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Bezzdna [24]
3 years ago
12

A car slows from 22m/s to 3.0m/a at a constant rate of 2.1m/s. How many seconds are required before the car is traveling at 3.0m

/s?
Physics
1 answer:
lilavasa [31]3 years ago
8 0

Answer:

t = 9.05 s

Explanation:

Given,

The initial velocity of the car, u = 22 m/s

The final velocity of the car, v = 3.0 m/s

The rate of change of speed of the car is 2.1 m/s,

                                      ∴ the acceleration, a = -2.1 m/s²

The acceleration and velocity is given by the relation

                                           a = (v - u)/t     m/s²

Therefore,

                                           t = (v-u)/a  s

Substituting the values in the above equation

                                          t = (3.0 - 22)/(-2.1)

                                            = 9.05 s

Hence, the time required before the car is traveling at 3.0m/s is t = 9.05 s

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Six new refrigerator prototypes are tested in the laboratory. For each refrigerator, the electrical power P needed for it to ope
Mandarinka [93]

Answer:

performance coefficient from largest to the smallest

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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

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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

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V×B means the cross product of the velocity and the magnetic field

NOTE:

i×i=j×j×k×k=0

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j×k=i.  k×j=-i

k×i=j.  i×k=-j

So, if the electron is moving southward, then, it implies that the velocity of it motion is southward, so the electron is in the positive z-direction

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Then,

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F •j = q ( V•k × B•x)

Let x be the unknown

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Then, the magnetic field point in the direction of positive x axis, which is towards the west

You can as well use the Fleming right hand rule

The thumb represent force

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