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3 years ago

What is electric potential

Physics

2 answers:

Assoli18 [71]3 years ago

3 0

Answer:

The electric potential is the amount of work energy needed to move a unit of electric charge from a reference point to the specific point in an electric field with negligible acceleration of the test charge to avoid producing kinetic energy or radiation by test charge.

SI unit: volt

Other units: statvolt

In SI base units: V = kg⋅m^2⋅A^−1⋅s^−3

Dimension: M L^2 T^−3 I^−1

Explanation:

Hope it is helpful....

Fofino [41]3 years ago

3 0

Explanation:

common symbol - V

SI unit - volt

other unit - statvolt

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Una tractomula se desplaza con rapidez de 69 km/h. Cuando el conductor ve una vaca atravesada enmedio de la carretera, acciona l

Anestetic [448]

Answer:

Los datos que tenemos:

Rapidez: 69km/h

Tiempo que tarda en frenar = 4s.

Distancia inicial entre la tracto-mula y la vaca = 25m

Ok, la ecuación de desaceleración es:

D = (sf - si)/t

sf = velocidad final = 0m/s

si = velocidad inicial = 69km/h

t = tiempo = 4s

D = -69km/h/4s

ok, 1h = 3600s

D = (-69km/s)*1/(4*3600s) = -0.0048 km/s^2

Entonces la ecuación de aceleración es:

a(t) = -0.0048 km/s^2

Para la velocidad, integramos sobre el tiempo

v(t) = (-0.0048 km/s^2)*t + v0

donde v0 es la velocidad inicial, en este caso v0 = 69km/3600s = 0.0191km/s

v(t) = (-0.0048 km/s^2)*t + 0.0191km/s

Para la posición volvemos a integrar sobre el tiempo, esta vez suponemos la posición inicial igual a cero.

p(t) = (1/2)*(-0.0048 km/s^2)*t^2 + 0.0191m/s*t

Ahora, si p(t=4s) < 25m, esto implica que la tracto-mula no impacto con la vaca.

p(4s) = (1/2)*(-0.0048 km/s^2)*(4s)^2 + 0.0191km/s*4s = 0.038km

y 1km = 1000m

0.038km = 0.038*1000m = 38m

Entonces si, atropello a la vaca.

4 0

4 years ago

A vibrating pendulum makes 20 complete vibrations in 12.2 seconds. Determine the vibrational period of the pendulum, along with

Hitman42 [59]

Answer:

And the same can be said of a pendulum vibrating about a fixed position or of a guitar ... along the time axis, it is possible to determine the time for one complete cycle. ... It takes 2.3 seconds to complete the first full cycle of vibration. ... Since the standard metric unit of time is the second, frequency has units of cycles/second.

Explanation:

Plz mark brainliest im only 12. and i have never been marked brainliest.

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3 years ago

Car 1 drives 20 mph to the south, and car 2 drives 30 mph to the north. From the frame of reference of car 1, what is the veloci

EleoNora [17]

We subtract the velocity of car 1 from the velocity of car 2:
$v=(30\ mph\ North)-(20\ mph\ South)$
$=(30\ mph\ North)+(20\ mph\ North)$
$=50\ mph\ North$

4 0

3 years ago

Read 2 more answers

A cannonball with a mass of 1.0 kilogram is fired horizontally from a 500.-kilogram cannon, initially at rest, on a horizontal,

vfiekz [6]

The impulse J is equal to the magnitude of the force applied to the cannonball times the time it is applied:
$J=F \Delta t$
But the impulse is also equal to the change in momentum of the cannonball:
$J=\Delta p$
If we put the two equations together, we find
$F \Delta t= \Delta p$
And since we know the magnitude of the average force and the time, we can calculate the change in momentum:
$\Delta p= F \Delta t=(8.0 \cdot 10^3 N)(1.0 \cdot 10^{-1} s)=800 kg m/s$

7 0

3 years ago

A 300 gg ball on a 70-cmcm-long string is swung in a vertical circle about a point 200 cmcm above the floor. The string suddenly

Ratling [72]

Answer:

the tension in the string an instant before it broke = 34 N

Explanation:

Given that :

mass of the ball m = 300 g = 0.300 kg

length of the string r = 70 cm = 0.7 m

At highest point, law of conservation of energy can be expressed as :

$\frac{1}{2} mv^2 = mgh\\\\v = \sqrt{2gh}\\\\v = \sqrt{2*(9.8 \ m/s^2)*(6.00 \ m - 2.00 \ m)}\\\\$

$v = 8.854 \ m/s$

The tension in the string is:

$T = \frac{mv^2}{r}\\\\T = \frac{(0.300 \ kg)*(8.854 \ m/s^2)}{0.70 \ m}\\\\T = 33.59 N\\\\T = 34 \ N$

Thus, the tension in the string an instant before it broke = 34 N

6 0

3 years ago

What is electric potential ​

What is electric potential