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hoa [83]
3 years ago
15

A parallel-plate capacitor is disconnected from a battery, and the plates are pulled a small distance farther apart. Does Q incr

ease, decrease, or stay the same?
Physics
1 answer:
kolezko [41]3 years ago
4 0

Answer:

Q stay the same

Explanation:

Charging of capacitor is done by battery . If battery is disconnected , charging will stop . There will not be any discharging as plates are separate . So pulling the plates apart will not affect the charge lying on the capacitor . It will decrease its capacity and increase its potential , keeping its charge constant.

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A 10 kg mass car initially at rest on a horizontal track is pushed by a horizontal force of 10 N magnitude. If we neglect the fr
vlada-n [284]

Answer:

50 m

Explanation:

F = ma

10 N = (10 kg) a

a = 1 m/s²

Given:

v₀ = 0 m/s

a = 1 m/s²

t = 10 s

Find: Δx

Δx = v₀ t + ½ at²

Δx = (0 m/s) (10 s) + ½ (1 m/s²) (10 s)²

Δx = 50 m

5 0
3 years ago
A mole of a monatomic ideal gas at point 1 (101 kPa, 5 L) is expanded adiabatically until the volume is doubled at point 2. Then
Paha777 [63]

Answer:

(a). Check attachment.

(b). 280.305 J.

(c). 31.81 kpa; 38.26K.

(d). 24.05K.

(e). 24.05k; 40kpa.

(f). -138.6J.

Explanation:

(a). Kindly check the attached picture for the diagram showing the four process.

1 - 2 = adiabatic expansion process.

2 - 3 = Isochoric process.

3 - 4 = isothermal process.

4 - 1 = isochoric process.

(b). Recall that the process from 1 to is an adiabatic expansion process.

NB: b = 5/3 for a monoatomic gas.

Then, the workdone = (1/ 1 - 1.66) [ (p1 × v1^b)/ v2^b × v2 - (p1 × v1)].

= ( 1/ 1 - 5/3) [ (101 × 5^5/3) × 10^1 -5/3] - 101 × 5.

Thus, the workdone = 280.305 J.

(c). P2 = P1 × V1^b/ V2^b = 101 × 5^5/3/ 10^5/3 = 31.81 kpa.

T2 = P2 × V2/ R × 1 = 31.81 × 10/ 8.324 = 38.36k.

(d). The process 2 - 3 is an Isochoric process, then;

T3 = T2/P2 × P3 = 38.26/ 31.82 × 20 = 24.05K.

(e). The process 3 - 4 Is an isothermal process. Then, the temperature at 4 will be the same temperature at 3. Tus, we have the temperature; point 3 = point 4 = 24.05k.

The pressure can be determine as below;

P4 = P3 × V3/ V4 = 20 × 10/ 5 = 200/ 5 = 40 kpa.

(f) workdone = xRT ln( v4/v3) = 1 × 8.314 × 24.05 × ln (5/10) = - 138.6 J

6 0
3 years ago
Si tenemos tres cubos del mismo tamaño (hierro, madera e icopor). ¿Qué diferencias puede encontrar entre ellos?
cricket20 [7]

Answer: La diferencia es el peso (o la masa), siendo que el cubo de hierro es el mas pesado, después viene el de madera y después el de icopor.

Explanation:

Ok, los 3 cubos tienen el mismo tamaño, lo que implica que tienen el mismo volumen.

Ahora es útil recordar la relación:

Densidad = masa/volumen.

Masa = densidad*volumen.

Nosotros sabemos que la densidad del hierro es mas grande que la de la madera, y la densidad de la madera es mas grande que la del icopor.

Entonces, por la relación anterior, y sabiendo que todos los cubos tienen el mismo volumen, podemos reconocer que el cubo de hierro tiene mayor masa, después viene el de madera y después viene el de icopor.

Y sabiendo que:

masa*gravedad = peso

podemos saber que el cubo mas pesado es el de hierro, después el de madera y después el de icopor.

Además de esta diferencia, también hay otras que no dependen tanto del tamaño del objeto, como pueden ser las capacidades caloríficas, el como reaccionan a campos eléctricos y cosas así que son triviales, pues son diferentes para casi todos los materiales.

5 0
3 years ago
A piece of paper looks white in both the noonday sun and under moonlight, even though there is less light being reflected off th
ddd [48]
Answer: Brightness consistency

There are three types of perceptual consistency
Types of Perceptual Constancy: Shape, Size, and Brightness Size constancy
Since the moon and sun affect light, brightness consistency is occurring.
Brightness constancy is our visual ability to perceive objects as having the same level of brightness even though the level of lighting changes.
7 0
3 years ago
Speed and Motion you went from the starting line to the finish line at different rates. If you repeated the activity while carry
Phantasy [73]

Answer:

It will cause kinetic energy to increase.

Explanation:

Given that Speed and Motion you went from the starting line to the finish line at different rates.

If you repeated the activity while carrying weights but keeping your times the same, the weight carried will add up to the mass of the body.

And since Kinetic energy K.E = 1/2mv^2

Increase in the mass of the body will definitely make the kinetic energy of the body to increase.

Since the time is the same, that means the speed V is the same.

Weight W = mg

m = W/g

The new kinetic energy will be:

K.E = 1/2(M + m)v^2

This means that there will be increase in kinetic energy.

3 0
3 years ago
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