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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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Read 2 more answers
Using a rope that will snap if the tension in it exceeds 387 N, you need to lower a bundle of old roofing material weighing 449
Anni [7]

Answer:

(a) a\approx1.4 m.s^{-2}

(b) v\approx 4.133 m.s^{-1}

Explanation:

Given:

  • Limiting tension of snapping of the rope, T= 387 N
  • Weight of the object to be lifted, w=449 N
  • ∴mass, \Rightarrow m= 45.8163 kg
  • height of letting down the weight, h = 6.1 m

(a)

Now,

The force to be compensated for  being on the verge of snapping:

(T-w) = 62 N

Therefore, we need to produce and acceleration equivalent to the above force.

∵F=m.a

62=45.8163\times a

a= \frac{62}{45.8163}

a\approx 1.4 m.s^{-2}

(b)

From the equation of motion ,we have:

v^{2} =u^{2} +2a.s....................(2)

where:

u= initial velocity= 0 (here, starting from rest)

v= final velocity = ?

a= 1.4 m.s^{-2}

s= displacement =h =6.1 m

Now, putting the values in eq. (2)

v^2= 0^2 + 2\times 1.4\times 6.1

v\approx 4.133 m.s^{-1}        is the velocity with which the body will hit the ground in the given conditions.

7 0
4 years ago
Current passes through a solution of sodium chloride. In 1.00 s , 2.68× 10 16 Na + ions arrive at the negative electrode and 3.9
Lady bird [3.3K]

Answer

a) charge of the sodium ion is,

q = n e

q = 2.68 x 10¹⁶ x 1.6 x 10⁻¹⁹

q = 4.288 x 10⁻³ C

charge of the chlorine ion is,

q' = n e

q' = 3.92 x 10¹⁶ x 1.6 x 10⁻¹⁹

q' = 6.272 x 10⁻³ C

the current

i = \dfrac{q}{t} + \dfrac{q'}{t}

i = \dfrac{4.288 \times 10^{-3}}{1} + \dfrac{6.272 \times 10^{-3}}{1}

i = 10.56 \times 10^{-3}

b) positive ion moves toward negative electrode hence direction of will be in the direction toward negative electrode.

5 0
3 years ago
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