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Airida [17]
3 years ago
7

A penny is placed on a rotating turntable. where on the turntable does the penny require the largest centripetal force to remain

in place?
Physics
1 answer:
Mama L [17]3 years ago
4 0

m = mass of the penny

r = distance of the penny from the center of the turntable or axis of rotation

w = angular speed of rotation of turntable

F = centripetal force experienced by the penny

centripetal force "F" experienced by the penny of "m" at distance "r" from axis of rotation is given as

F = m r w²

in the above equation , mass of penny "m"  and angular speed "w" of the turntable is same at all places. hence the centripetal force directly depends on the radius .

hence greater the distance from center , greater will be the centripetal force to remain in place.  

So at the edge of the turntable , the penny experiences largest centripetal force to remain in place.

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Describe You toss a ball to someone one meter away. Then you toss it to someone four meters away. How does your throw change?
krek1111 [17]

We have that the Throw has changed in distance the ball has traveled and the Force applied in trowing the ball and Possibly the time of travel

From the Question we are told that

You toss a ball to someone one meter away

You toss it to someone four meters away

Generally

When you toss toss a ball to someone one meter away and You toss it to someone four meters away the are  things that have changed

  • Force
  • Distance

Therefore

The Throw has changed in distance the ball has traveled and the Force applied in trowing the ball and Possibly the time of travel

For more information on this visit

brainly.com/question/12008506?referrer=searchResults

3 0
3 years ago
A physics student mounts two thin lenses along a single optical axis (the lenses are at right angles to the line connecting them
Tatiana [17]

Answer:

A)    q₂ = 75.98 cm, B)     q₂' = 115.38 cm, C)

Explanation:

A) This is an exercise in geometric optics, as the two lenses are separated by a greater distance than their focal lengths from each lens, they must be worked as independent lenses.

Lens 1. More to the left

let's use the constructor equation

         \frac{1}{f} = \frac{1}{p} + \frac{1}{q}

where f is the focal length, p and q are the distance to the object and the image, respectively,

We must assume a distance to the object to perform the calculation, suppose that the object is 50 cm from lens 1 that is further to the left of the system.

          \frac{1}{q_1} = \frac{1}{f} - \frac{1}{p}

         \frac{1}{q_1} = \frac{1}{14.8} - \frac{1}{50}  

          1 / q₁ = 0.04756

           q₁ = 21.0227 cm

this image is the object for the second lens that has f₂ = 14.8 cm

the distance must be measured from the second lens

          p₂ = 39.4 -q₁

          p₂ = 39.4 -21.0227

          p₂ = 18.38 cm

let's use the constructor equation

            1 / q₂ = 1 / f - 1 / p2

             

             \frac{1}{q_2} = \frac{1}{14.8} - \frac{1}{18.38}

            \frac{1}{q_2} = 0.01316

            q₂ = 75.98 cm

measured from the second lens

B) the position of the final image with respect to the first lens is

            q₂’= q₂ + 39.4

             q₂'= 75.98 +39.4

              q₂' = 115.38 cm

C) the magnification of a lens is

              m = - q / p

in this case the image measured from lens 2 is q2 = 75.98 cm

the distance to the object from the first lens is p1 = 50cm

          m = - 75.98 / 50

          m = -1.5 X

the negative sign indicates that the image is inverted

4 0
3 years ago
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