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3 years ago

All of the following are examples of ways to improve energy efficiency of heating systems, except _______.

Physics

1 answer:

Yuri [45]3 years ago

5 0

Answer:

D. Opening the front door in winter to filter and clean warm ai

Explanation:

All of the following are examples of ways to improve energy efficiency of heating systems, except _______.

A. Closing off rooms that aren’t in use.

B. Frequently cleaning the furnace filter.

C. Fixing cracks in the walls and floor.

D. Opening the front door in winter to filter and clean warm air

for an heating systems it is better to keep the front door shut so that all the heat energy does not escape outside.

when heat energy is within a closed system ,it can still be utilized to keep the inhabitants warm otherwise there will be an exchange with the cold from winter fall outside the air, and the heating system's efficiency will be reduced

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It is noted that a sample of 134La with a half-life of 6.5 minutes has an activity of 2.6 Ci. What was its activity 72 minutes a

stira [4]

Answer:

36//

Explanation:

(½)^t

(½)^72

72/2=36//

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2 years ago

The diffuser in a jet engine is designed to decrease the kinetic energy of the air entering the engine compressor without any wo

wariber [46]

Explanation:

Expression for energy balance is as follows.

$\Delta E_{system} = E_{in} - E_{out}$

or, $E_{in} = E_{out}$

Therefore,

$m(h_{1} \frac{v^{2}_{1}}{2}) = m (h_{2} \frac{V^{2}_{2}}{2})$

$h_{1} + \frac{V^{2}_{1}}{2} = h_{2} + \frac{V^{2}_{2}}{2}$

Hence, expression for exit velocity will be as follows.

$V_{2} = [V^{2}_{1} + 2(h_{1} - h_{2})^{0.5}$

= $V^{2}_{1} + 2C_{p}(T_{1} - T_{2})]^{0.5}$

As $C_{p}$ for the given conditions is 1.007 kJ/kg K. Now, putting the given values into the above formula as follows.

$V_{2} = V^{2}_{1} + 2C_{p}(T_{1} - T_{2})]^{0.5}$

= $[(350 m/s)^{2} + 2(1.007 kJ/kg K) (30 - 90) K \frac{1000 m^{2}/s^{2}}{1 kJ/kg}]^{0.5}$

= 40.7 m/s

Thus, we can conclude that velocity at the exit of a diffuser under given conditions is 40.7 m/s.

5 0

3 years ago

. A magnetic field has a magnitude of 0.078 T and is uniform over a circular surface whose radius is 0.10 m. The field is orient

Dmitriy789 [7]

Answer:

The magnetic flux through surface is $2.22 \times 10^{-3}$ Wb

Explanation:

Given :

Magnitude of magnetic field $B = 0.078$ T

Radius of circle $r = 0.10$ m

Angle between field and surface normal $\theta =$ 25°

From the formula of flux,

$\phi = B.A$

$\phi = BA\cos \theta$

Where $\theta =$ angle between magnetic field line and surface normal, $A =$ area of circular surface.

$A = \pi r^{2}$

$A = 3.14 \times (0.10) ^{2}$

$A = 0.0314$ $m^{2}$

Magnetic flux is given by,

$\phi = 0.078 \times 0.0314 \times \cos 25$

$\phi = 2.22 \times 10^{-3}$ Wb

Therefore, the magnetic flux through surface is $2.22 \times 10^{-3}$ Wb

6 0

3 years ago

What is carnot engine.express the relation for its efficiency?

patriot [66]

Answer:

Explanation:

The Carnot cycle is a special case of a thermodynamic cycle that produces an ideal gas and consists of two isothermal processes and two adiabatic processes. This cycle is a theoretical solution given by Sadi Karnot to refine heat engines for their efficient use.

The formula for the coefficient of efficiency is:

η = (Q₁ - Q₂) / Q₁ = (T₁ - T₂) / T₁

Where Q₁ is is the amount of heat of the heater supplied to the working body and Q₂ is the amount of heat that the working body transfers to the refrigerator according to this T₁ is the temperature of the heater T₂ is the temperature of the refrigerator.

This formula provides a theoretical limit for the maximum value of the coefficient of efficiency of heat engines.

God is with you!!!

6 0

3 years ago

Read 2 more answers

Suppose a 20-foot ladder is leaning against a building, reaching to the bottom of a second-floor window 15 feet above the ground

Papessa [141]

Answer:

The answer is β=0,85 rads

Explanation:

As the ladder is leaning against the building, we can imagine there´s a triangle where 20ft is the hypotenuse and 15ft is the maximum vertical distance between the ladder and the ground, it means, the leg opposite to β which is the angle we need

Let β(betha) be the angle between the ladder and the ground

We also know that $sin(betha)=(leg opposite)/(hypotenuse)$

In this case we will need to find β, this way:

$betha=sin^-1((15ft/20ft))$

Then β=48,6°

We also have that 2πrads is equal to 360°, in this way we find how much β is in radians:

$betha=(48,6°)*(2pirads/360°)$

then we find β=0,85rads

7 0

2 years ago