298 kg*m²

Total moment of intertia

I = 0.162 + 0.1.38 + 0.9129 + 0.0298

It = 1.2427 kg*m²

τ = α*2*I ,τ = 95%*180N= 171N

α = τ/2*I

α = 171N/2*1.2427

α = 68.8 rad/s²

7 0

3 years ago