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4 years ago

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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?

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 pon $T=\frac{1}{0.5Hz} =$ 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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If R is the total resistance for a parallel circuit with two resistors of resistance r1 and r2, then . Find the resistance, r1,

Goshia [24]

For a parallel circuit with two resistors, the total resistance is calculated from the expression:

1/R = 1/R1 + 1/R2

We are given the total resistance, R, which is 20 ohms and R2 which is 75 ohms. We calculate R1 as follows:

1/20 = 1/R1 + 1/75
1/R1 = 11/300
R1 = 27.27 ohms

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3 years ago

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A block is projected up a frictionless inclined plane with initial speed v0 = 1.72 m/s. The angle of incline is θ = 44.8°. (a) H

Snowcat [4.5K]

Explanation:

Given

initial velocity(v_0)=1.72 m/s

$\theta =44.8{\circ}$

using $v^2-u^2=2as$

Where v=final velocity (Here v=0)

u=initial velocity(1.72 m/s)

a=acceleration $(gsin\theta )$

s=distance traveled

$0-(1.72)^2=2(-9.81\times sin(44.8))s$

s=0.214 m

(b)time taken to travel 0.214 m

v=u+at

$0=1.72-gsin(44.8)\times t$

$t=\frac{1.72}{9.8\times sin(44.8)}$

t=0.249 s

(c)Speed of the block at bottom

$v^2-u^2=2as$

Here u=0 as it started coming downward

$v^2=2\times gsin(44.8)\times 0.214$

$v=\sqrt{2.985}$

v=1.72 m/s

3 0

3 years ago

Two particles, each of mass m, are initially at rest very far apart.Obtain an expression for their relative speed of approach at

PSYCHO15rus [73]

Answer:

$|\Delta v |=\sqrt{\frac{4Gm}{d} }$

Explanation:

Consider two particles are initially at rest.

Therefore,

the kinetic energy of the particles is zero.

That initial K.E. = 0

The relative velocity with which both the particles are approaching each other is Δv and their reduced masses are

$\mu= \frac{m_1m_2}{m_1+m_2}$

now, since both the masses have mass m

therefore,

$\mu= \frac{m^2}{2m}$

= m/2

The final K.E. of the particles is

$KE_{final}=\frac{1}{2}\times \mu\times \Delta v^2$

Distance between two particles is d and the gravitational potential energy between them is given by

$PE_{Gravitational}= \frac{Gmm}{d}$

By law of conservation of energy we have

$KE_{initial}+KE_{final}= PE_{gravitaional}$

Now plugging the values we get

$0+\frac{1}{2}\frac{m}{2}\Delta v^2= -\frac{Gmm}{d}$

$|\Delta v |=\sqrt{\frac{4Gm}{d} }$

$=\sqrt{\frac{Gm}{d} }$

This the required relation between G,m and d

5 0

3 years ago

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Answer:

Explanation:

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4 years ago

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An electron moving in a direction perpendicular to a uniform magnetic field at a speed of 1.6 107 m/s undergoes an acceleration

umka2103 [35]

Answer:

B = 0.024T positive z-direction

Explanation:

In this case you consider that the direction of the motion of the electron, and the direction of the magnetic field are perpendicular.

The magnitude of the magnetic force exerted on the electron is given by the following formula:

$F=qvB$ (1)

q: charge of the electron = 1.6*10^-19 C

v: speed of the electron = 1.6*10^7 m/s

B: magnitude of the magnetic field = ?

By the Newton second law you also have that the magnetic force is equal to:

$F=qvB=ma$ (2)

m: mass of the electron = 9.1*10^-31 kg

a: acceleration of the electron = 7.0*10^16 m/s^2

You solve for B from the equation (2):

$B=\frac{ma}{qv}\\\\B=\frac{(9.1*10^{-31}kg)(7.0*10^{16}m/s^2)}{(1.6*10^{-19}C)(1.6*10^7m/s)}\\\\B=0.024T$

The direction of the magnetic field is found by using the right hand rule.

The electron moves upward (+^j). To obtain a magnetic forces points to the positive x-direction (+^i), the direction of the magnetic field has to be to the positive z-direction (^k). In fact, you have:

-^j X ^i = ^k

Where the minus sign of the ^j is because of the negative charge of the electron.

Then, the magnitude of the magnetic field is 0.024T and its direction is in the positive z-direction

8 0

3 years ago