PHYSICS COLLEGE

Two resistors are to be combined in parallelto form an equivalent resistance of 400Ω. The resistors are takenfrom available stock on hand as acquired over the years. Readily available are two common resistorsrated at 500±50 Ωand two common resistors rated at 2000 Ω±5%. What isthe uncertainty in an equivalent 400 Ωresistance?(Hint: the equivalent resistance connected in parallel can be obtained by 1212TRRRRR=+)

Answers

Answer 1

Answer:

Answer:

Δ $R_(e)$ = 84 Ω, $R_(e)$ = (40 ± 8) 10¹ Ω

Explanation:

The formula for parallel equivalent resistance is

1 / $R_(e)$ = ∑ 1 / Ri

In our case we use a resistance of each

R₁ = 500 ± 50 Ω

R₂ = 2000 ± 5%

This percentage equals

0.05 = ΔR₂ / R₂

ΔR₂ = 0.05 R₂

ΔR₂ = 0.05 2000 = 100 Ω

We write the resistance

R₂ = 2000 ± 100 Ω

We apply the initial formula

1 / $R_(e)$ = 1 / R₁ + 1 / R₂

1 / $R_(e)$ = 1/500 + 1/2000 = 0.0025

$R_(e)$ = 400 Ω

Let's look for the error (uncertainly) of Re

$R_(e)$ = R₁R₂ / (R₁ + R₂)

R’= R₁ + R₂

$R_(e)$ = R₁R₂ / R’

Let's look for the uncertainty of this equation

Δ $R_(e)$ / $R_(e)$ = ΔR₁ / R₁ + ΔR₂ / R₂ + ΔR’/ R’

The uncertainty of a sum is

ΔR’= ΔR₁ + ΔR₂

We substitute the values

Δ $R_(e)$ / 400 = 50/500 + 100/2000 + (50 +100) / (500 + 2000)

Δ $R_(e)$ / 400 = 0.1 + 0.05 + 0.06

Δ $R_(e)$ = 0.21 400

Δ $R_(e)$ = 84 Ω

Let's write the resistance value with the correct significant figures

$R_(e)$ = (40 ± 8) 10¹ Ω

Related Questions

The acceleration due to gravity on Earth is 9.80 m/s2. If the mass of a honeybee is 0.000100 kilograms, what is the weight of this insect?
When you blow some air above a paper strip, the paper, rises. This is because 1.) the air above the paper moves faster and the pressure is lower 2.) the air above moves faster and the pressure is higher 3.) the air above the paper moves faster and the pressure remains constant 4.) the air above the paper moves slower and the pressure is higher 5.) the air above the paper moves slower and the pressure is lower
A. With what speed must a ball be thrown vertically from ground level to rise to a maximum height of 45m?b. How long will it be in the air?
How does simple machines make work easier?
A piece of soft iron is placed in a solenoid increasing the magnetic field in an arrangement that can be switched on and off. Such a device is calleda. a hysteresis loop.b. an electrosolenoid.c. a permanent magnet.d. a ferromagnet.

A 5.49 kg ball is attached to the top of a vertical pole with a 2.15 m length of massless string. The ball is struck, causing it to revolve around the pole at a speed of 4.654.65 m/s in a horizontal circle with the string remaining taut. Calculate the angle, between 0° and 90°, that the string makes with the pole. Take ????=9.81g=9.81 m/s2. angle: °

Answers

Answer: $\theta =45.73^(\circ)$

Explanation:

Given

Length of string =2.15 m

mass of ball =5.49 kg

speed of ball=4.65 m/s

Here

Tension provides centripetal acceleration

$T\cos\theta =mg$ -----1

$T\sin \theta =(mv^2)/(r)$ ------2

Divide 2 & 1

$tan\theta =(v^2)/(rg)$

$tan\theta =(4.65^2)/(2.15* 9.8)$

$tan\theta =1.026$

$\theta =45.73^(\circ)$

Four +2 µC point charges are at the corners of a square of side 2 m. Find the potential at the center of the square (relative to zero potential at infinity) for each of the following conditions.(a) All the charges are positive(b) Three of the charges are positive and one is negative(c) Two are positive and two are negative

Answers

Answer:

(a) 51428.59 J/C

(b) 25714.29 J/C

(c) 0 J/C

Explanation:

Parameters given:

Q1 = 2 * 10^-6 C

Q2 = 2 * 10^-6 C

Q3 = 2 * 10^-6 C

Q4 = 2 * 10^-6 C

=> Q1 = Q2 = Q3 = Q4 = Q

Side of the square = 2m

The center of the square is the midpoint of the diagonals, i.e. Using Pythagoras theorem:

BD² = 2² + 2²

BD² = 8

BD = √(8) = 2.8m

OD = 1.4m

(The attached diagram explains better)

Hence, the distance between the center and each point charge, r, is 1.4m.

