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3 years ago

The magnitude of the electric current is directly proportional to the _____ of the electric field.

2 answers:

8 0

Answer: The magnitude of the electric current is directly proportional to the potential difference of the electric field.

Explanation:

According to Ohm's law, the current flowing between two points in the circuit is directly proportional to the potential difference at a constant temperature.

The electric potential is the amount of work done in moving a unit positive charge from reference point to a particular point in a electric field. And the electric field is equal to the negative potential gradient.

Therefore, the magnitude of the electric current is directly proportional to the potential difference of the electric field.

Sedbober [7]3 years ago

3 0

Is proportional to the CHANGE.

hope this helps

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Make the following conversion: 34.9 cL = _____ hL

aliina [53]

We know that: 1 L = 100 cL. Or 1 cL = 0.01 L. Then we will make the conversion: 34.9 cL = 34.9 / 100 L = 0.349 L. Also: 1 hL = 100 L. 0.349 L = 0.349 / 100 hL = 0.00349 hL. This can be also written as: 3.49 * 10^(-3) hL ( in the scientific notation ). Answer: 3.49 cL = 0.00349 <span>hL </span>

8 0

4 years ago

The attraction or repulsion of two objects is called blank force what is the blank

AysviL [449]

Answer:

Coulomb Force

Explanation:

Coulomb force, also called electrostatic force or Coulomb interaction, attraction or repulsion of particles or objects because of their electric charge.

5 0

3 years ago

A ball on the end of a string is whirled around in a horizontal circle of radius 0.478 m. The plane of the circle is 1.02 m abov

m_a_m_a [10]

Answer:

Explanation:

We shall find first the velocity of ball at the time when string breaks. Let it be v . During its fall on the ground , 1.02 m below, we use the formula

h = 1/2 gt² where t is time of fall .

1.02 = 1/2 x 9.8 x t²

t²= .2081

t = .456

During this time it travels horizontally at distance of 2.5 m with uniform velocity of v

v x .456 = 2.5

v = 5.48 m /s

centripetal acceleration

= v² / r where r is radius of the circular path

= 5.48² / .478

= 62.82 m /s²

3 0

3 years ago

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 rigid tank contains nitrogen gas at 227 °C and 100 kPa gage. The gas is heated until the gage pressure reads 250 kPa. If the a

aleksley [76]

Answer:

T₂ =602 °C

Explanation:

Given that

T₁ = 227°C =227+273 K

T₁ =500 k

Gauge pressure at condition 1 given = 100 KPa

The absolute pressure at condition 1 will be

P₁ = 100 + 100 KPa

P₁ =200 KPa

Gauge pressure at condition 2 given = 250 KPa

The absolute pressure at condition 2 will be

P₂ = 250 + 100 KPa

P₂ =350 KPa

The temperature at condition 2 = T₂

We know that

$\dfrac{T_2}{T_1}=\dfrac{P_2}{P_1}\\T_2=T_1\times \dfrac{P_2}{P_1}\\T_2=500\times \dfrac{350}{200}\ K\\$

T₂ = 875 K

T₂ =875- 273 °C

T₂ =602 °C

5 0

4 years ago