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Maslowich
2 years ago
12

Question 24 1 pts Find the voltage in an extension cord having a 0.0600 22 resistance and through which a 5.00A current is flowi

ng: 500 V 30V O5V O 126V O 60V​
Physics
1 answer:
Neporo4naja [7]2 years ago
6 0

The voltage in the extension cord is 30 V.

The problem above can be solved using ohm's law

⇒ Formula:

V = IR.................. Equation 1

⇒ Where:

  • V = Voltage in the extension cord
  • I = Current flowing through the extension cord
  • R = Resistance of the extension cord.

From the question, I think there was a slight error in the value of the current given it suppose to be 500 A, and not 5.00 A

⇒ Given:

  • I = 500 A
  • R = 0.06 ohms

⇒ Substitute these values into equation 1

  • V = 500(0.06)
  • V = 30 V

Hence the voltage in the extension cord is 30 V

Learn more about voltage here: brainly.com/question/4429782

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Low blood pressure. The person could faint and have an irregular heartbeat.
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The​ time, t, required to drive a fixed distance varies inversely as the​ speed, r. It takes 3 hr at a speed of 14 ​km/h to driv
kenny6666 [7]

Answer:

time taken with speed 23 km/h will be 1.8 hours or 1 hour 48 minutes

Explanation:

Given:

Time is inversely proportional to the speed

mathematically,

t ∝ (1/r)

let the proportionality constant be 'k'

thus,

t = k/r

therefore, for case 1

time = 3 hr

speed = 14 km/hr

3 = k/14

also,

for case 2

let the time be = t

r = 23 km/h

thus,

we have

t = k/23

on dividing equation 2 by 1

we get

\frac{t}{3}=\frac{k/23}{k/14}

or

t=\frac{14\times3}{23}

or

t = 1.8 hr = or 1 hour 48 minutes ( 0.8 hours × 60 minutes/hour = 48 minutes)

4 0
3 years ago
X rays of wavelength 0.0169 nm are directed in the positive direction of an x axis onto a target containing loosely bound electr
mamaluj [8]

Answer:

a) 4.04*10^-12m

b) 0.0209nm

c) 0.253MeV

Explanation:

The formula for Compton's scattering is given by:

\Delta \lambda=\lambda_f-\lambda_i=\frac{h}{m_oc}(1-cos\theta)

where h is the Planck's constant, m is the mass of the electron and c is the speed of light.

a) by replacing in the formula you obtain the Compton shift:

\Delta \lambda=\frac{6.62*10^{-34}Js}{(9.1*10^{-31}kg)(3*10^8m/s)}(1-cos132\°)=4.04*10^{-12}m

b) The change in photon energy is given by:

\Delta E=E_f-E_i=h\frac{c}{\lambda_f}-h\frac{c}{\lambda_i}=hc(\frac{1}{\lambda_f}-\frac{1}{\lambda_i})\\\\\lambda_f=4.04*10^{-12}m +\lambda_i=4.04*10^{-12}m+(0.0169*10^{-9}m)=2.09*10^{-11}m=0.0209nm

c) The electron Compton wavelength is 2.43 × 10-12 m. Hence you can use the Broglie's relation to compute the momentum of the electron and then the kinetic energy.

P=\frac{h}{\lambda_e}=\frac{6.62*10^{-34}Js}{2.43*10^{-12}m}=2.72*10^{-22}kgm\\

E_e=\frac{p^2}{2m_e}=\frac{(2.72*10^{-22}kgm)^2}{2(9.1*10^{-31}kg)}=4.06*10^{-14}J\\\\1J=6.242*10^{18}eV\\\\E_e=4.06*10^{-14}(6.242*10^{18}eV)=0.253MeV

5 0
3 years ago
If the 0.09-mm-diameter tungsten filament in a light bulb is to have a resistance of 0.11 Ω at 20∘C, how long should it be?
skad [1K]

Answer:

Length L=0.0125 m

Explanation:

Given data

Resistance R=0.11Ω

Temperature T=20°C

Resistivity of tungsten at 20°C p=5.6×10⁻⁸Ω.m²

Radius of filament r=0.09/2 =0.045 mm =0.045×10⁻³m

To find

Length of filament L

Solution

The Cross section of area of filament is given as:

A=\pi r^{2}\\ A=\pi (0.045*10^{-3})^{2}\\A=6.3585*10^{-9}m^{2}

The Resistance of a material is given as:

R=p\frac{L}{A} \\L=\frac{AR}{p}\\ L=\frac{(6.3585*10^{-9})(0.11)}{5.6*10^{-8}}\\L=0.125m

6 0
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