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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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At 20 C, a steel rod of length 40.000 cm and a brass rod
balandron [24]

Answer:

a. stress in steel rod is -9.6 X 10⁷pa while stress in brass rod is -7.2 X 10⁷pa

b. new junction from steel rod is 40.0192cm while new junction from brass rod is 30.024cm

Explanation:

<u>Step 1:</u> <u>identify the given parameters and standard parameters</u>

⇒Steel rod: length of rod = 40.000 cm

                    Young modulus(Υ) = 20 X 10¹⁰ pa

                    coefficient of linear expansion(α) = 1.2 X 10⁻⁵ K⁻¹

                     stress in the rod =?

 ⇒Brass rod: length of rod = 30.000 cm

                    Young modulus(Υ) = 9 X 10¹⁰ pa

                    coefficient of linear expansion(α) = 2.0 X 10⁻⁵ K⁻¹

                     stress in the rod =?

--------------------------------------------------------------------------------------------

<u>Step 2:</u> <u>calculate the stress in each rod</u>

⇒Steel rod: stress in the rod = -Y*α*ΔT

                                                = (-20 X 10¹⁰ pa) (1.2 X 10⁻⁵ K⁻¹)(60-20)K

                                                = (-20 X 10¹⁰ pa) (1.2 X 10⁻⁵ K⁻¹)(40)K

                                                =  -9.6 X 10⁷ pa

--------------------------------------------------------------------------------------------

⇒Brass rod: stress in the rod = -Y*α*ΔT

                                                = (-9 X 10¹⁰ pa) (2.0 X 10⁻⁵ K⁻¹)(60-20)K

                                                = (-9 X 10¹⁰ pa) (2.0 X 10⁻⁵ K⁻¹)(40)K

                                                =  -7.2 X 10⁷ pa

--------------------------------------------------------------------------------------------

∴ stress in steel rod is -9.6 X 10⁷pa while stress in brass rod is -7.2 X 10⁷pa

--------------------------------------------------------------------------------------------

<u>Step 3:</u> <u>calculate the new length of each rod</u>

⇒Steel rod: New length = ΔL + L₀

                                   ΔL = α*L₀*ΔT

                                  ΔL = (1.2 X 10⁻⁵ K⁻¹)(40cm)(40)K

                                   ΔL = 1920 X 10⁻⁵ cm = 0.0192cm

                    New length = ΔL + L₀ = 0.0192cm + 40cm

                    New length = 40.0192cm

--------------------------------------------------------------------------------------------

⇒Brass rod: New length = ΔL + L₀

                                    ΔL = α*L₀*ΔT

                                   ΔL = (2.0 X 10⁻⁵ K⁻¹)(30cm)(40)K

                                   ΔL = 2400 X 10⁻⁵ cm = 0.024cm

                    New length = ΔL + L₀ = 0.024cm + 30cm

                    New length = 30.024cm

--------------------------------------------------------------------------------------------

Therefore, new junction from steel rod is 40.0192cm while new junction from brass rod is 30.024cm

8 0
3 years ago
An car is brought to rest in a distance of 652 m using a constant acceleration of -2.7 m/s2. What was the
Goshia [24]

Answer:

I’m so sorry I tried solving it but I don’t understand it can you explain the question a little bit more ty

Explanation:

7 0
3 years ago
If you walk 1.2 km north and then 1.6 km east, what are the magnitude and direction of your resultant displacement?
SVETLANKA909090 [29]

B

Assume north and east as two sides of a right angled triangle. magnitude of the distance is then given by the length of the hypotenuse which is \sqrt{a^2 + b^2}

where a = 1.2 km north

and b = 1.6 km east

magnitude = 2 km

Direction is given by the angle between them, that is atan(a/b) = 36.86 deg north of east = 53.1 deg east of north.

8 0
3 years ago
Which best describes electrical energy? a. energy released by a chemical reaction b. energy produced by flow of electric charge
inessss [21]
<span> energy produced by flow of electric charge describes a electrical energy
because movement of electric charge do effect the work on system 
so correct option is B 
hope it helps</span>
4 0
3 years ago
Read 2 more answers
Lasers are now used in eye surgery. Given the wavelength of a certain laser is 514 nm and the power of the laser is 1.1 W, how m
Leno4ka [110]

Answer: 1.593*10^{17} photons released if the laser is used 0.056 s during the surgery

                           

Explanation:

First, you have to calculate the energy of each photon according to Einstein's theoty, given by:

                            E =\frac{hc}{\lambda}

Where \lambda is the wavelength, h is the Planck's constant and  h is the speed of light

              h = 6.626*10^{-34} \frac{m^{2} kg }{s}  -> Planck's constant

              c = 3*10^{8} \frac{m}{s}  -> Speed of light

So, replacing in the equation:

                E =\frac{ 6.626*10^{-34} \frac{m^{2} kg }{s}*3*10^{8} \frac{m}{s}}{514*10^-9 m}

Then, the energy of each released photon by the laser is:

                E = 3.867*10^{-19} \frac{J}{photons}

After, you do the inverse of the energy per phothon and as a result, you will have the number of photons in a Joule of energy:

               \frac{1}{3.867*10^{-19}} = 2.586*10^{18} \frac{photons}{J}

The power of the laser is 1.1 W, or 1.1 J/s, that means that you can calculate how many photons the laser realease every second:

              2.586*10^{18}\frac{photons}{J} * 1.1 \frac{J}{s} = 2.844*10^{18} \frac{photons}{s}

And by doing a simple rule of three, if 2.844*10^{18} photons are released every second, then in 0.056 s:

            0.056 s*2.844*10^{18} \frac{photons}{s} = 1.593*10^{17} photons are released during the surgery

5 0
3 years ago
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