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Kipish [7]
3 years ago
9

A step up transformer used in an automobile has a potential difference across the primary of 1800 V and a potential difference a

cross the secondary of 12 V. What is the ratio of the number of turns of wire on the primary to the number of turns on the secondary?
Physics
1 answer:
Zepler [3.9K]3 years ago
5 0
Please don't use a car's ignition coil as an example. Let's just say it's an ordinary transformer. If you connect 1800 volts AC to one side of the transformer and you get 12 volts out of the other side, then the turns of wire are in the same ratio as the voltages ... 1800/12 = 150.
A car coil doesn't work like an ordinary transformer. In a car, you put 12-volt pulses into one side, and you get voltage out of the other side that's high enough to fire spark plugs and ignite gasoline.
So you've actually got the primary and secondary windings labeled in reverse in the question, and you're actually using it as a step-DOWN transformer.
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You are in your car at rest when the traffic light turns green. You place your coffee cup on the horizontal dash and hit the gas
umka21 [38]

Answer:

(d) Negative.

Explanation:

let's test each at a time.

(a) It can't be 0, because cup would slide back other wise.

(b) Positive, well if forward is positive, than the work done against the forward acceleration must be negative , so it can't be positive.

(c) Equal to non-conservative work done by the car's engine.

well no, because work done by car's engine dosen't go all of it into getting car to move, so it can't be that.

(d) negative, this look like it, because work that friction does must be nagative to counteract positive thrust of car which is positive and in forward direction.

(d) this can't be true.

So the answer is (d) negative.

3 0
3 years ago
It is 5.5 km from your home to the physics lab. As part of your physical fitness program, you could run that distance at 10 km/h
stealth61 [152]

Explanation:

Displacement = 5 km

A.

Converting km/h to m/s,

10 km/h * 1000 m/1 km * 1 h/3600 s

= 25/9 m/s

Remember,

700 watt = 700 J/s

Velocity = displacement/time

Time = 5000/(25/9)

= 1800 s

Energy = power * time

= 700 * 1800

= 1,260,000

= 1260 kJ

B.

Converting km/h to m/s,

3 km/h * 1000 m/1 km * 1 h/3600 s

= 5/6 m/s

290 watt = 290 J/s

Velocity = displacement/time

Time = 5000/(5/6)

= 6000 s

Energy = power * time

= 290 * 6000

= 1,740,000

= 1740 kJ

C.

Walking burns more energy; 1,740,000 joules. It burns more because you walk for a greater period of time.

6 0
3 years ago
With what speed must you approach a source of sound to observe a 25% change in frequency?
insens350 [35]
Sound source is at rest, you are moving with velocity v, f = frequency, c = speed of sound:

f = f0(1 + v/c)

115 = 100(1 + v/343)
115 = 100 + 100v/343
15 = 100v/343
v = 15*343/100
<span> v = 51,45 m/s </span>
5 0
3 years ago
a 42.3 kg girl and a 7.93 kg sled are on the surface of a frozen lake, 15.0m apart and linked by a rope, but not moving yet. the
ycow [4]

Answer:

they meet from the girl's original position at: 2.37 (meters)

Explanation:

We need to use the Newton's law, exactly the second law that relate force, mass and acceleration as: F=m*a with this we can get both accelerations; solving for acceleration a=\frac{F}{m}. Now a_{girl}=\frac{5.76}{42.3}=0.14 (m/s^{2}) anda_{sled}=\frac{5.76}{7.93}=0.73(m/s^{2}). Then knowing that they both travel at the same time and assuming that the distance among the girl and the sled is: 15.0-x, so, x=\frac{1}{2}*a_{girl}*t^{2} and15.0-x=\frac{1}{2}*a_{sled}*t^{2}, solving for the time we get:t=\sqrt{\frac{2x}{a_{girl} } } and t=\sqrt{\frac{2*(15.0-x)}{a_{sled} } } with this equations we solving for the x that is the distance between the girl and the sled after the apply the force and we get:\sqrt{\frac{2x}{a_{girl}}} = \sqrt{\frac{2*(15.0-x)}{a_{sled} }. Finally we get:\frac{x}{a_{girl} }=\frac{(15.0-x)}{a_{sled} } and replacing the values we have got:\frac{x}{0.14} =\frac{(15.0-x)}{0.73} so 5.33*x=15-x so x=2.37 (meters).

5 0
3 years ago
Example of centrifugal force <br>(wrong answer will get reported)(right answer will get brainliest)​
Lera25 [3.4K]

Vehicle driving round a curve

3 0
3 years ago
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