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3 years ago

14

What is the dimensional formula of force and torque

2 answers:

andreyandreev [35.5K]3 years ago

6 0

Answer:

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LuckyWell [14K]3 years ago

5 0

Answer:

Units. Torque has the dimension of force times distance, symbolically T−2L2M. Although those fundamental dimensions are the same as that for energy or work, official SI literature suggests using the unit newton metre (N⋅m) and never the joule. The unit newton metre is properly denoted N⋅m.

Dimension: M L2T−2

In SI base units: kg⋅m2⋅s−2

Other units: pound-force-feet, lbf⋅inch, ozf⋅in

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3 years ago

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a) What is the average useful power output of a person who does6.00×10^6Jof useful work in 8.00 h? (b) Working at this rate, how

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Answer:

208.33 W

141.26626 seconds

Explanation:

E = Energy = $6\times 10^6\ J$

t = Time taken = 8 h

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g = Acceleration due to gravity = 9.81 m/s²

h = Height of platform = 1.5 m

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4 years ago

A ray of light incident in water strikes the surface separating water from air making an angle of 10 ° with the normal to the su

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Answer:

a

$\theta _2 = 13^o$

b

$\theta _1 =32.94^o$

c

$\theta_c = 53.05^o$

Explanation:

From the question we are told that

The angle of incidence is $\theta_1 = 10^o$

The refractive index of water is $n_1 = 1.3$

Generally Snell's law is mathematically represented as

$n_1 sin(\theta_1) = n_2 sin(\theta_ 2)$

Here $n_2$ is the refractive index of air with value $n_2 = 1$

$\theta_2$ is the angle of refraction

So

$\theta _2 = sin^{-1}[\frac{n_1 * sin(\theta _1)}{n_2} ]$

=> $\theta _2 = sin^{-1}[\frac{1.3 * sin(10)}{1} ]$

=> $\theta _2 = 13^o$

Given that the angle should not be greater than $\theta _2 =45^o$ then the angle of incidence will be

$\theta _1 = sin^{-1}[\frac{n_2 * sin(\theta _2)}{n_1} ]$

=> $\theta _1 = sin^{-1}[\frac{1 * sin(45)}{1.3} ]$

=> $\theta _1 =32.94^o$

Generally for critical angle is mathematically represented as

$\theta_c = sin^{-1}[\frac{n_2}{n_1} ]$

=> $\theta_c = sin^{-1}[\frac{1}{1.3} ]$

=> $\theta_c = 53.05^o$

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