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SVETLANKA909090 [29]
2 years ago
12

A heat pump is used to maintain a house at 22°C by extracting heat from the outside air on a day when the outside air temperatur

e is 2°C. The house is estimated to lose heat at a rate of 110,000 kJ/h, and the heat pump consumes 5 kW of electrical power when running. Is this heat pump powerful enough to do the job?
Physics
1 answer:
Ilia_Sergeevich [38]2 years ago
4 0

Answer:

It is enough

Explanation:

To develop the problem it is necessary to take into account the concepts related to the coefficient of performance of a pump.

The two ways in which the performance coefficient can be expressed are given by:

COP_p = \frac{T_H}{T_H-T_L}

Where,

T_H =High Temperature

T_L = Low Temperature

And the other way is,

COP_p = \frac{\dot{Q}}{W}

Where \dot{Q} is heat rate and W the power consumed.

We have all our terms in Celsius, so we calculate the temperature in Kelvin

T_H = 22+273 = 295K

T_L = 2+273 = 275k

The rate at which heat is lost is:

\dot{Q} = 110000kJ/h

The power consumed by the heat pump is

\dot{W} = 5kW

And the coefficient of performance is

COP_p = \frac{T_H}{T_H-T_L}

COP_p = \frac{295}{295-275}

COP_p = 14.75

With this value we can calculate the Power required,

COP_p = \frac{\dot{Q}}{W}

14.75 = \frac{110000}{W}

W = \frac{110000}{3600*14.75}

W = 2.07kW

<em>The power consumed is consumed is 5kW which is more than 2.07kW so this heat pump powerful enough.</em>

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GuDViN [60]

Answer:

The total charge on the rod is 47.8 nC.

Explanation:

Given that,

Charge = 5.0 nC

Length of glass rod= 10 cm

Force = 840 μN

Distance = 4.0 cm

The electric field intensity due to a uniformly charged rod of length L at a distance x on its perpendicular bisector

We need to calculate the electric field

Using formula of electric field intensity

E=\dfrac{kQ}{x\sqrt{(\dfrac{L}{2})^2+x^2}}

Where, Q = charge on the rod

The force is on the charged bead of charge q placed in the electric field of field strength E

Using formula of force

F=qE

Put the value into the formula

F=q\times\dfrac{kQ}{x\sqrt{(\dfrac{L}{2})^2+x^2}}

We need to calculate the total charge on the rod

Q=\dfrac{Fx\sqrt{(\dfrac{L}{2})^2+x^2}}{kq}

Put the value into the formula

Q=\dfrac{840\times10^{-6}\times4.0\times10^{-2}\sqrt{(\dfrac{10.0\times10^{-2}}{2})^2+(4.0\times10^{-2})^2}}{9\times10^{9}\times5.0\times10^{-9}}

Q=47.8\times10^{-9}\ C

Q=47.8\ nC

Hence, The total charge on the rod is 47.8 nC.

6 0
2 years ago
What is the acceleration of a ball rolling down a ramp that starts from rest and travels 0.9 m in 3 s?
cupoosta [38]
Given:
u = 0, initial velocity
s 0.9 m, distance traveled.
t = 3 s, the time taken.

Let a =  the acceleration. Then
s = ut + (1/2)*a*t²
(0.9 m) = 0.5*(a m/s²)*(3 s)²
0.9 = 4.5a
a = 0.2 m/s²

Answer: 0.2 m/s²
3 0
2 years ago
Lizette works in her school’s vegetable garden. Every Tuesday, she pulls weeds for 15 minutes. Weeding seems like a never-ending
Gnoma [55]

Answer:

Constant or Controlled variables: Same concentration of vinegar solution, same quantity of vinegar, same type of weed etc

Explanation:

In an experiment, certain variables are kept unchanged or constant for both the experimental group and control group in order not to influence the outcome of the experiment. These variables are called CONTROLLED VARIABLES or CONSTANTS.

In the case of this experiment where Lizette is testing the effect of vinegar on weed, the variable that should be kept the same (controlled variables) for the control group of weeds and the sprayed weeds include Same concentration of vinegar solution, Same quantity of vinegar, same type of weed.

8 0
2 years ago
Read 2 more answers
At the circus, a 100-kilogram clown is fired 15 meters per second from a 500-kilogram cannon. What is the recoil speed of the ca
Ahat [919]
Recoil speed= 3m/s, method shown in photo

4 0
2 years ago
An 888.0 kg elevator is moving downward with a velocity of 0.800 m/s. It decelerates uniformly and comes to a stop in a distance
bagirrra123 [75]

Answer:

The value of tension on the cable T = 1065.6 N

Explanation:

Mass = 888 kg

Initial velocity ( u )= 0.8 \frac{m}{sec}

Final velocity ( V ) = 0

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Now use third law of motion V^{2} = u^{2} - 2 a s

Put all the values in above formula we get,

⇒ 0 = 0.8^{2} - 2 × a ×0.2667

⇒ a = 1.2 \frac{m}{sec^{2} }

This is the deceleration of the box.

Tension in the cable is given by T = F = m × a

Put all the values in above formula we get,

T = 888 × 1.2

T = 1065.6 N

This is the value of tension on the cable.

5 0
2 years ago
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