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RoseWind [281]
3 years ago
12

20.0 Ω resistor and a 2.50 μF capacitor are connected in parallel with signal generator. The signal generator produces a sinusoi

dal voltage with an amplitude 3.00 V at a frequency of f=2.48e3 Hz What is the maximum current provided by the power supply?‘
Physics
1 answer:
Free_Kalibri [48]3 years ago
6 0

Answer:

0.15A

Explanation:

The parameters given are;

R=20.0 Ω

C= 2.50 μF

V= 3.00 V

f= 2.48×10^-3 Hz

Xc= 1/2πFc

Xc= 1/2×3.142 × 2.48×10^-3 × 2.5 ×10^-6

Xc= 25666824.1

Z= 1/√(1/R)^2 +(1/Xc)^2

Z= 1/√[(1/20)^2 +(1/25666824.1)^2]

Z= 1/√(2.5×10^-3) + (1.5×10^-15)

Z= 20 Ω

But

V=IZ

Where;

V= voltage

I= current

Z= impedance

I= V/Z

I= 3.00/20

I= 0.15A

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In New Hampshire, people consumed an average of 39.8 gallons of beer in 2018, and the price per gallon was about $7.61. Suppose
sladkih [1.3K]

Answer:

Price elasticity of demand = 0.25

Explanation:

The detailed steps and appropriate derivation is as shown in the attached file.

8 0
3 years ago
Octane is an important component of gasoline. Octane will ignite at a temperature of 428°F. What is this temperature on the Cels
xxMikexx [17]
The requirement is to change the °F to °C scale. That is;

T(°C) = [T(°F) - 32]/1.8

Substituting;

T(°C) = [428 - 32]/1.8 = 396/1.8 = 220 °C.

Therefore, 428°F will be 220°C on Celsius scale.
5 0
4 years ago
A 60kg student traveling in a 1000kg car with a constant velocity has a kinetic energy of 1.2 x 10^4 J. What is the speedometer
777dan777 [17]

Answer:

17.64 km/h

Explanation:

mass of car, m = 1000 kg

Kinetic energy of car, K = 1.2 x 10^4 J

Let the speed of car is v.

Use the formula for kinetic energy.

K = \frac{1}{2}mv^{2}

By substituting the values

1.2\times 10^{4} = \frac{1}{2}\times 1000\times v^{2}

v = 4.9 m/s

Now convert metre per second into km / h

We know that

1 km = 1000 m

1 h = 3600 second

So, v = \left (\frac{4.9}{1000}   \right )\times \left ( \frac{3600}{1} \right )

v = 17.64 km/h

Thus, the reading of speedometer is 17.64 km/h.

4 0
3 years ago
an athlete runs 5.4 laps around a circular track that is 400.0 m long. If this takes 540 s, what is the average velocity of the
Slav-nsk [51]

Answer:

Average velocity is 0.296 m/s.

Average speed is 4.0  m/s.

Explanation:

Given:

Distance of the circular track is, D=400.0\ m

Number of laps ran is, n=5.4

Time taken for the run is, t=540\ s

Now, total distance covered in 5.4 laps = D_T=D\times n=400\times 5.4=2160\ m

Also, since the path is a circle, the final position of the athlete after 5.4 laps will be 0.4 of 400 m ahead of the starting point.

Distance covered in 0.4 laps is, \textrm{Displacemet}=0.4\times 400=160\ m

Therefore, the displacement of the athlete will be 160 m as the athlete is 160 m ahead of the starting point and displacement depends on the initial and final points only.

Now, average velocity is given as:

v_{avg}=\frac{\textrm{Displacemet}}{t}=\frac{160}{540}=0.296\ m/s

Average speed is the ratio of total distance covered to total time taken.

So, average speed = \frac{D_T}{t}=\frac{2160}{540}=4\ m/s

6 0
4 years ago
In order to measure current in a circuit, the multi-meter needs to be placed in __A__ , while when measuring voltage, the multi-
lakkis [162]

Answer:

A- series B- parallel

Explanation:

In order to measure current in a circuit, the multi-meter needs to be placed in series with the circuit while when measuring voltage, the multi-meter needs to be placed in parallel with the circuit.

It should be however noted that the same current flows in a series connected circuit and same voltage flows through loads connected in parallel. The ammeter is placed in series with the load to ensure that same value of currents is flowing in both the ammeter and loads(since same current flows in series connected circuit elements and all the amount of voltage must be made to appear on the load for the current to be measured accurately.

Voltmeter is connected in parallel to the load due to high value of current possessed by the voltmeter. The parallel connection will cause the current flowing through the voltmeter to reduce to zero so that it won't have effect (increase) on the amount of current initially on the resistor thereby measuring the exact amount of voltage on the load.

3 0
4 years ago
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