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

Three point charges are positioned on the x axis. If the charges and corresponding positions are +32 µC at x = 0, +20 µC at x =

amid [387]

Answer:

Magnitude of the electrostatic force on the +32 µC charge, $F_{net} = 12 N$

Explanation:

Let q₁ = +32 µC, x₁ = 0

q₂ = +20 µC, x₂ = 40 cm = 0.4 m

q₃ = -60 µC, x₃ = 60 cm = 0.6 m

Let magnitude of the electrostatic force on the +32 µC charge due to the + 20 µC charge = F₁ (i.e force on q₁ due to q₂)

$F_{2} = \frac{kq_{1}q_{2} }{x_{2}^2 }$

$F_{2} = \frac{9 * 10^{9} * 32 * 10^{-6} * 20 * 10^{-6} }{0.4^2 }\\F_{2} = 36 N$

Let magnitude of the electrostatic force on the +32 µC charge due to the -60 µC charge = F₂ (i.e force on q₁ due to q₃)

$F_{3} = \frac{kq_{1}q_{3} }{x_{3}^2 }$

$F_{3} = -\frac{9 * 10^{9} * 32 * 10^{-6} * 60 * 10^{-6} }{0.6^2 }\\F_{3} =-48 N$

The electrostatic force on the 32 µC charge, $F_{net} = |F_{2} + F_{3}|$

$F_{net} =| 36 + (-48)| \\F_{net} =|- 12 N| \\ F_{net} = 12 N$

7 0

1 year ago

An instructor’s laser pointer produces a beam of light with a circular cross section of diameter 0.900 mm and a total power outp

il63 [147K]

Answer:

$E_0 = 2180.53N/C.$

$B_0 = 7.27*10^{-6}T.$

$U_{avg} = 4.2*10^{-5}J/m^3.$

Explanation:

The Intensity $I$ of the beam is

$I = P /A$

The diameter of the beam is 0.900mm; therefore, the area is

$A = \pi( \dfrac{0.900*10^{-3}}{2} )^2$

$A = 6.36*10^{-7}m^2$

and since $P = 4.00*10^{-3}W$ , the intensity of the beam is

$I = \dfrac{4.00*10^{-3}W}{6.36*10^{-7}m^2}$

$\boxed{I = 6289.3W/m^2.}$

Now, the intensity $I$ is related to $E_0$ by the relation

$I = \dfrac{E_0^2}{2\mu_0 c}$

solving for $E_0$ we get

$E_0 = \sqrt{2\mu_0 c I}$

putting in the numbers we get:

$E_0 = \sqrt{2(1.26*10^{-6}) (3*10^8) (6289.3)}$

$\boxed{E_0 = 2180.53N/C.}$

The amplitude of magnetic field $B_0$ is related to $E_0$ by

$B_0 = \dfrac{E_0}{c}$

putting in numerical values we get:

$B_0 = \dfrac{2180.53}{3*10^8}$

$\boxed{B_0 = 7.27*10^{-6}T. }$

The average energy density of the laser light is

$U_{avg} = \epsilon_0 E_0^2$

$U_{avg} = (8.85*10^{-12}) (2180.53)^2$

$\boxed{U_{avg} = 4.2*10^{-5}J/m^3.}$

8 0

2 years ago

A car is driving east at 120. km/h from Toronto to Ottawa. The distance between the two cities is 425.5 km, how long will it tak

Ket [755]

Answer:

The time it will take for the driver to reach Ottawa is 3 hours 32 minutes and 45 seconds

Explanation:

The given parameters are;

Speed of the car = 120 km/h

Distance from Toronto to Ottawa = 425.5 km

The formula for speed is given as follows;

Speed = Distance/Time

Therefore, to find the time duration it takes from Toronto to Ottawa, we have;

Time duration = Distance from Toronto to Ottawa/(Speed of the car)

The time duration = 425.5/120 = 3.54583 hours = 212.75 min = 12765 seconds

The time it takes from Toronto to Ottawa while driving at 425.5 km/h = 12765 seconds.

4 0

1 year ago

The mass of the Sun is 2 × 1030 kg, and the mass of Saturn is 5.68 × 1026 kg. The distance between Saturn and the Sun is 9.58 AU

Maurinko [17]

The formula for computing the orbital time period of a body is given as:

T² = 4π²r³ / GM

where T is the time period, r is the distance between the two bodies, G is the gravitational constant and M is the mass of the body that is being orbited. If we compute this time using SI units, the working is:

9.58 AU is 1.43 x 10¹² meters

T = √[(4*π²*(1.43 x 10¹²)³) / (6.67 × 10⁻¹¹ * 2 x 10³⁰)]
T = 9.30 x 10⁸ seconds which is approximately 29 years

Using the astronomical units, distance is in astronomical units and the mass is in solar masses. In these conditions, the ratio:
4π²/G = 1 so

T² = a³ (since the solar mass of the sun is 1)

T = √(9.58)³
T = 27 years

3 0

1 year ago

Read 2 more answers

Yamel is heating up leftover mashed potatoes from Thanksgiving. She forgot to put gravy on them. So she puts the cold gravy on t

monitta

Answer:

The correct answer is - option A. The mashed potatoes will transfer heat into the gravy.

Explanation:

In this case, where Yamel heats the mashed potatoes but forgets to heat the gravy and put the cold gravy on the hot mashed potatoes. Heat always transfers from the high-temperature object to the low-temperature object. So the hot mashed potatoes will transfer the heat to the gravy according to option A. Cold is not a form of heat but the condition of absence of heat or very low temperature.

Thus, the correct answer is - option A. The mashed potatoes will transfer heat into the gravy.

3 0

1 year ago