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JulsSmile [24]
3 years ago
10

Which is not a characteristic of a heat pump?

Physics
2 answers:
Monica [59]3 years ago
8 0
Hello :)

The answer would be C. It requires no work.

Heat pumps transfer heat to cooler areas to keep things cold, and are most common in refrigerators and air conditioners.
kramer3 years ago
5 0
The answer is c. it requires no works
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The membrane that surrounds a certain type of living cell has a surface area of 5.3 x 10-9 m2 and a thickness of 1.1 x 10-8 m. A
kotykmax [81]

Answer:

2.1\times 10^{-12} c

Explanation:

We are given that

Surface area of membrane=5.3\times 10^{-9} m^2

Thickness of membrane=1.1\times 10^{-8} m

Assume that membrane behave like a parallel plate capacitor.

Dielectric constant=5.9

Potential difference between surfaces=85.9 mV

We have to find the charge resides on the outer surface of membrane.

Capacitance between parallel plate capacitor is given by

C=\frac{k\epsilon_0 A}{d}

Substitute the values then we get

Capacitance between parallel plate capacitor=\frac{5.9\times 8.85\times 10^{-12}\times 5.3\times 10^{-9}}{1.1\times 10^{-8}}

C=0.25\times 10^{-12}F

V=85.9 mV=85.9\times 10^{-3}

Q=CV

Q=0.25\times 10^{-12}\times 85.9\times 10^{3}=2.1\times 10^{-12} c

Hence, the charge resides on the outer surface=2.1\times 10^{-12} c

5 0
3 years ago
Read 2 more answers
When Karl Kaveman adds chilled grog to his new granite mug, he removes 10.9 kJ of energy from the mug. If it has a mass of 625 g
mrs_skeptik [129]

Answer:

3°C

Explanation:

We can that heat Q=mc_p dT

Where m is the mass c_p = specific heat capacity

dT = Temperature difference

here we have given m=625 g =.625 kg

specific heat of granite =0.79 J/(g-K) = 0.79 KJ/(kg-k)

T_1 =25°C

T_2 we have to find

we have also given Q=10.9 KJ

10.9=0.625×0.79×(25-T_2)

25-T_2 =22

T_2=3°C

7 0
3 years ago
Read 2 more answers
How do you calculate elastic potential energy
Nimfa-mama [501]
U=1/2kx2

This image sums it up
5 0
3 years ago
In the equation for the gravitational force between two objects, which quantity must be squared?
Alex787 [66]

Answer:

The distance between the two objects must be squared.

Explanation:

Gravitational force always act between two objects that have mass. The gravitational force is a weak force and attractive in nature.

The force of pull depends on the masses of the two objects and the distance between them.

The formula to calculate gravitational force between two objects having masses 'm' and 'M' and separated by a distance 'd' is given as:

F_g=\frac{GmM}{d^2}

Where, 'G' is called the universal gravitational constant and its value is equal to 6.674\times10^{-11}\ m^3 kg^{-1} s^{-2}.

Now, from the above formula, it is clear that, the force of gravitation is inversely proportional to the square of the distance between the two objects.

Thus, the quantity that must be squared in the equation of gravitational force between two objects is the distance 'd'.

7 0
3 years ago
2.- Si una cámara fotográfica emite un pulso de sonido para enfocar un objeto, determinar
uranmaximum [27]

Answer:

a. El tiempo de recorrido es 5.882\times 10^{-3} segundos para un objeto localizado a un metro de distancia de la cámara fotográfica.

b. El tiempo de recorrido es 0.118 segundos para un objeto localizado a un metro de distancia de la cámara fotográfica.

Explanation:

El sonido es un tipo de onda mecánica, que es un tipo de onda que necesita de un medio material para propagarse. En este caso, entendemos que el sonido se propaga a través del aire atmosférico hasta llegar a su destino y devolverse a rapidez constante. Entonces, podemos estimar el tiempo (t), medido en segundos, a partir de la siguiente fórmula:

t = \frac{2\cdot x_{s}}{v_{s}}

Donde:

x_{s} - Distancia entre la cámara fotográfica y el objeto, medida en metros.

v_{s} - Rapidez del sonido en el aire atmosférico, medida en metros por segundo.

A continuación, calculamos el tiempo de recorrido:

a. (x_{s} = 1\,m, v_{s} = 340\,\frac{m}{s})

t = \frac{2\cdot (1\,m)}{340\,\frac{m}{s} }

t = 5.882\times 10^{-3}\,s

El tiempo de recorrido es 5.882\times 10^{-3} segundos para un objeto localizado a un metro de distancia de la cámara fotográfica.

b. (x_{s} = 20\,m, v_{s} = 340\,\frac{m}{s})

t = \frac{2\cdot (20\,m)}{340\,\frac{m}{s} }

t = 0.118\,s

El tiempo de recorrido es 0.118 segundos para un objeto localizado a un metro de distancia de la cámara fotográfica.

8 0
3 years ago
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