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Gre4nikov [31]
3 years ago
12

The charge stored in a parallel plate capacitor is proportional to the _____(potential, current) and across _____ (resistance, c

apacitance) the plates.
Physics
1 answer:
Aleksandr-060686 [28]3 years ago
4 0

Answer:

The charge stored in a parallel plate capacitor is proportional to the potential, and capacitance across the plates.

Explanation:

The capacitance of a parallel plate capacitor is given by:

C=\frac{Q}{V}

where

Q is the charge stored

V is the potential difference across the capacitor

Re-arranging the formula, we get

Q=CV

From this expression, we see that the charge stored (Q) is proportional to both the potential (V) and the capacitance (C).

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6.
Vikentia [17]

Answer:

Explanation:

Givens

vi = 0

a = 9.81

d = 4.50 m

vf = ?

Formula

vf^2 = vi^2 + 2 * a * d

Solution

Substitute the knowns into the formula

vf^2 =0 +  2 * 9.81 * 4.50

vf^2 = 88.29                          Take the square root of both sides.

sqrt(vf^2) = sqrt(88.29)    

vf = 9.40 m/s

3 0
3 years ago
¿Qué trabajo hace una fuerza de 110 N cuando mueve su punto de aplicación 20 mt en su misma dirección? *
lys-0071 [83]

Answer:

W = 2.200 Joules

Explanation:

Datos (data):

  • Fuerza [force] (F) = 110 N
  • Metros [meters] (m) = 20 m
  • Trabajo [work] (W) = ?

Usar la fórmula (use formula):

  • \boxed{\bold{W = F*d}}

Reemplazar (replace):

  • \boxed{\bold{W = 110\ N*20\ m}}

Resolver la multiplicación, recuerda que 1 N * 1 m = 1 J (resolve the multiplication, remember that 1 N * 1 m = 1 J:

  • \boxed{\boxed{\bold{W =2.200\ J}}}

Greetings.

7 0
3 years ago
The light rays in the illustration below do not properly focus at the focal point. this problem occurs with?
tankabanditka [31]

Answer:

The problem occurs with all spherical mirrors.

Spherical mirrors are practical up to about inches in diameter.

Reflecting telescopes use spherical mirrors for apertures up to about 4 ".

Larger aperture telescopes use parabolic mirrors to obtain sharp focus.

7 0
3 years ago
outward from a wall just above floor level. A 1.5 kg box sliding across a frictionless floor hits the end of the spring and comp
sweet [91]

Answer:

v = 0.489 m/s

Explanation:

It is given that,

Mass of a box, m = 1.5 kg

The compression in the spring, x = 6.5 cm = 0.065 m

Let the spring constant of the spring is 85 N/m

We need to find the velocity of the box (v) when it hit the spring. It is based on the conservation of energy. The kinetic energy of spring before collision is equal to the spring energy after compression i.e.

\dfrac{1}{2}mv^2=\dfrac{1}{2}kx^2

v=\sqrt{\dfrac{kx^2}{m}} \\\\v=\sqrt{\dfrac{85\times (0.065)^2}{1.5}} \\\\v=0.489\ m/s

So, the speed of the box is 0.489 m/s.

3 0
4 years ago
In your own words, describe conservation of mass. Use examples to support your answer.
olganol [36]

The law of conservation of mass states that mass in an isolated system is neither created nor destroyed by chemical reactions or physical transformations. According to the law of conservation of mass, the mass of the products in a chemical reaction must equal the mass of the reactants.

4 0
4 years ago
Read 2 more answers
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