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2 years ago

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A heat pump absorbs heat from the cold outdoors at 3Â°c and supplies heat to a house at 20Â°c at a rate of 30,000 kj/h. if the p

ower consumed by the heat pump is 3 kw, the coefficient of performance of the heat pump is

1 answer:

Mazyrski [523]2 years ago

6 0

A heat pump absorbs heat from the cold outdoors at 3 C and supplies heat to a
house at 20 C at a rate of 30,000 kJ/h. If the power consumed by the heat pump
<span>is 3 kW, find the coefficient of performance of the heat pump.</span>

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A student releases a block of mass m from rest at the top of a slide of height h1. The block moves down the slide and off the en

Nina [5.8K]

Answer:

B) d = √ ( 4 h₂ ( H - 2h₂))

Explanation:

A) 1) If the height of the slide is very small, there is no speed to leave the table, therefore do not recreate almost any horizontal distance

2) If the height is very small downwards, it touches the earth a little and the horizon is small,

B) to find an equation for horizontal distance (d)

We must maximize the speed at the bottom of the slide let's use energy

Starting point Higher

Em₀ = U = m g h₁

Final point. Lower (slide bottom)

Emf = K + U = ½ m v² + m gh₂

As there is no friction the energy is conserved

mgh₁ = ½ m v² + mgh₂

v² = 2 g (h₁-h₂)

This is the speed with which the block leaves the table, bone is the horizontal speed (vₓ)

The distance traveled when leaving the table can be searched with kinematics, projectile launch

x = v₀ₓ t

y = t - ½ g t²

The height is the height of the table (y = h₂), as it comes out horizontally the vertical speed is zero

t = √ 2h₂ / g

We substitute in the other equation

d = √ (2g (h₁-h₂)) √ 2h₂ / g

d = √ (4 h₂ (h₁-h₂))

H = h₁ + h₂

h₁ = H -h₂

d = √ ( 4 h₂ ( H - 2h₂))

Explanation:

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2 years ago

What is a limitation of the electron cloud model theory that a law about electrons would not have?

Drupady [299]

Electrons couldn't orbit the nucleus like miniature planets~!

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2 years ago

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Anna pushes a box with a force of 8.00 newtons. She generates a power of 3.00 watts. How much time does it take for Anna to move

QveST [7]

Power is the energy in a system per time. It will have units of Watts which is equal to joules per second. It can be expressed as:

P = E / t

where E = Force x distance

P = Fd / t
t = Fd / P
t = 8 (9.72) / 3.0
t = 25.92 s

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2 years ago

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A rocket starting from its launch pad is subjected to a uniform acceleration of 100 meters/second2. Determine the time needed to

balandron [24]

The velocity is the integral of acceleration. If acceleration is 100 m/s^2 then velocity is:

$v= \int\limits^{}_{}100 \, dt=100t$

So to know the velocity at any time, t, we just put t in seconds into this equation. To know at what time we get to a certain velocity, we set this equation equal to that velocity and solve for t:

$100t = 1000 \\ \\ t= \frac{1000}{100} =10s$

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2 years ago

You are a member of a geological team in Central Africa. Your team comes upon a wide river that is flowing east. You must determ

Dovator [93]

Answer:

(a). The width of the river is 90.5 m.

The current speed of the river is 3.96 m.

(b). The shortest time is 15.0 sec and we would end 59.4 m east of our starting point.

Explanation:

Given that,

Constant speed = 6.00 m/s

Time = 20.1 sec

Speed = 9.00 m/s

Time = 11.2 sec

We need to write a equation for to travel due north across the river,

Using equation for north

$v^2-c^2=\dfrac{w^2}{t^2}$

Put the value in the equation

$6.00^2-c^2=\dfrac{w^2}{(20.1)^2}$

$36-c^2=\dfrac{w^2}{404.01}$ ....(I)

We need to write a equation for to travel due south across the river,

Using equation for south

$v^2-c^2=\dfrac{w^2}{t^2}$

Put the value in the equation

$9.00^2-c^2=\dfrac{w^2}{(11.2)^2}$

$81-c^2=\dfrac{w^2}{125.44}$ ....(II)

(a). We need to calculate the wide of the river

Using equation (I) and (II)

$45=\dfrac{w^2}{125.44}-\dfrac{w^2}{404.01}$

$45=w^2(0.00549)$

$w^2=\dfrac{45}{0.00549}$

$w=\sqrt{\dfrac{45}{0.00549}}$

$w=90.5$

We need to calculate the current speed

Using equation (I)

$36-c^2=\dfrac{(90.5)^2}{(20.1)^2}$

$36-c^2=20.27$

$c^2=20.27-36$

$c=\sqrt{15.73}$

$c=3.96\ m/s$

(b). We need to calculate the shortest time

Using formula of time

$t=\dfrac{d}{v}$

$t=\dfrac{90.5}{6}$

$t=15.0\ sec$

We need to calculate the distance

Using formula of distance

$d=vt$

$d=3.96\times15.0$

$d=59.4\ m$

Hence, (a). The width of the river is 90.5 m.

The current speed of the river is 3.96 m.

(b). The shortest time is 15.0 sec and we would end 59.4 m east of our starting point.

3 0

2 years ago