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sleet_krkn [62]
3 years ago
9

Ethan and Rebecca are riding on a merry-go-round. Ethan rides on a horse at the outer rim of the circlar platform, twice as far

from the center of the circular platform as Rebecca, who rides on an inner horse. When the merry-go-round is rotating at a constant angular speed, describe Ethans tangential speed?
Physics
1 answer:
xz_007 [3.2K]3 years ago
3 0

Answer:

Ethan's tangential speed is twice as Rebecca's.

Explanation:

Let r be the distance from the center of the platform to Rebecca's horse, v_R Rebecca's tangential speed, v_E the Ethan's tangential speed and \omega the merry-go-round angular speed. The distance from the center of the platform to Ethan's horse is 2r. The angular speed is related to the tangential speed by the equation:

\omega =\frac{v}{R}

Since the angular speed is the same for Ethan and Rebecca, we have that:

\frac{v_E}{2r}=\frac{v_R}{r}

Now, solving for v_E we get:

v_E=2v_R

It means that Ethan's tangential speed is twice as Rebecca's.

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Two thin concentric spherical shells of radii r1 and r2 (r1 < r2) contain uniform surface charge densities V1 and V2, respect
Lyrx [107]

Answer:

Answer is explained in the explanation section below.

Explanation:

Solution:

We know that the Electric field inside the thin hollow shell is zero, if there is no charge inside it.

So,

a)  0 < r < r1 :

We know that the Electric field inside the thin hollow shell is zero, if there is no charge inside it.

Hence, E = 0 for r < r1

b)  r1 < r < r2:

Electric field =?

Let, us consider the Gaussian Surface,

E x 4 \pi r^{2}  = \frac{Q1}{E_{0} }

So,

Rearranging the above equation to get Electric field, we will get:

E = \frac{Q1}{E_{0} . 4 \pi. r^{2}   }

Multiply and divide by r1^{2}

E = \frac{Q1}{E_{0} . 4 \pi. r^{2}   } x \frac{r1^{2} }{r1^{2} }

Rearranging the above equation, we will get Electric Field for r1 < r < r2:

E= (σ1 x r1^{2}) /(E_{0} x r^{2})

c) r > r2 :

Electric Field = ?

E x 4 \pi r^{2}  = \frac{Q1 + Q2}{E_{0} }

Rearranging the above equation for E:

E = \frac{Q1+Q2}{E_{0} . 4 \pi. r^{2}   }

E = \frac{Q1}{E_{0} . 4 \pi. r^{2}   } + \frac{Q2}{E_{0} . 4 \pi. r^{2}   }

As we know from above, that:

\frac{Q1}{E_{0} . 4 \pi. r^{2}   } =  (σ1 x r1^{2}) /(E_{0} x r^{2})

Then, Similarly,

\frac{Q2}{E_{0} . 4 \pi. r^{2}   } = (σ2 x r2^{2}) /(E_{0} x r^{2})

So,

E = \frac{Q1}{E_{0} . 4 \pi. r^{2}   } + \frac{Q2}{E_{0} . 4 \pi. r^{2}   }

Replacing the above equations to get E:

E = (σ1 x r1^{2}) /(E_{0} x r^{2}) + (σ2 x r2^{2}) /(E_{0} x r^{2})

Now, for

d) Under what conditions,  E = 0, for r > r2?

For r > r2, E =0 if

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4 0
3 years ago
Does 3.60 x 10 ⁻² have 2 significant figures
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Answer:

Number of Significant Figures: 2

The Significant Figures are 3 6

Explanation:

= 3.60 × 102

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= 3.60e2

(scientific e notation)

= 360 × 100

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Answer:

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