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Marizza181 [45]

3 years ago

8

A 3.0 X 10^3 grams rock swings in a circle with a diameter of 1000 cm. Given the constant speed around the circle is 17.90 mph (

a) what is the centripetal acceleration. *

1 answer:

marissa [1.9K]3 years ago

3 0

Answer:

a = 12.8 m/s^2

Explanation:

To find the centripetal acceleration you use the following formula:

$a_c=\frac{v^2}{r}$ (1)

v: tangential speed of the rock = 17.90mph

r: radius of the orbit = 1000cm/2 = 500cm = 0.5m

You first change the units of the tangential velocity in order to replace the values of v and r in the equation (1):

$17.90mph*\frac{1609.34m}{1\ m}*\frac{1\ h}{3600\ s}=8m/s$

Then, you use the equation (1):

$a_c=\frac{(8m/s)^2}{5m}=12.8\frac{m}{s^2}$

hence, the centripetal acceleration is 12.8 m/s^2

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3 years ago

I NEED HELP PLEASE, THANKS! :)

Zina [86]

Answer:

charge C = greatest net force

charge B = the smallest net force

ratio = 9 : 1

Explanation:

we know that in Electrostatic Forces, when 2 charges are at same sign then they repel each other and if they are different signed charges then they attract each other

so as per Coulomb's formula of Electrostatic Forces

F = $\frac{k\ q_1\ q_2}{r^2}$ .....................1

and here k is 9 × $10^9$ N.m²/c² and we consider each charge at distance d

so two charge force at A to B is

F1 = $\frac{k\ q^2}{d^2}$

and force between charges at A to C, at 2d distance

F1 = $\frac{k\ q^2}{(2d)^2}$ = $\frac{k\ q^2}{4d^2}$

force between charges at A to D, 3d distance

F1 = $\frac{k\ q^2}{(3d)^2}$ = $\frac{k\ q^2}{9d^2}$

so

Charge a It receives force to the left from b and c and to the right from d

so at a will be

F(a) = -F1 - F2 + F3 ....................2

put here value

F(a) = $-\frac{k\ Q^2}{d^2}-\frac{k\ Q^2}{4d^2}+\frac{k\ Q^2}{9d^2}$

solve it

F(a) = $\frac{k\ q^2}{d^2}(-1-\frac{1}{4}+\frac{1}{9})$

F(a) = $-\frac{41}{36}\ F1$ = 1.13 F1

and

Charge b It receives force to the right from a and d and to the left from c

F(b) = F1 - F1 + F2 ....................3

F(b) = $\frac{k\ q^2}{d^2}-\frac{k\ q^2}{d^2}+\frac{k\ q^2}{4d^2}$

F(b) = $\frac{1}{4} \ F1$ = 0.25 F1

and

Charge c It receives forces to the right from all charges.

F(c) = F2 + F 1 + F 1 ....................4

F(c) = $\frac{k\ q^2}{4d^2}+\frac{k\ q^2}{d^2}+\frac{k\ q^2}{d^2}$

F(c) = $\frac{9}{4} \ F1$ = 2.25 F1

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Charge d It receives forces to the left from all charges

F(d) = - F3 - F2 -F 1 ....................5

F(d) = $-\frac{k\ q^2}{9d^2}-\frac{k\ q^2}{4d^2}-\frac{k\ q^2}{d^2}$

so

F(d) = $-\frac{49}{36} \ F1$ = 1.36 F1

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ratio = $\frac{2.25}{0.25}$

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If you want to learn more, you can read:

brainly.com/question/14514975

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