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Yanka [14]
3 years ago
10

An object moving at a constant velocity will always have a

Physics
2 answers:
SVETLANKA909090 [29]3 years ago
8 0

Answer:

It will always have a zero acceleration

sergejj [24]3 years ago
6 0
When an object is moving with constant velocity, it does not change direction nor speed

0 acceleration
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A circuit contains two resistors in series. The voltage drop across the first is 10 V. The voltage drop across the second is als
iragen [17]

The voltage provided by the power supply = 20 V

<h3>Resistors in series </h3>

The Voltage across resistors in series in a circuit is equal to the sum of the voltage drops across each resistor.

Therefore for two resistors in series the total voltage provided by the power supply is equivalent to the summation of the voltage drops acroos each resistor ( i.e 10 V + 10 V = 20 V )

Hence we can conclude that the voltage provided by the power supply is 20 V

Learn more about reisitors in series : brainly.com/question/11657573

3 0
2 years ago
An electric fence displays a warning sign about the voltage and amperage of the fence. How would amperage and voltage affect the
Lorico [155]
V = volts 
<span>I = amps </span>
<span>P = rate or energy transfer (power) </span>
<span>and </span>
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4 0
3 years ago
Heellppppppppppp!!!!
alukav5142 [94]

Answer:

B, the internet serves to provide people with more insightful explanations on things that they have not experienced yet but want to find out more on.

3 0
3 years ago
What are 2 factors affect air resistance on an object
inessss [21]

friction and the density of the air.

5 0
3 years ago
A 75-hp (shaft output) motor that has an efficiency of 91.0 percent is worn out and is to be replaced by a high- efficiency moto
daser333 [38]

Convert the shaft ouput from HP to kW

Shaft output = 75HP = 55.93kW

 

1st: Finding for the power consumption based on 55.93kW output

Power consumption (Old) = 55.93kW / .910 = 61.46kW

Power consumption (New) = 55.93kW / .954 = 58.63kW

 

2nd: Total power used in kWh:

Power Used = Power consumption * load factor * Hours:

Power (Old) = 61.46kW * .75 * 4368 = 201343 kWh

Power (New) = 58.63kW * .75 * 4368 = 192072 kWh

Energy saved = 201343 kWh - 192072 kWh = 9,271 kWh

 

3rd: Calculating for the price:

Price = kW-Hr * $/kWh

Price (Old) = 201343kWh * $0.08/kWh = $16107.44

Price (New) = 192072 kWh * $0.08/kWh = $15365.76

Cost saved = $16107.44 - $15365.76 =  $741.68/yr

 

4th: Setting up the cost equation:

Cost over time, F(t) = Motor_Cost + (Price * Number of Years, t)

Cost (Old) = 5449 + 16107.44*t

Cost (New) = 5520 + 15365.76*t

Equate the two to find for t when they cost equally:

5449 + 16107.44*t = 5520 + 15365.76*t

16107.44*t = 15365.76*t +71

16107.44*t - 15365.76*t = 71

741.68*t = 71

t = 71 / 741.68 = .095 years = 35 days

So the payback period is after 35 days.

6 0
3 years ago
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