WHICH SHOULD NOT BE PART OF SCIENTIFIC INQUIRY

The coefficient of performance of a residential heat pump is 1.6. Calculate the heating effect in kJ/s this heat pump will produ

grin007 [14]

Answer:

8 KJ/ s

Explanation:

Heat pumps Transfer thermal energy through absorbing of heat that comes from cold region and then release to warmer area by utilizing external power.

The coefficient of performance known as COP provide the ratio of both heating and cooling that are supplied to required work.

✓QH=The rate at which heat is produced = ?

✓COP= Coefficient of performance of a residential heat pump = 1.6

✓ W(in)= power consumption= 5KW

QH=The rate at which heat is produced=[Coefficient of performance of a residential heat pump] × [power consumption]

= 1.6 × 5KW

=8 KJ/ s

7 0

3 years ago

A ball is thrown straight up from the edge of the roof of a building. A second ball is dropped from the roof 1.15 s later. Ignor

pogonyaev

Answer:

a) v₀ = 9.2 m/s

b) y₀ = 7.9 m

Explanation:

The position of the balls is given by the equation:

$y =- \frac{1}{2} gt^2 + v_0 t + y_0$

where:

acceleration g = 9.8 m/s²

time t

initial velocity v₀

initial height y₀

a) lets divide (a) in two parts:

1.part: How long will it take the second ball to fall down?

$v_0 = 0, y = 0\\0=- \frac{1}{2} gt^2 + y_0\\ t = \sqrt{\frac{2y_0}{g}}$

2. part: At time t from part1 + 1.15s, the first ball should land on the ground.

$y = 0, y_0 = 19.6, t = \sqrt{\frac{2y_0}{g}} + 1.15\\ 0 =- \frac{1}{2} gt^2 + v_0t + y_0$

This leaves only one unknown: v₀

$v_0 =\frac{1}{t}(\frac{1}{2} gt^2 - y_0)\\ v_0 = 9.2 \frac{m}{s}$

b)again, lets divide in two parts

1.part: Where will ball1 be relative to ball2 in 1.15s:

$t = 1.15s, v_0 = 8.6 m/s\\y= -\frac{1}{2} gt^2 + v_0t + y_0\\ \delta y = y - y_0 =v_0t -\frac{1}{2} gt^2$

and how fast will it go:

$v' = -gt + v_0$

2.part: Now we can plug in to the equation for the position of the two balls. Let's start with the second ball first:

$0 = -\frac{1}{2} gt^2 + y_0\\ y_0 = \frac{1}{2} gt^2$

Now let's use this result in the equation for the first ball:

$0 = - \frac{1}{2} gt^2 + v't + y_0 + \delta y = - \frac{1}{2} gt^2 + v't + \frac{1}{2} gt^2 + \delta y\\ 0 = v't + \delta y\\ t =- \frac{\delta y}{v'} \\ y_0 = \frac{1}{2} g(\frac{\delta y}{v'})^2\\ y_0 = 7.9m$

3 0

3 years ago

Please ANSWER IMPORTANT

valentina_108 [34]

Answer:

dont know my guy

Explanation:

5 0

3 years ago

Read 2 more answers

Look at the triangle below.

slega [8]

Answer:

a = 32.48538491

b = 25.59883264

Step-by-step explanation:

To find the length of a, use the sine function:

$sin\alpha = \frac{opposite}{hypotenuse}$

$sin 38 = \frac{20}{a} \\a = \frac{20}{sin38}\\a = 32.48538491$

To find the length of b, use tangent:

$tan\alpha = \frac{opposite}{adjacent}$

$tan 38 = \frac{20}{b} \\b = \frac{20}{sin38}\\b = 25.59883264$

7 0

3 years ago

Is there water on Uranus? And what temperature is it

stiks02 [169]

It atmosphere is compsosed promarly of hyrogen and helium

5 0

3 years ago