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Sholpan [36]
4 years ago
8

Suppose you watch a leaf bobbing up and down as ripples pass it by in a pond. You notice that it does two full up and down bobs

each second. What is true of the ripples on the pond?
Physics
1 answer:
otez555 [7]4 years ago
4 0

Answer:

The frequency of the ripples is 2Hz, and their period is 0.5 seconds.

Explanation:

Since the ripples on the pond are making the leaf oscillate up and down at a rate of two times per second, we can calculate the period T and the frequency f of the ripples on the ponT=\frac{1}{0.5Hz} =d.

The frequency, by definition, is the number of waves per unit of time. In this case, we have two waves per second, so the frequency is 2s⁻¹, or 2Hz.

The period is the inverse of the frequency, so

T=\frac{1}{2Hz} =0.5s

Then, the period is equal to 0.5 seconds.

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3 years ago
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A typical laboratory centrifuge rotates at 4000 rpm. Testtubes have to be placed into a centrifuge very carefully because ofthe
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Answer:

A)  a_c = 1.75 10⁴ m / s², B) a = 4.43 10³ m / s²

Explanation:

Part A) The relation of the test tube is centripetal

               a_c = v² / r

the angular and linear variables are related

              v = w r

we substitute

               a_c = w² r

let's reduce the magnitudes to the SI system

              w = 4000 rpm (2pi rad / 1 rev) (1 min / 60s) = 418.88 rad / s

               r = 1 cm (1 m / 100 cm) = 0.10 m

let's calculate

              a_c = 418.88² 0.1

               a_c = 1.75 10⁴ m / s²

part B) for this part let's use kinematics relations, let's start looking for the velocity just when we hit the floor

as part of rest the initial velocity is zero and on the floor the height is zero

                v² = v₀² - 2g (y- y₀)

                v² = 0 - 2 9.8 (0 + 1)

                v =√19.6

                v = -4.427 m / s

now let's look for the applied steel to stop the test tube

                v_f = v + a t

                0 = v + at

                a = -v / t

                a = 4.427 / 0.001

                a = 4.43 10³ m / s²

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3 years ago
If the force is applied to a car, then its acceleration will ___________ because of Newton's _____________.
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3 years ago
What can directly lead to unconformity on an exposed rock?
madreJ [45]
The answer is B. Hope I helped!
8 0
3 years ago
Suppose that the metal cylinder in the last problem has a mass of 0.10 kg and the coefficient of static friction between the sur
andrew11 [14]

Answer:

The speed maximum speed is 0.49ms^{-1}

Explanation:

The centrifugal force always acts on the cylinder and move away the rotating platform from the  rotational axis. so the centripetal force provide by the frictional force:

Therefore

\frac{mv^{2} }{r} =u_{s} mg        

coefficient of static friction: u_{s} =0.12

mass of the cylinder: m=0.10kg\

distance of the cylinder from the turntable: r=0.20m

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3 years ago
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