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:

ΔR_(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

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

The uncertainty of a sum is

      ΔR’= ΔR₁ + ΔR₂

We substitute the values

     ΔR_(e) / 400 = 50/500 + 100/2000 + (50 +100) / (500 + 2000)

     ΔR_(e) / 400 = 0.1 + 0.05 + 0.06

     ΔR_(e) = 0.21 400

     ΔR_(e) = 84   Ω

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

    R_(e) = (40 ± 8) 10¹   Ω


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A pressure antinode in a sound wave is a region of high pressure, while a pressure node is a region of low pressure.True
False

Answers

A pressure antinode in a sound wave is not a region of high pressure, while a pressure node is not a region of low pressure.

The answer is false

Final answer:

A pressure antinode in a sound wave is indeed a region of high pressure, while a pressure node is a region of low pressure. These definitions hold true for all types of waves.

Explanation:

That's true. In terms of sound waves, a pressure antinode is a region of high pressure, while a pressure node is a region of low pressure. This is true for all types of waves, not only sound waves. In essence, a wave moves through a medium (in case of a sound wave, that medium is typically air) by creating areas of high and low pressure - the high pressure areas are called antinodes, and the low pressure areas are called nodes.

Learn more about Sound Waves here:

brainly.com/question/31375531

#SPJ2

1. Hoai Nguyen, a Physics 2A student, drop a soccer ball from the roof of the new science building. The ball strikes the ground in 3.30 s later. You may ignore air resistance, so the ball is in free fall. How tall, in meters, is the building? 2. How fast was the ball moving right before hitting the floor?

3. Thu Tran, another Physics 2A student, grabs the ball and kicks it straight up to Hoai Nguyen, who is still up on the building rooftop. Assuming that the ball is kicked at 0.50 m above the ground and it goes on a vertical path, what is the minimum velocity required for the ball to make it to the building rooftop? Ignore air resistance. (Hint: the ball will pass the rooftop level with a higher speed...)

Answers

Answer:

1. 53.415 m

2. 32.373 m/s

3. 30.82 m/s

Explanation:

Let g = 9.81 m/s2. We can use the following equation of motion to calculate the distance traveled by the ball in 3.3s, and the velocity it achieved

1.s = gt^2/2 = 9.81*3.3^2/2 = 53.415 m

2.v = gt = 9.81*3.3 = 32.373 m/s

3. If the ball is kicked at 0.5 m above the ground then the net distance between the ball and the roof top is

53.415 - 0.5 = 48.415 m

For the ball to at least make it to the roof top at speed v = 0 m/s. We can use the following equation of motion to calculate the minimum initial speed

v^2 - v_0^2 = 2g\Delta s

where v = 0 m/s is the final velocity of the ball when it reaches the rooftop, v_0 is the initial velocity, \Delta s = 48.415 is the distance traveled, g = -9.81 is the gravitational acceleration with direction opposite with velocity

0 - v_0^2 = 2*(-9.81)*48.415

v_0^2 = 950

v_0 = √(950) = 30.82 m/s

A 120-V rms voltage at 60.0 Hz is applied across an inductor, a capacitor, and a resistor in series. If the peak current in this circuit is 0.8484 A, what is the impedance of this circuit

Answers

Answer:

200 \Omega

Explanation:

The computation of the impedance of the circuit is shown below:

Provided that

RMS voltage = 120 v

Frequency = 60.0 Hz

RMS current = 0.600 A

Based on the above information, the formula to compute the impedance is

Z=(V_(max))/(I_(peak))

where,

V_(max) = √(2) * V_(rms)

= √(2) * 120

= 169.7 V

And, I_Peak = 0.8484

Now placing these above values to the formula

So, the impedance of the circuit is

= (169.7)/(0.8484)

= 200 \Omega

Consider a vertical elevator whose cabin has a total mass of 800 kg when fully loaded and 150 kg when empty. The weight of the elevator cabin is partially balanced by a 400-kg counterweight that is connected to the top of the cabin by cables that pass through a pulley located on top of the elevator well. Neglecting the weight of the cables and assuming the guide rails and the pulleys to be frictionless, determine (a) the power required while the fully loaded cabin is rising at a constant speed of 1.2 m/s and (b) the power required while the empty cabin is descending at a constant speed of 1.2 m/s. What would your answer be to (a) if no counterweight were used? What would your answer be to (b) if a friction force of 800 N has developed between the cabin and the guide rails?

Answers

Answer:

Part a)

P = 4.71 * 10^3 Watt

Part b)

P = 2.94 * 10^3 W

Part c)

P = 9.4 * 10^3 W

Part d)

P = 3.9 * 10^3 W

Explanation:

Part a)

When cabin is fully loaded and it is carried upwards at constant speed

then we will have

net tension force in the rope = mg

T = (800)(9.81)

T = 7848 N

now it is partially counterbalanced by 400 kg weight

so net extra force required

F = 7848 - (400 * 9.81)

F = 3924 N

now power required is given as

P = Fv

P = 3924 (1.2)

P = 4.71 * 10^3 Watt

Part b)

When empty cabin is descending down with constant speed

so in that case the force balance is given as

F + (150 * 9.8) = (400 * 9.8)

F = 2450 N

now power required is

P = F.v

P = (2450)(1.2)

P = 2.94 * 10^3 W

Part c)

If no counter weight is used here then for part a)

F = 7848 N

now power required is

P = F.v

P = 7848 (1.2)

P = 9.4 * 10^3 W

Part d)

Now in part b) if friction force of 800 N act in opposite direction

then we have

F + (150 * 9.8) = 800 +(400 * 9.8)

F = 3250 N

now power is

P = (3250)(1.2)

P = 3.9 * 10^3 W

What would the position of arrows on a target need to be to illustrate measurements that are neither accurate nor precise

Answers

Answer:

The position of the arrows will not be on the target i.e. outside the bull's eye, neither will they be close to one another (widely scattered).

Explanation:

Accuracy refers to the closeness of a measurement to an actual or accepted value while precision refers to the closeness of measurements to one another.

Using archery as an illustration of precision and accuracy, measurements (arrows) that are neither accurate not precise are those arrows that will be far away or outside the bull's eye region (target) of the board and also far apart from one another.

In a nutshell, the arrows will be distant from the bull's eye or target (not accurate) and also distant from one another (not precise).

The sun does not normally affect the tides to a noticeable degree. however, under certain circumstances the gravitational pull of the sun does affect the tides. under what two (2) conditions.the sun does not normally affect the tides to a noticeable degree. however, under certain circumstances the gravitational pull of the sun does affect the tides. under what two (2) conditions.

Answers

The answers are :
1)  when the sun, moon, and earth are in a line only

2)  when the gravitational forces of the Moon and the Sun are
     perpendicular to one
another with respect to the Earth.  

Answer:

High and low tides are result of combined effect of gravitational pull of the sun and the moon. When the two align in a straight line, the range of tides is maximum. This happens on new moon and full moon day.

On the other hand, when the sun and the moon align at right angles, the effect of gravity is minimum and the range of the tides is minimum.