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Mandarinka [93]
3 years ago
5

A certain corner of a room is selected as the origin of a rectangular coordinate system. If a fly is crawling on an adjacent wal

l at a point having coordinates (2.3, 1.2), where the units are meters, what is the distance of the fly from the corner of the room?
Physics
1 answer:
Helga [31]3 years ago
5 0

Answer:

2.59 m

Explanation:

Coordinates of origin = (0, 0)

Coordinates of Point p where the fly reach = (2.3 m, 1.2 m)

Use the distance formula of coordinates to find the distance between the origin and the point P.

d=\sqrt{\left ( x_{2}-x_{1} \right )^{2}+\left ( y_{2}-y_{1} \right )^{2}}

d=\sqrt{\left ( 2.3- 0 \right )^{2}+\left ( 1.2-0 \right )^{2}}

d = 2.59 m

Thus, the distance between the origin and the point P is 2.59 m.

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If you place a glass cylinder in Wessin Oil, the view of the cylinder nearly disappeared when the eyedropper was full of Wessin
Viefleur [7K]

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A. the indices of refraction matched

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For example, try putting a pencil in a jar full of water. If you look at the pencil from above, it would look as though the pencil has bent in the water. That happens due to the refraction of light. It occurs because as light rays enter water, they slow down, as the speed of light in water is lower than the speed of light in air. The magnitude of how much a medium refracts a light ray is determined by the index

3 0
3 years ago
Suppose that a rectangular toroid has 1,500 windings and a self-inductance of 0.060 H. If the height of the rectangular toroid i
V125BC [204]

Answer:

Current in the toroid will be I=17.32\times 10^{-3}A

Explanation:

We have given number of winding in rectangular toroid N = 1500

Self inductance of toroid L = 0.06 H

Magnetic energy stored in toroid E=9\times 10^{-16}J

We have to find the current in the toroid

Magnetic energy stored is equal to E=\frac{1}{2}Li^2

9\times 10^{-6}=\frac{1}{2}\times 0.06\times I^2

I=17.32\times 10^{-3}A

So current in the toroid will be I=17.32\times 10^{-3}A

5 0
4 years ago
A car is designed to get its energy from a rotating flywheel with a radius of 1.50 m and a mass of 430 kg. Before a trip, the fl
fiasKO [112]

Answer:

a

  KE  =  7.17 *10^{7} \ J

b

 t = 6411.09 \ s

Explanation:

From the question we are told that

    The radius of the flywheel is  r =  1.50 \ m

      The mass of the flywheel is m  = 430 \ kg

          The rotational speed of the flywheel is w  =  5,200 \ rev/min = 5200 *  \frac{2 \pi }{60} =544.61 \ rad/sec

      The power supplied by the motor is  P  =  15.0 hp =  15 * 746 =  11190 \ W

         

     Generally the moment of inertia of the flywheel is  mathematically represented as

       I  = \frac{1}{2} mr^2

substituting values

       I  = \frac{1}{2}  ( 430)(1.50)^2

       I  = 483.75 \  kgm^2

The kinetic energy that is been stored is  

       KE  =  \frac{1}{2} * I * w^2

substituting values

        KE  =  \frac{1}{2} * 483.75 * (544.61)^2

        KE  =  7.17 *10^{7} \ J

Generally power is mathematically represented as

          P =  \frac{KE}{t}

=>      t =  \frac{KE}{P}

substituting the value

        t = \frac{7.17 *10^{7}}{11190}

        t = 6411.09 \ s

3 0
3 years ago
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