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3 years ago

8

Cody is using a sling shot and rock to knock down a can that he has placed on a tree stump. His first attempt fails because he r

ealizes the rock didn't go far enough. How does Cody increase the amount of potential energy of the rock to make it reach the can and knock it down?
A. Cody needs to stretch the sling shot back more.

B. Cody needs to get a bigger rock.

C. There is no way that Cody can knock the can down.

2 answers:

Archy [21]3 years ago

3 0

it is believed to be the letter a

Hunter-Best [27]3 years ago

3 0

Answer:

c or a Hope's that helps but I think it's c

Explanation:

byeeeee

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A light bulb is connected to a 120.0-V wall socket. The current in the bulb depends on the time t according to the relation I =

kykrilka [37]

a) Frequency: 62.7 Hz

b) Resistance: $186.3\Omega$

c) Average power: 38.6 W

Explanation:

a)

An alternating current can be written in the form

$I(t)=I_0 sin(\omega t)$

where

$I_0$ is the maximum current in the bulb

$\omega=2\pi f$ (1) is the angular frequency, where f is the frequency

t is the time

Here the current in the bulb is

$I(t)=(0.644 A)(sin (394 rad/s\cdot t)$

This means that

$\omega=394 rad/s$

And therefore by using formula (1), we can find the frequency:

$f=\frac{\omega}{2\pi}=\frac{394}{2\pi}=62.7 Hz$

b)

The resistance of the bulb's filament can be found by using Ohm's law:

$V=RI$

where

V is the voltage

R is the resistance

I is the current

Here we have:

$V_0 = 120.0 V$ (maximum voltage)

$I_0 = 0.644 A$ (maximum current)

Therefore, the resistance of the bulb is

$R=\frac{V_0}{I_0}=\frac{120}{0.644}=186.3 \Omega$

c)

In this circuit, the average power delivered to the light bulb is equal to the average power dissipated by the resistor.

For an AC circuit, the average power dissipated on a resistor is given by

$P=I_{rms}^2 R$

where

$I_{rms}$ is the rms current

The rms current can be calculated as

$I_{rms}=\frac{I_0}{\sqrt{2}}$

where in this case

$I_0 =0.644 A$ is the peak current

Here the resistance is

$R=186.3 \Omega$

Therefore, the average power is:

$P=(\frac{I_0}{\sqrt{2}})^2R=(\frac{0.644}{\sqrt{2}})^2(186.3)=38.6 W$

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3 years ago

Please help I will mark brainliest!!

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Heat makes things expand, so in hot water the ballon will get bigger

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