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3 years ago

8

A ball tied on a string rotates in a circular path as shown above. The only forces acting on the ball at any point are the weigh

t and of the string. What is the equation for the net centripetal force at point C?

1 answer:

Harlamova29_29 [7]3 years ago

6 0

Answer:

the third one T-W

Explanation:

the direction of the Tension and weight are opposite

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The amount of dissolved oxygen in water may decrease because of the

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<span>The amount of dissolved oxygen in water may decrease because of the increase in organic matter in the water. <span>Aquatic organisms breathe and use oxygen. Large amounts of oxygen are consumed by the decomposition of bacteria (when there are large amounts of dead matter to decompose, there will be a significant number of bacteria). Examples: dead organic matter (algae), wastewater, garden waste, oils and fats, all this results in a decrease in dissolved oxygen in the water.</span></span>

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The condition of earths atmosphere at a given time and place is the weather.

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3 years ago

Light travels 186 000 miles per second.how many miles dose light travel in one year

Troyanec [42]

(186,000 mi/sec) x (3,600 sec/hr) x (24 hr/da) x (365 da/yr)

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3 years ago

A sinusoidally oscillating current I ( t ) with an amplitude of 9.55 A and a frequency of 359 cycles per second is carried by a

UNO [17]

Answer:

$P_{avg} = 6.283*10^{-9} \ W$

Explanation:

Given that;

I₀ = 9.55 A

f = 359 cycles/s

b = 72.2 cm

c = 32.5 cm

a = 80.2 cm

Using the formula;

$\phi = \frac{\mu_o Ic }{2 \pi} In (\frac{b+a}{b})$

where;

$E= \frac{d \phi}{dt}$

$E = \frac{\mu_o}{2 \pi}c In (\frac{b+a}{a}) I_o \omega cos \omega t$

$E_{rms} = \frac { {\frac{\mu_o \ c}{2 \pi} In (\frac{b+a}{a}) I_o (2 \pi f)}}{\sqrt{2}}$

Replacing our values into above equation; we have:

$E_{rms} = \frac { {\frac{4 \pi*10^{-7}*0.325}{2 \pi} In (\frac{72.2+80.2}{80.2}) *9.55 (2 \pi *359)}}{\sqrt{2}}$

$E_{rms} = \frac {8.98909588*10^{-4} }{\sqrt{2}}$

$E_{rms} = 6.356*10^{-4} \ V$

Then the $P_{avg$ is calculated as:

$P_{avg} = \frac{E^2}{R}$

$P_{avg} = \frac{(6.356*10^{-4})^2}{64.3}$

$P_{avg} = 6.283*10^{-9} \ W$

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3 years ago

(a) Calculate the buoyant force on a 2.00-L helium balloon.

iragen [17]

Answer:

0.0239364 N

0.0057879 N

Explanation:

$\rho$ = Density of the gas

g = Acceleration due to gravity = 9.81 m/s²

V = Volume

Mass of rubber = 1.5 g

Buoyant force is given by

$F_b=\rho gV\\\Rightarrow F_b=1.22\times 9.81\times 2\times 10^{-3}\\\Rightarrow F_b=0.0239364\ N$

The buoyant force is 0.0239364 N

Net vertical force is given by

$F_n=F_b-W_{He}-W_{r}\\\Rightarrow F_n=0.0239364-0.175\times 2\times 10^{-3}\times 9.81-1.5\times 10^{-3}\times 9.81\\\Rightarrow F_n=0.0057879\ N$

The net vertical force is 0.0057879 N

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3 years ago