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3 years ago

5

What is the current in a circuit that has a resistance of 75Ω and a voltage drop of 120V across the cell? (I'll give brainliest!

)

1 answer:

Agata [3.3K]3 years ago

3 0

Resistance = (voltage) divided by (current)

= (120 V) / (0.5 Amp)

= 240 ohms .

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A double-slit experiment is set up using red light (λ = 706 nm). A first order bright fringe is seen at a given location on a sc

Elanso [62]

Answer:

λ = 470.66 nm

Explanation:

for bright fringe $y_m = \frac{m\lambda D}{d}$

D= distance between slit and screen

d= distance between the slits

for first order bright fringe m = 1,

$y_1 = \frac{1\lambda D}{d}$

$y_1 = {706*D}{d}$

for dark fringe,we have

$y_m = {(m + 1/2)\lambda D}{d}$

Now to get the dark fringes at the same location we should have;

(706)D/d = (m + 1/2)λD/d

put m = 1

(1 + 1/2)λ = (706)

λ = 470.66 nm

6 0

4 years ago

I need help on B. please help me

yuradex [85]

I think unbalanced forces is one of the reasons

3 0

3 years ago

Suppose that a parallel-plate capacitor has circular plates with radius R = 25 mm and a plate separation of 4.7 mm. Suppose also

ozzi

Answer:

a) $B_{max} = 1.784*10^{-12}$

Explanation:

Given paraeters are:

R = 25 cm

d = 4.7 mm

f = 60 Hz

$V_m$ = 160 V

a) $V = V_msin(2\pi ft)$

Where $f = 60$ Hz and $V_m = 160$ V

$E =V/d= \frac{V_msin(2\pi ft)}{d}$

For $r = R$

$A = \pi R^2$

Since $\Phi_E = EA$

$\Phi_E=\frac{\pi R^2V_msin(2\pi ft) }{d}$

From Ampere's Law:

$\int B.ds = \mu_0\epsilon_0\frac{d\Phi_E}{dt} + \mu_0I_{encl}$ where $I_{encl}=0$

So at $r = R$ ,

$B.2\pi R = \mu_0\epsilon_0\frac{d\Phi_E}{dt}\\B.2\pi R = \mu_0\epsilon_0\frac{2\pi^2fR^2V_mcos(2\pi ft)}{d}\\B = \frac{\mu_0\epsilon_0\pi fRV_mcos(2\pi ft)}{d}$

For maximum B, cos(2πft) = 1. Hence,

$B_{max}=\frac{\mu_0\epsilon_0\pi fRV_m}{d}=\frac{4\pi*10^{-7}*8.85*10^{-12}*\pi*60*0.025*160}{4.7*10^{-3}}=1.784*10^{-12}$ T

b) From r = 0 to r = R = 0.025 m, by Ampere's Law, the equation will be:

$B = \frac{\mu_0\epsilon_0\pi fR^2V_m}{rd}$

From r = R = 0.025 m to r = 0.1 m, by Ampere's Law, the equation will be:

$B = \frac{\mu_0\epsilon_0\pi fRV_m}{d}$

The plot is given in the attachment.

5 0

4 years ago

Stars begin burning helium to carbon when the temperature rises in the core. This temperature increase is caused by a. gravitati

swat32

When the core temperature rises, stars start burning helium to carbon. Hydrogen and helium are fused in a shell surrounding the core, raising the temperature.

The most prevalent method of generating energy in the cosmos is hydrogen fusion. The gravitational force is utilised to fuse hydrogen at the centres of our Sun and other stars, where it has a density more than 70 times greater than that of steel. Thus, all of the heat and light on earth comes from fusion. Helium is a chemical element with the atomic number 2 and the symbol He. The first member of the noble gas group in the periodic table, it is a colourless, odourless, tasteless, inert, monatomic gas that is not harmful.

Learn more about gravitational force here

brainly.com/question/12528243

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4 0

1 year ago

*Urgent* I WILL GIVE BRAINLIEST

Butoxors [25]

Answer:

walking to school

Explanation:

Driving a car to school

, and taking the bus to school both take up energy, unlike walking to school.

unless ur talking about energy, counting energy you produce and use to complete things, then it would be the 3rd one, taking the bus to school.

4 0

3 years ago