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ch4aika [34]
3 years ago
15

a light bulb has a resistance of 360 . what is the current in the bulb when it has a potential difference of 120 v across it? 0.

33 a 3 a 480 a 43,200 a
Physics
2 answers:
xenn [34]3 years ago
7 0
Presume we are looking for the current:

V = IR

120 = I*360

120/360 = I

1/3 = I

I = 1/3 = 0.333..

Current ≈ 0.33 Ampere.
tekilochka [14]3 years ago
6 0

Answer:

i = 0.33 A

Explanation:

As per ohm's law we know that voltage across any ohmic resistor is product of resistance and current through it

So it is given as

V = i R

now we know that

V = 120 Volts

R = 360 ohm

so here we have

120 = 360 \times i

i = \frac{1}{3} A

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How is force related to math
dybincka [34]

Answer:

Newton's second law of motion describes the relationship between force and acceleration. They are directly proportional. If you increase the force applied to an object, the acceleration of that object increases by the same factor. In short, force equals mass times acceleration.

Explanation:

8 0
3 years ago
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A ball is thrown horizontally from a window that is 15.4 meters high at a speed of 3.01 m/s. How far will the ball go before hit
tia_tia [17]

The distance travelled by the ball that is thrown horizontally from a window that is 15.4 meters high at a speed of 3.01 m/s is 5.34 m

s = ut + 1 / 2 at²

s = Distance

u = Initial velocity

t = Time

a = Acceleration

Vertically,

s = 15.4 m

u = 0

a = 9.8 m / s²

15.4 = 0 + ( 1 / 2 * 9.8 * t² )

t² = 3.14

t = 1.77 s

Horizontally,

u = 3.01 m / s

a = 0 ( Since there is no external force )

s = ( 3.01 * 1.77 ) + 0

s = 5.34 m

Therefore, the distance travelled by the ball before hitting the ground is 5.34 m

To know more about distance travelled

brainly.com/question/12696792

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7 0
1 year ago
A car initially at rest, accelerates at a constant rate of 4.0 m/s for 6s. How fast will the car be traveling at 6s
Katen [24]
It will be traveling exactly 24 miles per hour <span />
7 0
3 years ago
Please Help!<br> _______________
marishachu [46]

Answer:

constant at the speed of light

8 0
3 years ago
Consider an electron with charge −e and mass m orbiting in a circle around a hydrogen nucleus (a single proton) with charge +e.
alexandr1967 [171]

Answer:

v=\sqrt{k\frac{e^2}{m_e r}}, 2.18\cdot 10^6 m/s

Explanation:

The magnitude of the electromagnetic force between the electron and the proton in the nucleus is equal to the centripetal force:

k\frac{(e)(e)}{r^2}=m_e \frac{v^2}{r}

where

k is the Coulomb constant

e is the magnitude of the charge of the electron

e is the magnitude of the charge of the proton in the nucleus

r is the distance between the electron and the nucleus

v is the speed of the electron

m_e is the mass of the electron

Solving for v, we find

v=\sqrt{k\frac{e^2}{m_e r}}

Inside an atom of hydrogen, the distance between the electron and the nucleus is approximately

r=5.3\cdot 10^{-11}m

while the electron mass is

m_e = 9.11\cdot 10^{-31}kg

and the charge is

e=1.6\cdot 10^{-19} C

Substituting into the formula, we find

v=\sqrt{(9\cdot 10^9 m/s) \frac{(1.6\cdot 10^{-19} C)^2}{(9.11\cdot 10^{-31} kg)(5.3\cdot 10^{-11} m)}}=2.18\cdot 10^6 m/s

7 0
3 years ago
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