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3 years ago

6

SAT QUESTION // A capacitor is connected to a variable power supply with potential V. Determine the effect on the charge stored

on the capacitor if the potential of the power supply is cut in half.

1 answer:

Illusion [34]3 years ago

6 0

Answer:

decrease by a factor of 2

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ZanzabumX [31]

Answer:

a) El valor de la densidad es 0.79 $\frac{g}{cm^{3} }$ o 790 $\frac{g}{cm^{3} }$

b) El peso especifico es 7749.9 $\frac{N}{m^{3} }$

Explanation:

a) La densidad se define como la propiedad que tiene la materia, ya sean sólidos, líquidos o gases, para comprimirse en un espacio determinado. En otras palabras, la densidad es una magnitud que permite medir la cantidad de masa que hay en determinado volumen de una sustancia. Entonces, la expresión para el cálculo de la densidad es el cociente entre la masa de un cuerpo y el volumen que ocupa:

$d=\frac{m}{V}$

En este caso:

masa= 237 g= 0,237 kg (siendo 1000 g= 1 kg)
volumen= 300 cm³= 0,0003 m³ (siendo 1 cm³= 0,000001 m³)

Reemplazando:

$d=\frac{237 g}{300cm^{3} }$ → d=0.79 $\frac{g}{cm^{3} }$

$d=\frac{0,237 kg}{0,0003m^{3} }$ → d=790 $\frac{g}{cm^{3} }$

<u><em>El valor de la densidad es 0.79 </em></u> $\frac{g}{cm^{3} }$ <u><em> o 790 </em></u> $\frac{g}{cm^{3} }$ <u><em></em></u>

b) El peso específico es la relación existente entre el peso y el volumen que ocupa una sustancia en el espacio.

Entonces, en este caso, siendo el peso:

P= m*g= 0,237 kg* 9,81 $\frac{m}{s^{2} }$ = 2,32497 N

el peso especifico es calculado como:

$Pe=\frac{Peso}{Volumen}= \frac{2,32497N}{0,0003 m^{3} }$

Pe= 7749.9 $\frac{N}{m^{3} }$

<u><em>El peso especifico es 7749.9</em></u> $\frac{N}{m^{3} }$ <u><em></em></u>

8 0

3 years ago

What can be said about the sign of the work done by the force f⃗ 1??

Tju [1.3M]

I did not get your question properly

5 0

3 years ago

According to the exercise principle of balance, a workout should __________.

xxMikexx [17]

<span>a. address all components of fitness

</span>According to the exercise principle of balance, a workout should address all components of fitness.
<span>NOT:
</span>b. focus on balance and coordination
<span>c. target specific areas for improvement </span>
<span>d. vary exercise activities over time</span>

6 0

3 years ago

Read 2 more answers

What is a concave lens?

ozzi

A concave lens is a lens that has at least one of its surfaces or both surfaces curved inwards. Due to this reason, this lens diverges the light that falls on it and hence is also called a diverging lens. The concave lens is thinner in the middle compared to its edges. These are used in flashlights, binoculars, telescopes, etc.
Please see attached image for reference.

5 0

3 years ago

A person 1.8m tall stands 0.75m from a reflecting globe in a garden.

Maru [420]

Answer:

1. The image of the person is 1.41 m, virtual and formed at the back of the surface of the globe.

2. The person's image is 3.38 m tall.

Explanation:

From the given question, object distance, u = 0.75 m, object height = 1.8 m, radius of curvature of the reflecting globe, r = 8 cm = 0.08 m.

f = $\frac{r}{2}$ = $\frac{0.08}{2}$ = 0.04 m

1. The image distance, v, can be determined by applying mirror formula:

$\frac{1}{f}$ = $\frac{1}{u}$ + $\frac{1}{v}$

$\frac{1}{0.04}$ = $\frac{1}{0.75}$ + $\frac{1}{v}$

$\frac{4}{100}$ - $\frac{75}{100}$ = $\frac{1}{v}$

$\frac{1}{v}$ = $\frac{4 - 75}{100}$

= - $\frac{71}{100}$

⇒ v = - $\frac{100}{71}$

= - 1.41 m

The image of the person is 1.41 m, virtual and formed at the back of the surface of the globe.

2. $\frac{image distance}{object distance}$ = $\frac{image height}{object height}$

$\frac{1.41}{0.75}$ = $\frac{v}{1.8}$

v = $\frac{2.538}{0.75}$

= 3.384

v = 3.38 m

The person's image is 3.38 m tall.

6 0

3 years ago