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faltersainse [42]
3 years ago
14

On a cold day, a heat pump absorbs heat from the outside air at 14°F (−10°C) and transfers it into a home at a temperature of 86

°F (30°C). Determine the maximum κ of the heat pump.
Physics
1 answer:
Leto [7]3 years ago
7 0

Answer:

6.575

Explanation:

T1 = 30C = 30 + 273 = 303 K

T2 = - 10 C = - 10 + 273 = 263 K

The coefficient of performance of heat pump

k = T2 / (T1 - T2)

k = 263 / (303 - 263) = 6.575

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EX
Serjik [45]

Given the distance traveled and time elapsed, the average speed of the train is approximately 26.944m/s.

<h3>What is the average speed of the train?</h3>

Speed is simply referred to as distance traveled per unit time.

Mathematically, Speed = Distance ÷ time.

Given the data in the question;

  • Distance traveled = 221miles
  • Elapsed time = 3 hours and 40 minutes

First we convert miles to meters and Hours minutes to seconds.

221 miles = ( 221 × 1609.344 )m = 355665.024 meters

3 hours and 40 minutes = ( 3×60×60)s + ( 40×60)s

= 10800s + 2400s

= 13200s

Now, determine the average speed.

Speed = Distance ÷ time

Speed = 355665.024m / 13200s

Speed = 26.944m/s

Given the distance traveled and time elapsed, the average speed of the train is approximately 26.944m/s.

Learn more about speed here: brainly.com/question/7359669

#SPJ1

4 0
1 year ago
The design speed of a multilane highway is 60 mi/hr. What is the minimum stopping sight distance that should be provided on the
kicyunya [14]

Answer:

Part a: When the road is level, the minimum stopping sight distance is 563.36 ft.

Part b: When the road has a maximum grade of 4%, the minimum stopping sight distance is 528.19 ft.

Explanation:

Part a

When Road is Level

The stopping sight distance is given as

SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}

Here

  • SSD is the stopping sight distance which is to be calculated.
  • u is the speed which is given as 60 mi/hr
  • t is the perception-reaction time given as 2.5 sec.
  • a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
  • G is the grade of the road, which is this case is 0 as the road is level

Substituting values

                              SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35-0)}\\SSD=220.5 +342.86 ft\\SSD=563.36 ft

So the minimum stopping sight distance is 563.36 ft.

Part b

When Road has a maximum grade of 4%

The stopping sight distance is given as

SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}

Here

  • SSD is the stopping sight distance which is to be calculated.
  • u is the speed which is given as 60 mi/hr
  • t is the perception-reaction time given as 2.5 sec.
  • a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
  • G is the grade of the road, which is given as 4% now this can be either downgrade or upgrade

For upgrade of 4%, Substituting values

                              SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35+0.04)}\\SSD=220.5 +307.69 ft\\SSD=528.19 ft

<em>So the minimum stopping sight distance for a road with 4% upgrade is 528.19 ft.</em>

For downgrade of 4%, Substituting values

                              SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35-0.04)}\\SSD=220.5 +387.09 ft\\SSD=607.59ft

<em>So the minimum stopping sight distance for a road with 4% downgrade is 607.59 ft.</em>

As the minimum distance is required for the 4% grade road, so the solution is 528.19 ft.

3 0
3 years ago
A positive point charge Q is located at x=a and a negative point charge −Q is at x=−a. A positive charge q can be placed anywher
Scilla [17]

Answer:

\vec{F}_x = \frac{2KQqa}{(a^2 + y^2)^{3/2}} \^x

Explanation:

The Coulomb's Law gives the force by the charges:

\vec{F} = K\frac{q_1q_2}{r^2}\^r

Let us denote the positon of the charge q on the y-axis as 'y'.

The force between 'Q' and'q' is

F_1 = K\frac{Qq}{x^2 + y^2}\\F_1_x = F_1\cos(\theta)

where Θ is the angle between F_1 and x-axis.

F_1_x = K\frac{Qq}{x^2 + y^2}(\frac{x}{\sqrt{x^2 + y^2}}) = \frac{KQqa}{(a^2 + y^2)^{3/2}}

whereas

F_2_x = K\frac{-Qq}{a^2 + y^2}(-\frac{a}{\sqrt{a^2 + y^2}}) = \frac{KQqa}{(a^2 + y^2)^{3/2}}

Finally, the x-component of the net force is

\vec{F}_x = \frac{2KQqa}{(a^2 + y^2)^{3/2}} \^x

8 0
3 years ago
Describe the three times acceleration can happen
Elden [556K]

Answer:

There are three ways an object can accelerate: a change in velocity, a change in direction, or a change in both velocity and direction.

Explanation:

7 0
2 years ago
Read 2 more answers
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sp2606 [1]
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8 0
3 years ago
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