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GenaCL600 [577]

3 years ago

12

While David was riding his bike around the circular cul-de-sac by his house, he wondered if the constant circular motion was hav

ing any effect on his tires. What would be the best way for David to investigate this?
A.
Measure the circumference of the tire before and after riding.
B.
Measure the total distance traveled on his bike and divide this by how long it took him.
C.
Measure the wear on his treads before and after riding a certain number of laps.
D.
Time how long it takes him to ride 5 laps around his cul-de-sac.

1 answer:

RSB [31]3 years ago

6 0

Answer:

C.

Measure the wear on his treads before and after riding a certain number of laps.

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A uniformly charged sphere has a potential on its surface of 450 V. At a radial distance of 8.1 m from this surface, the potenti

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Answer:

The radius of the sphere is 4.05 m

Explanation:

Given;

potential at surface, $V_s$ = 450 V

potential at radial distance, $V_r$ = 150

radial distance, l = 8.1 m

Apply Coulomb's law of electrostatic force;

$V = \frac{KQ}{r} \\\\V_s = \frac{KQ}{r} \\\\V_r = \frac{KQ}{r+ l}$

$450 = \frac{KQ}{r} ------equation (i)\\\\150 = \frac{KQ}{r+8.1} ------equation (ii)\\\\divide \ equation (i)\ by \ equation \ (ii)\\\\\frac{450}{150} = (\frac{KQ}{r} )*(\frac{r+8.1}{KQ} )\\\\3 = \frac{r+8.1}{r} \\\\3r = r + 8.1\\\\2r = 8.1\\\\r = \frac{8.1}{2} \\\\r = 4.05 \ m$

Therefore, the radius of the sphere is 4.05 m

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A 5.00 kg rock whose density is 4300 kg/m3 is suspended by a string such that half of the rock's volume is under water. You may

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Answer:

The tension on the string is $T = 43.302 \ N$

Explanation:

From the question we are told that

The mass of the rock is $m_r = 5.00 \ kg = 5000 \ g$

The density of the rock is $\rho = 4300 \ kg/m^3 = 4.3 g/dm^3$

Generally the volume of the rock is mathematically evaluated as

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substituting values

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The volume of the rock immersed in water is

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substituting values

$V_w = \frac{1162.7 }{2}$

$V_w = 581.4 \ dm^3$

mass of water been displaced by the this volume is

$m_w = V_w$ According to Archimedes principle

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The weight of the water displace is

$W _w = m_w * g$

$W _w = 0.5814 * 9.8$

$W _w = 5.698 \ N$

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$W_r = 5.0 * 9.8$

$W_r = 49.0 \ N$

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$T = 49.0 - 5.698$

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