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marishachu
1 month ago
6

Two Resistances R1 = 3 Ω and R2 = 6 Ω are connected in series with an ideal battery supplying a voltage of ∆ = 9 Volts. Sketch t

his circuit diagram. Now, replace the two resistors with an equivalent resistance R connected to the same battery. Sketch this circuit. (a) What is current I in R? (b) What is the potential difference V across R? Using this information, answer the following questions about the original, two-resistor circuit. (c) What is the current I1 in R1? (d) What is the current I2 in R2? (e) What is the potential difference V1 across R1? (f) What is in the potential difference V2 across R2? (g) How are V1 and V2 related to the battery voltage? Comparing the two circuits: (h) How are I1 and I2 related to I? (i) How are V1 and V2 related to ∆?
Two Resistances R1 = 3 Ω and R2 = 6 Ω are connected in parallel with an ideal battery supplying a voltage of ∆ =
9 Volts. Now, replace the two resistors with an equivalent resistance R connected to the same battery. Sketch this circuit. (a) What is current I in R? (b) What is the potential difference V across R? Using this information, answer the following questions about the original, two-resistor circuit. (c) What is the current I1 in R1? (d) What is the current I2 in R2? (e) What is the potential difference V1 across R1? (f) What is in the potential difference V2 across R2? (g) How are V1 and V2 related to the battery voltage? Comparing the two circuits: (h) How are I1 and I2 related to I? (i) How are V1 and V2 related to ∆?

Physics
1 answer:
Ostrovityanka [3.2K]1 month ago
4 0
For the requested solution, please refer to the attachment.
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A solid conducting sphere carrying charge q has radius a. It is inside a concentric hollow conducting sphere with inner radius b
Softa [3030]

Response:

Clarification:

Refer to the diagram indicating the charges on the specified sphere (see attachment).

The electric field at the stated positions is

E(r) = 0 for r≤a.  Equation 1

E(r) = kq/r² for a<r<b.   Equation 2

E(r) = 0 for b<r<c.      Equation 3

E(r) = kq/r² for r>c.    Equation 4.

We understand that electric potential correlates with the electric field through

V = Ed

A. To compute the potential at the outer surface of the hollow sphere (r=c), we determine that the electric field there is

E = kQ / r²

Then,

V = Ed,

At d = r = c

Thus,

Vc = (kQ / c²) × c

Vc = kQ / c

As a result, the total charge Q consists of +q, -q, and +q

Hence, Q = q - q + q = q

V = kq / c

B. To calculate the potential at the inner surface of the hollow sphere (r=b), we have

V = kQ/r

V = kQ / b,   noting that r = b

So, Q = q

V = kq / b

C. At r = a

Following from equation 1:

E(r) = 0 for r≤a.  Equation 1

The electric field at the surface of the solid sphere is 0, E = 0N/C

Thus,

V = Ed = 0 V

Consequently, the electric potential at the solid sphere's surface is 0.

D. At r = 0

The electric potential can be determined by

V = kq / r

As r approaches 0,

V = kq / 0

V approaches infinity.

8 0
1 month ago
The position of an object is given by x = at3 - bt2 + ct,where a = 4.1 m/s3, b = 2.2 m/s2, c = 1.7 m/s, and x and t are in SI un
serg [3582]

Answer:

The response to your inquiry is: 15 m/s²

Explanation:

Equation    x = at³ - bt² + ct

a = 4.1 m/s³

b = 2.2 m/s²

c = 1.7 m/s

First we calculate x at t = 4.1 s

x = 4.1(4.1)³ - 2.2(4.1)² + 1.7(4.1)

x = 4.1(68.921) - 2.2(16.81) + 6.97

x = 282.58 - 36.98 + 6.98

x = 252.58 m

Now we calculate speed

v = x/t = 252.58/ 4.1 = 61.6 m/s

Finally

acceleration = v/t = 61.6/4.1 = 15 m/s²

6 0
1 month ago
A communications satellite orbiting the earth has solar panels that completely absorb all sunlight incident upon them. The total
Ostrovityanka [3204]

Answer:

0.000047N

Explanation:

We know that

intensity (I) = P/ A

Where

P= power

A= Area

Thus, the power absorbed can be calculated as:

Power = Intensity x Area

This equals = 1.4 x 10^3 x(10)

Thus,

14000 Watts = 14 kWatt

However, the radiation pressure can be defined as

time-averaged intensity divided by the speed of light in a vacuum

So,

P = (1.4 x 1000)/c

Also,

F= P x A

Thus,

((1.4 x 1000)/(3 x10^8)) x 10

This results in

=0.000046666N

Rounded to two significant figures gives us

=0.000047 N

3 0
1 month ago
A 230mg piece of gold is hammered into a sheet measuring 23cm x 17cm. What is the thickness of the sheet in meters? Density of g
kicyunya [3294]
The sheet's thickness is calculated to be t= 0.0003 mm. The mass provided is m = 230 mg, equivalent to 0.23 g. The area of the sheet is A= 23 x 17 cm², totaling A= 391 cm². Given the density of gold is ρ = 19.32 g/cm³, we assume the sheet's thickness is t cm. From the equation Mass = Density x Volume, we know m = ρ A t. Substituting the values results in: 0.23 = 19.32 × 391 × t, leading to t = 0.000030 cm, or equivalently, 0.0003 mm as the final thickness.
7 0
8 days ago
A 50-kg load is suspended from a steel wire of diameter 1.0 mm and length 11.2 m. By what distance will the wire stretch? Young'
Ostrovityanka [3204]

Answer:

3.5 cm

Explanation:

mass, m = 50 kg

diameter = 1 mm

radius, r = half the diameter = 0.5 mm = 0.5 x 10^-3 m

L = 11.2 m

Y = 2 x 10^11 Pa

Cross-sectional area of the wire = π r² = 3.14 x 0.5 x 10^-3 x 0.5 x 10^-3

= 7.85 x 10^-7 m^2

Let the change in length of the wire be ΔL.

The equation for Young's modulus is given by

Y =\frac{mgL}{A\Delta L}

\Delta L =\frac{mgL}{A\times Y}

ΔL = 0.035 m = 3.5 cm

Thus, the wire stretches by 3.5 cm.

5 0
1 month ago
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