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IceJOKER [234]
2 years ago
8

Please help 9.2.1 project in science just ned an example​

Physics
1 answer:
olga55 [171]2 years ago
3 0

Answer:

Give me what kind of example you need please so I can help you. Put it in the comments.

Explanation:

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Why are atoms in a covalent bond usually a certain distance away from each other?
ANTONII [103]
I think it is because the electrons repel each other

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3 years ago
Explain the process of refraction of light​
brilliants [131]

The process of refraction of light​ occurs when light rays bends when travelling between media of different densities.

What is refraction of light?

Refraction of light is the bending of light rays or the change in the direction of light rays as it travels between media of different densities.

Light waves travel faster in media of less density than media of more density.

The change in density of the media makes light waves to be bend towards or away from the normal.

For example, when light travels from less dense air to more dense water, the light rays are bent towards the normal. However, when light rays travel from water to air, the light rays are refracted away from the normal.

In conclusion, refraction of light waves occur when light crosses the boundary of media of different densities.

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4 0
1 year ago
Two pulleys, one with a radius 2R and one with a radius of R, are welded together and can rotate about the same axis. A block is
Arisa [49]

Answer:

speed will double

Explanation:

Tangential velocity is given by

v=wr

where r is radius and w is rotational velocity. When r is increased to 2R, keeping roational velocity constant, the tangential velocity doubles. Hence unwinding will take place at a twice the rate and the final velocity will be double

6 0
2 years ago
A road with a radius of 75.0 m is banked so that a car can navigate the curve at a speed of 15.5 m/s without any friction. When
kirza4 [7]

Answer: 0.683

Explanation: The relationship between a car moving along a curve and the frictional force is given below as

us×g = v²/r

Where us = coefficient of static friction =?, g = acceleration due to gravity = 9.8 m/s², v = 22.4 m/s and r = 75m

By substituting the parameters, we have that

us×9.8 = 22.4²/ 75

us ×9.8 = 501.76/75

us ×9.8 = 6.690

us = 6.690/9.8 = 0.683

8 0
2 years ago
Captain Hook is ghting Peter Pan, and they are about to step onto a tightrope strung horizontally between two maststhat are 16 m
andreyandreev [35.5K]

Complete Question

Captain Hook is fighting  Peter Pan, and they are about to step onto a tightrope strung horizontally between two masts that are 16 m apart. When Pan and Hook are standing exactly halfway between the masts, the rope makes a 3 anglewith the horizontal. The rope has a diameter of 0.02 m and a Youngs modulus of 35 GPa.

What is the combined mass,in kilograms, of Peter Pan and Captain Hook?

Answer:

Their combined mass is m= 161.2kg

Explanation:

A sketch that describes the question is shown on the first uploaded image  

  From the question we are told that

          The distance apart is d_A = 16m

          The angle the rope makes is \theta = 3^o

          The diameter of the rope is d = 0.02m

          The Young modulus is  Y = 35Pa

From the diagram we see that the elongation of the rope can be  mathematically evaluated as

         \Delta L = 2x - 16

And applying  SOHCATOH rule    x = \frac{8}{cos \theta}

   Substituting values

                              x = \frac{8}{cos (3)}

                                = 8.01m

      And   \Delta L = \frac{16}{cos 3}  -16

                     \Delta L = 0.02196m

The Tension on the rope can be mathematically represented as

               T = Y A * \frac{\Delta L}{L}

Where A is the area and is mathematically represented as

              A = \frac{\pi}{4} d^2

 Substituting values

            A = \frac{\pi}{4} (0.02)^2

Now Substituting values into the formula for the tension on the rope

          T = (35*10^9) * \frac{\pi}{4} (0.02)^2 * \frac{(0.02196)}{16}

             =15093.4 N

From the diagram we can mathematically evaluate the the weight of peter and hook as

              W = 2T sin \theta

Where W = mg

Now substituting this into the equation and making m the subject

                   m = \frac{2Tsin \theta}{g}

Substituting values

                m = \frac{2* 15093.4 sin(3)}{9.8}

                    m= 161.2kg

Note  SOHCATOH is

                         Sin \theta = \frac{opposite}{hypotenuse}\\ Cos \theta = \frac{adjacent }{hypotenuse} \\Tan \theta = \frac{opposite}{adjacent}      

8 0
3 years ago
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