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Svetllana [295]
2 years ago
10

Six new refrigerator prototypes are tested in the laboratory. For each refrigerator, the electrical power P needed for it to ope

rate and the maximum heat energy that can be removed per second QC,max/Δt from its interior are given.
Part A
Rank these refrigerators on the basis of their performance coefficient.
Rank from largest to smallest. To rank items as equivalent, overlap them.
Part B
Thsee six refrigerators are placed in six identical sealed rooms. Rank the refrigerators on the basis of the rate at which they raise the temperature of the room.
Rank from largest to smallest. To rank items as equivalent, overlap them.
1) P= 750 W, Qc,max/deltaT= 1500 J/s
2) P= 400 W, Qc,max/deltaT= 1200 J/s
3) P= 500 W, Qc,max/deltaT= 2000 J/s
4) P= 250 W, Qc,max/deltaT= 1000 J/s
5) P= 500 W, Qc,max/deltaT= 1500 J/s
6) P= 1000 W, Qc,max/deltaT= 3000 J/s
Physics
1 answer:
Mandarinka [93]2 years ago
5 0

Answer:

performance coefficient from largest to the smallest

P= 500 W, Qc,max/deltaT= 2000 J/s

P= 250 W, Qc,max/deltaT= 1000 J/s

P= 750 W, Qc,max/deltaT= 1500 J/s

) P= 400 W, Qc,max/deltaT= 1200 J/s

P= 500 W, Qc,max/deltaT= 1500 J/s

P= 1000 W, Qc,max/deltaT= 3000 J/s.

the rate at which they raise the temperature of the room.

2.1.P= 1000 W, Qc,max/deltaT= 3000 J/s

P= 500 W, Qc,max/deltaT= 2000 J/s

P= 750 W, Qc,max/deltaT= 1500 J/s

P= 500 W, Qc,max/deltaT= 1500 J/s

P= 400 W, Qc,max/deltaT= 1200 J/s

P= 250 W, Qc,max/deltaT= 1000 J/s

Explanation:

A refrigerator is a device that uses work to remove heat energy from a cold reservoir and deposit it into a hot reservoir. .A good refrigerator (with a large performance coefficient) will remove a large amount of heat energy from the cold reservoir for a small amount of work input

The performance coefficient  of a refrigerator is defined as the ratio of the heat energy removed from the cold reservoir  to the work  input to the refrigerator:

k=QC/W

power is defined as work per unit time

1.k=1500/750=2

2. 1200/400=3

3.2000/500=4

4.1000/250=4

5.1500/500=3

6.3000/1000=3

performance coefficient from largest to the smallest

P= 500 W, Qc,max/deltaT= 2000 J/s

P= 250 W, Qc,max/deltaT= 1000 J/s

P= 750 W, Qc,max/deltaT= 1500 J/s

) P= 400 W, Qc,max/deltaT= 1200 J/s

P= 500 W, Qc,max/deltaT= 1500 J/s

P= 1000 W, Qc,max/deltaT= 3000 J/s

2, Rate at which they raise the temperature of the room.

rate at which temperature rises in the inner chamber of the refrigerator is proportional to the rate of energy used to dispel heat from the refrigerator

1.P= 1000 W, Qc,max/deltaT= 3000 J/s

P= 500 W, Qc,max/deltaT= 2000 J/s

P= 750 W, Qc,max/deltaT= 1500 J/s

P= 500 W, Qc,max/deltaT= 1500 J/s

P= 400 W, Qc,max/deltaT= 1200 J/s

P= 250 W, Qc,max/deltaT= 1000 J/s

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Answer:

maximum amplitude  = 0.13 m

Explanation:

Given that

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We know that time period of simple harmonic motion given as

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0.74=\dfrac{2\pi}{\omega}

ω = 8.48 rad/s

ω=angular frequency

Lets take amplitude = A

The maximum acceleration given as

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a= ω² A

10= 8.48² A

A=0.13 m

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2 years ago
A block spring system oscillates on a frictionless surface with an amplitude of 10\text{ cm}10 cm and has an energy of 2.5 \text
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Answer:

The energy of the system is 15 J.

Explanation:

Given that,

Energy E = 2.5 J

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Using formula of mechanical energy of the system

E=\dfrac{1}{2}kA^2

Put the value into the formula

2.5=\dfrac{1}{2}k\times(10\times10^{-2})^2

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If the block is replaced by a block with twice the mass of the original block

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We need to calculate the energy

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Put the value into the formula

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A small glass bead charged to 5.0 nCnC is in the plane that bisects a thin, uniformly charged, 10-cmcm-long glass rod and is 4.0
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}}

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Determine a formula for the maximum height h that a rocket will reach if launched vertically from the Earth's surface with speed
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Initially, the energies are:

U_{i}=-\frac{G M_{\varepsilon} m}{r_{e}} \\
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At final point, the energies are:

U_{f}=-\frac{G M_{\varepsilon} m}{r_{e}+h} \\
K_{f}=\frac{1}{2} m(0)^{2}=0

Using conservation law of energy,

-\frac{G M_{e} m}{r_{e}}+\frac{1}{2} m v_{0}^{2} &=-\frac{G M_{e} m}{r_{\varepsilon}+h} \\
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The equation is further simplified as:

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& h=\frac{r_{e}^{2} v_{0}^{2}}{2 G M_{e}-r_{e} v_{0}^{2}}

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