in a thermal power plant, heat from the flue gases is recovered in (A) chimney (B) de-super heater (C) economizer (D) condenser

If R is the total resistance of three resistors, connected in parallel, with resistances R1, R2, R3, then 1 R = 1 R1 + 1 R2 + 1

ANTONII [103]

Answer:

$\Delta R_{max} = 0.46 ohm$

Explanation:

Resistance is given of different values

now we have

$R_1 = 64 ohm$

$R_2 = 20 ohm$

$R_3 = 8 ohm$

possible error in all resistance is 0.5 %

now we know that

$\Delta R_1 = 0.005\times 64 = 0.32 ohm$

$\Delta R_2 = 0.005 \times 20 = 0.1 ohm$

$\Delta R_2 = 0.005 \times 8 = 0.04 ohm$

Now the maximum possible error when all resistance is connected in series

$\Delta R_{max} = \Delta R_1 + \Delta R_2 + \Delta R_3$

$\Delta R_{max} = 0.32 + 0.1 + 0.04$

$\Delta R_{max} = 0.46 ohm$

7 0

2 years ago

Use Wien’s Law to calculate the peak wavelength of Betelgeuse, based on the temperature found in Question #8. Note: 1 nanometer

kodGreya [7K]

The peak wavelength of Betelgeuse is 828 nm

Explanation:

The relationship between surface temperature and peak wavelength of a star is given by Wien's displacement law:

$\lambda=\frac{b}{T}$

where

$\lambda$ is the peak wavelength

T is the surface temperature

$b=2.898\cdot 10^{-3} m\cdot K$ is Wien's constant

For Betelgeuse, the surface temperature is approximately

T = 3500 K

Therefore, its peak wavelength is:

$\lambda=\frac{2.898\cdot 10^{-3}}{3500}=8.28\cdot 10^{-7} m = 828 nm$

Learn more about wavelength:

brainly.com/question/5354733

brainly.com/question/9077368

#LearnwithBrainly

8 0

2 years ago

The leaning tower of Pisa is about 56 meters tall. A ball released from the top takes 3.4 seconds to reach the ground. The final

geniusboy [140]

S=56, u=0, v=33, a=?, t=3.4

v=u+at
33=3.4 a
a = 9.7m/s^2

7 0

2 years ago

Read 2 more answers

Consider a 4-mg raindrop that falls from a cloud at a height of 2 km. When the raindrop reaches the ground, it won't kill you or

docker41 [41]

Answer:

The work done by the air resistance is -0.0782 J

Explanation:

Hi there!

The energy of the raindrop has to be conserved, according to the law of conservation of energy.

Initially, the raindrop has only gravitational potential energy:

PE = m · g · h

Where:

PE = potential energy.

m = mass of the raindrop.

g = acceleration due to gravity (9.8 m/s²)

h = height.

Let´s calculate the initial potential energy of the drop:

(convert 4 mg into kg: 4 mg · 1 kg / 1 × 10⁶ mg = 4 × 10⁻⁶ kg)

PE = 4 × 10⁻⁶ kg · 9.8 m/s² · 2000 m

PE = 0.0784 J

When the drop starts falling, some of the potential energy is converted into kinetic energy and some energy is dissipated by the work done by the air resistance. On the ground all the initial potential energy has been either converted into kinetic energy or dissipated by the resistance of the air:

initial PE = final KE + W air

Where:

KE = kinetic energy.

W air = work done by the air resistance.

The kinetic energy when the raindrop reaches the ground is calculated as follows:

KE = 1/2 · m · v²

Where:

m = mass

v = velocity

Then:

KE = 1/2 · 4 × 10⁻⁶ kg · (10 m/s)²

KE = 2 × 10⁻⁴ J

Now, we can calculate the work done by the air resistance:

initial PE = final KE + W air

0.0784 J = 2 × 10⁻⁴ J + W air

W air = 0.0784 J - 2 × 10⁻⁴ J

W air = 0.0782 J

Since the work is done in the opposite direction to the displacement, the work is negative, then, the work done by the air resistance is -0.0782 J.

5 0

2 years ago

A 0.20-kg object attached to the end of a string swings in a vertical circle (radius = 80 cm). at the top of the circle the spee

gayaneshka [121]

Answer:

Tension in the string at this position: 3.1 N.

Explanation:

Convert the radius of the circle to meters:

$r = 80\;\text{cm} = 0.80\;\text{m}$ .

What's the net force on the object?

The object is in a circular motion. As a result,

$\displaystyle \Sigma F = \frac{m\cdot v^{2}}{r}$ ,

where

$\Sigma F$ is the net force on the object,
$m$ is the mass of the object,
$v$ is the velocity of the object, and
$r$ is the radius of the circular motion.

For this object,

$\displaystyle \Sigma F = \frac{0.20\times {4.5}^{2}}{0.80} = 5.0625\;\text{N}$ .

The output unit of net force should be standard if the unit for mass, velocity, and radius are all standard. The net force shall always point towards the center. In this case the net force points downwards.

What are the forces on this object?

There are two forces on the object at this moment:

Weight, $W$ , which points downwards. $W = m\cdot g = 0.20\times 9.81 = 1.962\;\text{N}$ .
Tension, $T$ , which also points downwards. The size of the tension force needs to be found.

What's the size of the tension force?

Gravity and tension points in the same direction. The size of their resultant force is the sum of the two forces. In other words,

$\Sigma F = T + W$ .

$T = \Sigma F - W = 5.0625 - 1.962 = 3.1$ .

All three values in this question are given with two sig. fig. Round the value of $T$ to the same number of significant figures.

4 0

2 years ago