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enot [183]
2 years ago
12

A scientist testing to see how a cat “always” lands on their feet drops a cat upside down out of a third story window at a heigh

t of 20 meters.  How long does the cat have to right itself?
Physics
1 answer:
makkiz [27]2 years ago
7 0

<u>Answer: </u>

Cat has 2.02 seconds to right itself.

<u>Explanation: </u>

Initial height of cat from ground = 20 meter.

We have equation of motion , s= ut+\frac{1}{2} at^2, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

In this the velocity of cat in vertical direction = 0 m/s, acceleration = acceleration due to gravity =  9.8 m/s^2, we need to calculate time when s = 20 meter.

Substituting

20=0*t+\frac{1}{2} *9.8*t^2\\ \\ t = 2.02 seconds

So, cat has 2.02 seconds to right itself.

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True or False<br><br> The greater the speed of an object, the less kinetic energy it possesses.
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That is true because if the object is moving at Forceful speeds than it will lose more of its kinetic energy
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3 years ago
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Two shuffleboard disks of equal mass, one orange and the other yellow, are involved in an elastic, glancing collision. The yello
klemol [59]

Answer:

3.62m/s and 2.83m/s

Explanation:

Apply conservation of momentum

For vertical component,

Pfy = Piy

m* Vof (sin38) - m*Vgf (sin52) = 0

Divide through by m

Vof(sin38) - Vgf(sin52) = 0

Vof(sin38) = Vgf(sin52)

Vof (sin38/sin52) = Vgf

0.7813Vof = Vgf

For horizontal component

Pxf= Pxi

m* Vof (cos38) - m*Vgf (cos52) = m*4.6

Divide through by m

Vof(cos38) + Vgf(cos52) = 4.6

Recall that

0.7813Vof = Vgf

Vof(cos38) + 0.7813 Vof(cos52) = 4.6

0.7880Vof + 0.4810Vof = 4.

1.269Vof = 4.6

Vof = 4.6/1.269

Vof = 3.62m/s

Recall that

0.7813Vof = Vgf

Vgf = 0.7813 * 3.62

Vgf = 2.83m/s

3 0
3 years ago
A thin double convex glass lens with an index of 1.56 while surrounded by air has a 10 cm focal length. If it is placed under wa
bearhunter [10]

Explanation:

Formula which holds true for a leans with radii R_{1} and R_{2} and index refraction n is given as follows.

          \frac{1}{f} = (n - 1) [\frac{1}{R_{1}} - \frac{1}{R_{2}}]

Since, the lens is immersed in liquid with index of refraction n_{1}. Therefore, focal length obeys the following.  

            \frac{1}{f_{1}} = \frac{n - n_{1}}{n_{1}} [\frac{1}{R_{1}} - \frac{1}{R_{2}}]  

             \frac{1}{f(n - 1)} = [\frac{1}{R_{1}} - \frac{1}{R_{2}}]

and,       \frac{n_{1}}{f(n - n_{1})} = \frac{1}{R_{1}} - \frac{1}{R_{2}}

or,          f_{1} = \frac{fn_{1}(n - 1)}{(n - n_{1})}

              f_{w} = \frac{10 \times 1.33 \times (1.56 - 1)}{(1.56 - 1.33)}

                          = 32.4 cm

Using thin lens equation, we will find the focal length as follows.

             \frac{1}{f} = \frac{1}{s_{o}} + \frac{1}{s_{i}}

Hence, image distance can be calculated as follows.

       \frac{1}{s_{i}} = \frac{1}{f} - \frac{1}{s_{o}} = \frac{s_{o} - f}{fs_{o}}

              s_{i} = \frac{fs_{o}}{s_{o} - f}

             s_{i} = \frac{32.4 \times 100}{100 - 32.4}

                       = 47.9 cm

Therefore, we can conclude that the focal length of the lens in water is 47.9 cm.

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The correct option is


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How is water introduced into rivers and streams?
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Rain fall which is called run off
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