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DIA [1.3K]
3 years ago
11

A heat pump with a COP of 3.15 is used to heat an air-tight house. When running, the heat pump consumes 5 kW of power. If the te

mperature in the house is 7°C when the heat pump is turned on, how long will it take for the heat pump to raise the temperature of the house to 22°C? Assume the entire mass within the house (air, furniture, etc.) is equivalent to 1500 kg of air. The constant volume specific heat of air at room temperature is cv = 0.718 kJ/kg·°C.
Physics
2 answers:
cricket20 [7]3 years ago
6 0

Answer:

1026s, 17.1m suitability aye

Explanation:

Jet001 [13]3 years ago
5 0

Answer: 1026s, 17.1m

Explanation:

Given

COP of heat pump = 3.15

Mass of air, m = 1500kg

Initial temperature, T1 = 7°C

Final temperature, T2 = 22°C

Power of the heat pump, W = 5kW

The amount of heat needed to increase temperature in the house,

Q = mcΔT

Q = 1500 * 0.718 * (22 - 7)

Q = 1077 * 15

Q = 16155

Rate at which heat is supplied to the house is

Q' = COP * W

Q' = 3.15 * 5

Q' = 15.75

Time required to raise the temperature is

Δt = Q/Q'

Δt = 16155 / 15.75

Δt = 1025.7 s

Δt ~ 1026 s

Δt ~ 17.1 min

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A skydiver of mass 80.0 kg jumps from a slow-moving aircraft and reaches a terminal speed of 50.0 m/s. (a) What is her accelerat
kirill [66]

Answer:

6.22²

Explanation:

Given that

Mass of the skydiver, m = 80 kg

Terminal speed of the skydiver, v(f) = 50 m/s

Speed of the skydiver, v(i) = 30 m/s

Acceleration of the skydiver, a = ?

To solve this, we use the formula

W - k v² = ma, where

W = weight of the skydiver

k = constant

v = speed of the skydiver

m = mass of the skydiver

So, if we substitute the values into it we have

W = mg = 80 * 9.8 = 784 N

784 - k 50² = 80 *0

784 - 2500k = 0

784 = 2500k

k = 0.3136

Now, we use this value of k to find the needed acceleration using the same formula at a speed of 30 m/s

784 - 0.3136 * 30² = 80 * a

784 - 0.3136 * 900 = 80a

784 - 282.24 = 80a

497.76 = 80a

a = 497.76 / 80

a = 6.22 m/s²

Thus, we can conclude that the acceleration when the speed of the skydiver is 30 m/s, is 6.22 m/s²

4 0
3 years ago
Passenger side, rear-view mirrors on automobiles are generally ____.
Lesechka [4]
I believe it’s C. Plane .
8 0
3 years ago
Two long, straight wires both carry current to the right, are parallel, and are 25 cm apart. Wire one carries a current of 2.0 A
Artemon [7]

Answer:

x = 7.14 meters

Explanation:

It is given that,

Current in wire 1, I_1=2\ A

Current in wire 2, I_2=5\ A  

Distance between parallel wires, r = 25 cm

Let at P point the net magnetic field equal to 0. The magnetic field at a point midway between the is given by :

B=\dfrac{\mu_oI}{2\pi r}

Let the distance is x from wire 1. So,

\dfrac{I_1}{r}=\dfrac{I_2}{(0.25-r)}

\dfrac{2}{r}=\dfrac{5}{(25-r)}

x=\dfrac{50}{7}\ m

x = 7.14 meters

So, the magnetic field will be 0 at a distance of 7.14 meters from wire 1. Hence, this is the required solution.

6 0
3 years ago
I drop a meter stick , and catch it as it falls. If it fell exactly 22 cm, how much did it take me to catch the meter stick (wha
statuscvo [17]
D=Vot+1/2at^2

In this case, there is no initial y velocity so the term Vot=0 so d=1/2at^2

acceleration=acceleration due to gravity=-9.8m/s^2

It falls - 22cm or -0.22m

We have - 0.22=1/2(-9.8)t^2

t^2=(-0.44)/(-9.8)

t=sqrt[0.44/9.8]
3 0
3 years ago
A. A PH202 student lives next to a construction site and sees a crane with a wrecking ball demolish the building next door. The
netineya [11]

Answer:

The length of cable is 12.5 m

Explanation:

Since, the wrecking ball completes 1 oscillation, in the same time, as it takes for the rock to complete 5 oscillations.

Therefore,

Time Period of Wrecking Ball = 5 (Time Period of Rock)

Since,

Time Period of  Pendulum = 2π√(L/g)

Therefore,

2π√(L₁/g) = 5[2π√(L₂/g)]

√L₁ = 5√L₂

Squaring on both sides:

L₁ = 25 L₂

where,

L₁ = Length of Cable = ?

L₂ = Length of string = 0.5 m

Therefore,

L₁ = 25 (0.5 m)

<u>L₁ = 12.5 m</u>

8 0
3 years ago
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