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professor190 [17]

3 years ago

11

A purse at radius 2.30 m and a wallet at radius 3.45 m travel in uniform circular motion on the floor of a merry-go-round as the

ride turns. They are on the same radial line. At one instant, the acceleration of the purse is . At that instant and in unit-vector notation, what is the acceleration of the wallet?

1 answer:

ivolga24 [154]3 years ago

4 0

Answer:

The acceleration of the wallet is $3\hat{i}+6\hat{j}$

Explanation:

Given that,

Radius of purse r= 2.30 m

Radius of wallet r'= 3.45 m

Acceleration of the purse $a=2\hat{i}+4.00\hat{j}$

We need to calculate the acceleration of the wallet

Using formula of acceleration

$a=r\omega^2$

Both the purse and wallet have same angular velocity

$\omega=\omega'$

$\sqrt{\dfrac{a}{r}}=\sqrt{\dfrac{a'}{r'}}$

$\dfrac{a}{r}=\dfrac{a'}{r'}$

$\dfrac{a'}{a}=\dfrac{r'}{r}$

$\dfrac{a'}{a}=\dfrac{3.45}{2.30}$

$\dfrac{a'}{a}=\dfrac{3}{2}$

$a'=\dfrac{3}{2}\times(2\hat{i}+4.00\hat{j})$

$a'=3\hat{i}+6\hat{j}$

Hence, The acceleration of the wallet is $3\hat{i}+6\hat{j}$

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Answer:

Part A:

The proton has a smaller wavelength than the electron.

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Part B:

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Case for the proton:

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(1.67x10^{-27}Kg)(6.55x10^{6}m/s)}$

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<em>Part B </em>

For part b, the wavelength of the electron and proton for that energy will be determined.

First, it is necessary to find the velocity associated to that kinetic energy:

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$2KE = mv^{2}$

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