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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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the net external force on the 24-kg mower is stated to be 51 N. If the force of friction opposing the motion is 24 N, what force
-Dominant- [34]

Answer:

PART A)

External force will be 75 N

PART B)

distance moved will be 1.125 m

Explanation:

PART A)

Given that net force on the mower is

F_{net} = 51 N

now we also know that friction force due to ground is given as

F_f = 24 N

now we have

F_{net} = F_{ext} - F_f

51 = F_{ext} - 24

F_{ext} = 75 N

so external force will be 75 N

PART B)

deceleration due to friction when external force is removed from it

a = \frac{F_f}{m}

a = \frac{24}{24} = 1 m/s^2

now we can find the distance by kinematics

v_f^2 - v_i^2 = 2 a d

0 - 1.5^2 = 2(-1)d

d = 1.125 m

so the distance moved will be 1.125 m

6 0
3 years ago
Two parallel plates of area 2.34*10-3 M2 have 7.07*10-7C of charge placed on them. A6.62*10-5C charge q1 is placed between the p
ira [324]

Answer:

The force exerted on the q_1 is  F =  2.25*10^{3} \ N

Explanation:

From the question we are told that

     The area is  A =  2.34*10^{-3} \ m^2

     The magnitude of charge placed on them is  q =  7.07 * 10^{-7} C

     The charge placed between the plate is q_1 = 6.62 *10^{-5} C

   

The electric field generated around the plate  is mathematically represented as

           E =  \frac{q}{A \epsilon_o}

Substituting values

          E =  \frac{7.07*10^{-7}}{2.34*10^{-3} * 8.85 *10^{-12}}

         E = 34*10^{6} \ V/m

The force exerted the charge q_1 is  mathematically represented as

        F =  q_1 * E

Substituting values  

        F =  6.62 *10^{-5} * 34*10^{6}

        F =  2.25*10^{3} \ N

8 0
3 years ago
When Woods hits a 0.04593 kg golf ball, the golf ball is usually traveling around 281 kilometers per hour. What average force do
Semenov [28]

Answer:

128.9 N

Explanation:

The force exerted on the golf ball is equal to the rate of change of momentum of the ball, so we can write:

F=\frac{\Delta p}{\Delta t}

where

F is the force

\Delta p is the change in momentum

\Delta t=0.030 s is the time interval

The change in momentum can be written as

\Delta p = m(v-u)

where

m = 0.04593 kg is the mass of the ball

u = 0 is the initial velocity of the ball

v=281 km/h =78.1 m/s is the final velocity of the ball

Substituting into the original equation, we find the force exerted on the golf ball:

F=\frac{m(v-u)}{\Delta t}=\frac{(0.04593)(78.1-0)}{0.030}=128.9 N

7 0
3 years ago
Billiard ball A (0.35 kg) is struck such that it moves at 10 m/s toward a second identical ball (Ball B) initially at rest. Afte
timurjin [86]

Answer:

Explanation:

We shall apply law of conservation of momentum during the collision of ball A and B .

Total momentum before collision of A and B = .35 x 10 = 3.5 kg m/s

Let the velocity of B after collision be v .

Total momentum after collision  = .35 x 2 + .35v

According to law of conservation of momentum

.35 x 2 + .35v  = 3.5

.35 v = 2.8

v = 8 m /s .

The direction of B will be same as direction of A .

3 0
3 years ago
How long does it take the projectile to reach its highest point in its trajectory?​
NeTakaya

Answer:

It takes about 88 seconds for the cannonball to reach its maximum height (ignoring air resistance).

Explanation:

3 0
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