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icang [17]
3 years ago
12

A roundabout in a fairground requires an input power of 2.5 kW when operating at a constant angular velocity of 0.47 rad s–1 . (

a) Show that the frictional torque in the system is about 5 kN m. (3) 1. (b) When the power is switched off, the roundabout decelerates uniformly because the
Physics
1 answer:
natita [175]3 years ago
5 0

Answer:

Explanation:

Since the roundabout is rotating with uniform velocity ,

input power = frictional power

frictional power = 2.5 kW

frictional torque x angular velocity = 2.5 kW

frictional torque x .47 = 2.5 kW

frictional torque = 2.5 / .47 kN .m

= 5.32 kN . m

= 5 kN.m

b )

When power is switched off , it will decelerate because of frictional torque .

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Answer:

The  the intensity at an 11° angle to the axis in terms of the intensity of the central maximum is  

   I_c = \frac{I}{I_o}  =8.48 *10^{-8}

Explanation:

From the question we are told that

   The  width of the slit is  D  =  0.3 \ mm =  0.3 *10^{-3} \ m

    The  wavelength is  \lambda =  254 \ nm =  254 *10^{-9} \ m

     The angle is  \theta  =  11^o

The intensity of at 11^o to the axis in terms of the intensity of the central maximum. is mathematically represented as

        I_c = \frac{I}{I_o}  = [ \frac{sin \beta  }{\beta }] ^2

Where \beta is mathematically represented as

        \beta  =  \frac{D sin (\theta ) *  \pi}{\lambda }

substituting values

      \beta  =  \frac{0.3 *10^{-3} sin (11 ) * 3.142}{254 *10^{-9} }

     \beta  =  708.1 \ rad

So

  I_c = \frac{I}{I_o}  = [ \frac{sin (708.1)  }{(708.1)}] ^2

   I_c = \frac{I}{I_o}  =8.48 *10^{-8}

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Kamal, Kamala, Krishna invert a sum of RS= 30,000 on a business in the ratio 2 : 3 : 5. Find the each of their investment.​
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Here's the <em><u>solution</u></em>,

Let the <em><u>shares</u></em> be 2x, 3x and 5x according to their shared <em><u>investment</u></em> ratio,

now, we know :

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=》10x = 30000

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So, value of x = 3000

hence, the <em><u>share</u></em> of each <em><u>shareholder</u></em> is :

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i hope it helped.....

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