A) what is the resistance of a lightbulb that uses an average power of 25.0 w when connected to a 60.0-hz power source having a maximum voltage of 170 v?

Answers

Since you have not included the experiment situation and the data, of course, i can not give you specific details about it. but i can you with this question.1) i will assume that the motion is projectile (parabolic) motion.2) yes, the horizontal distance depends on the horizontal velocity as, naming vx the horizontal velocity and t the time, it is: x = vx × t3) now, observe that the flight time depends on the vertical velocity as, naming voy the initial vertical velocity, t the time and y the vertical distance before reaching the ground: y = yo + voy×t - gt² / 2.the higher the initial vertical velocity (voy) the longer the object will be flying and so t will be higher in the equation for the horizontal distance.that explains the relationship between the horizontal distance and the horizontal velocity.

answersysvol bloat;

To solve this problem we will start considering the relationship between rms voltage and the maximum voltage. Once the RMS voltage is obtained, we will proceed to find the system current through the given power and the voltage found. Finally by Ohm's law we will find the resistance of the system

The relation of RMS Voltage is,

V_{rms} = \frac{1}{\sqrt{2}}*V_{max}

Note that this conversion is independent from the Frequency.

V_{rms} = \frac{1}{\sqrt{2}}*(170)

V_{rms}= 120.20V

Now the current is,

I =\frac{P}{V}

I = \frac{(25 W)}{(120.20V)}

I = 0.2079A

By Ohm's Law

V = IR \rightarrow R = \frac{V}{I}

Replacing,

R = \frac{ (120.20 V)}{ ( 0.2079 A)}

R = 578.16 \Omega

Therefore the resistance of this lightbulb is578.16 \Omega



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