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Sholpan [36]
3 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]3 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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The neutral wire is often confused with ground wire, but in reality, they serve two distinct purposes. Neutral wires carry currents back to power source to better control and regulate voltage. Its overall purpose is to serve as a path to return energy.

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A bicyclist accelerates from a stop to a speed of 12m/s in 3 seconds.what is her acceleration
AVprozaik [17]

Answer:

4m/s2

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a=12-0/3

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2 years ago
Who invented abstract geometry
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3 years ago
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A skateboarder with a mass of 45 kilograms is riding on a skateboard with a mass of 2.5 kilograms. What should be the velocity o
klasskru [66]
So momentum is just velocity times mass, this means Momentum = Velocity x Mass.
We can rearrange this to be Velocity = Momentum/Mass.

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Velocity = 264/(45+2.5)
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5 0
3 years ago
Read 2 more answers
A thin spherical spherical shell of radius R which carried a uniform surface charge density σ. Write an expression for the volum
ozzi

Answer:

Explanation:

From the given information:

We know that the thin spherical shell is on a uniform surface which implies that both the inside and outside the charge of the sphere are equal, Then

The volume charge distribution relates to the radial direction at r = R

∴

\rho (r) \  \alpha  \  \delta (r -R)

\rho (r) = k \  \delta (r -R) \ \  at \ \  (r = R)

\rho (r) = 0\ \ since \ r< R  \ \ or  \ \ r>R---- (1)

To find the constant k, we  examine the total charge Q which is:

Q = \int \rho (r) \ dV = \int \sigma \times dA

Q = \int \rho (r) \ dV = \sigma \times4 \pi R^2

∴

\int ^{2 \pi}_{0} \int ^{\pi}_{0} \int ^{R}_{0} \rho (r) r^2sin \theta  \ dr \ d\theta \ d\phi = \sigma \times 4 \pi R^2

\int^{2 \pi}_{0} d \phi* \int ^{\pi}_{0} \ sin \theta d \theta * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2

(2 \pi)(2) * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2

Thus;

k * 4 \pi  \int ^{R}_{0}  \delta (r -R) * r^2dr = \sigma \times  R^2

k * \int ^{R}_{0}  \delta (r -R)  r^2dr = \sigma \times  R^2

k * R^2= \sigma \times  R^2

k  =   R^2 --- (2)

Hence, from equation (1), if k = \sigma

\mathbf{\rho (r) = \delta* \delta (r -R)  \ \  at   \ \  (r=R)}

\mathbf{\rho (r) =0 \ \  at   \ \  rR}

To verify the units:

\mathbf{\rho (r) =\sigma \ *  \ \delta (r-R)}

↓         ↓            ↓

c/m³    c/m³  ×   1/m            

Thus, the units are verified.

The integrated charge Q

Q = \int \rho (r) \ dV \\ \\ Q = \int ^{2 \ \pi}_{0} \int ^{\pi}_{0} \int ^R_0 \rho (r) \ \ r^2 \ \  sin \theta  \ dr \ d\theta \  d \phi  \\ \\  Q = \int ^{2 \pi}_{0} \  d \phi  \int ^{\pi}_{0} \ sin \theta  \int ^R_{0} \rho (r) r^2 \ dr

Q = (2 \pi) (2) \int ^R_0 \sigma * \delta (r-R) r^2 \ dr

Q = 4 \pi  \sigma  \int ^R_0  * \delta (r-R) r^2 \ dr

Q = 4 \pi  \sigma  *R^2    since  ( \int ^{xo}_{0} (x -x_o) f(x) \ dx = f(x_o) )

\mathbf{Q = 4 \pi R^2  \sigma  }

6 0
3 years ago
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