Electric Potential, V = kQ/r

k = Coulombs constant

(a) If all charges are positive:

V(Total) = V1 + V2 + V3 + V4

V1 = Potential due to Q1

V2 = Potential due to Q2

V3 = Potential due to Q3

V4 =Potential due to Q4

Since Q1 = Q2 = Q3 = Q4 = Q

=> V1 = V2 = V3 = V4

=> V(Total) = 4V1

V = (4 * 9 * 10^9 * 2 * 10^-6)/1.4

V = 51428.59J/C

(b) If 3 charges are positive and 1 is negative:

Since Q1 = Q2 = Q3 = Q

and Q4 = -Q

The total potential becomes:

V(Total) = V1 + V2 + V3 - V4

Since V1, V2, V3 and V4 have the same value,

V(Total) = V1 + V2

V(Total) = 2V1

V(Total) = (2 * 9 * 10^9 * 2 * 10^-6)/1.4

V(Total) = 25174.29 J/C

Since Q1 = Q2 = Q

and Q3 = Q4 = -Q

The total potential becomes:

V(Total) = V1 + V2 - V3 - V4

Since V1, V2, V3 and V4 have the same value,

V(Total) = 0 J/C

According to the World Flying Disk Federation, the world distance record for a flying disk throw in the men’s 85-years-and-older category is held by Jack Roddick of Pennsylvania, who on July 13, 2007, at the age of 86, threw a flying disk for a distance of 54.0 m. If the flying disk was thrown horizontally with a speed of 13.0 m/s, how long did the flying disk remain aloft? (Jack Roddick was also a physics teacher! Read more about him at

Answers

Answer:

t = 4.15 seconds

Explanation:

It is given that,

Distance traveled by a flying disk, d = 54 m

The speed at which it was thrown, v = 13 m/s

We need to find the time for which the flying disk remain aloft. Let the distance is d. We know that, speed is equal to the distance covered divided by time. So,

$t=(d)/(v)\n\nt=(54\ m)/(13\ m/s)\n\nt=4.15\ s$

Hence, for 4.15 seconds the flying disk remain aloft.

Singing that is off-pitch by more than about 1% sounds bad. How fast would a singer have to be moving relative to the rest of a band to make this much of a change in pitch due to the Doppler effect

Answers

Answer:

-3.396 m/s or 3.465 m/s

Explanation:

v = Speed of sound in air = 343 m/s

$v_s$ = Relative speed of the singer

f = Observed frequency

f' = Actual frequency

1% change can mean $f=1.01f'$

From the Doppler effect equation we have

$f=f'(v)/(v+v_s)\n\Rightarrow 1.01f'=f'(v)/(v+v_s)\n\Rightarrow 1.01=(343)/(343+v_s)\n\Rightarrow v_s=(343)/(1.01)-343\n\Rightarrow v_s=-3.396\ m/s$

The velocity is -3.396 m/s

when $f=0.99f'$

$f=f'(v)/(v+v_s)\n\Rightarrow 0.99f'=f'(v)/(v+v_s)\n\Rightarrow 0.99=(343)/(343+v_s)\n\Rightarrow v_s=(343)/(0.99)-343\n\Rightarrow v_s=3.46464646465\ m/s$

The velocity is 3.465 m/s

A ball is thrown into the air with 100 J of kinetic energy, which is transformed to gravitational potential energy at the top of its trajectory.When it returns to its original level after encountering air resistance, its kinetic energy is __________.

A) more than 100 J.

B) Not enough information given.

C) less than 100 J.

D) 100 J.

Answers

To solve this problem we could apply the concepts given by the conservation of Energy.

During the launch given in terms of kinetic energy and reaching the maximum point of the object, the potential energy of the body is conserved. However, part of all this energy is lost due to the work done by the friction force due to friction with the air, therefore

$E_T = PE + KE -W_f$

The potential and kinetic energy are conserved and are the same PE = KE and this value is equivalent to 100J, therefore

$E_T = 100-W_f$

The kinetic energy will ultimately be less than 100J, so the correct answer is C.

Enter your answer in the provided box. In water conservation, chemists spread a thin film of certain inert materials over the surface of water to cut down on the rate of evaporation of water in reservoirs. This technique was pioneered by Benjamin Franklin three centuries ago. Franklin found that 0.10 mL of oil could spread over the surface of water of about 32.0 m2 in area. Assuming that oil forms a monolayer (that is, a layer that is only one molecule thick) estimate the length of each oil molecule in nanometers. Assume that oil molecules are roughly cubic. (1 nm = 1 × 10−9 m)

Answers

Answer:

≅3.2 nm

Explanation:

Using the converter units as know for this case that:

1 ml is 1 cubic centimeter ⇒ 0.1 ml is 0.1 cubic centimeters

32.0 m² so :

32.0 m² *100 *100 cm² ⇒ 0.1 / ( 32.0 * 100 *100 ) = 100,000,000 * 0.1 / (32.0 * 100 * 100 ) nm

v = 100/32.0 nm = 3.125 nm thick.

v ≅3.2 nm

As oil is one molecule thick and the molecules are cubic, length of each oil is 3.2 nm

Other Questions

Tonya picks up a leaf from the ground and holds it at arm’s length. She lets go, and the leaf falls to the ground. What force pulled the leaf to the ground?

If Jim could drive a Jetson's flying car at a constant speed of 490 km/hr across oceans and space, approximately how long (in millions of years, in 106 years) would he take to drive to a nearby star that is 4.5 light-years away? Use 9.461 × 1012 km/light-year and 8766 hours per year (365.25 days). unanswered

A spring has a spring constant of 1350 N/m. You place the spring vertically with one end on the floor. You then drop a 1.3 kg book onto it from a height of 0.8 m above the top of the spring. Find the maximum distance the spring will be compressed. Express your answer with the appropriate mks units.

A 1.7 kg model airplane is flying north at 12.5 m/s initially, and 25 seconds later is observed heading 30 degrees west of north at 25 m/s. What is the magnitude of the average net force on the airplane during this time interval?