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ira [324]
3 years ago
11

A digital speedometer constantly reads zero mph. Technician A says the problem may be the vehicle speed sensor. Technician B say

s the problem may be the throttle position sensor. Who is correct
Physics
1 answer:
uysha [10]3 years ago
4 0

Answer:

Technician A

Explanation:

The vehicle speed sensor is located at the right side of the transmission near the output shaft, it is also known as the output speed sensor.it is what tells your vehicle control unit how fast it is going. The vehicle speedometer and odometer rely on the information from this sensor.

With a faulty speed sensor ,you might not be able to know how fast your car is going.

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a boy throw a ball upward from the top of a cliff 73m high.Calculate the time in which ball will fall on the ground and the velo
LekaFEV [45]

The ball rises for v/g seconds; which equals 14.7/9.8=1.5 seconds . After this time, it’s height will be:


h(t)=g/2(1.5)²+14.7(1.5)


=-4.9 x 2.25 + 22.05


=11.025m


The ball then falls for 49+11.025=60.025m, which takes:


g/2t²=60.025


t²=12.25


t=3.5 secs


Total time: 1.5+3.5=5 seconds

3 0
3 years ago
A tree falls in a forest. How many years must pass before the 14C activity in 1.03 g of the tree's carbon drops to 1.02 decay pe
Illusion [34]

Answer:

t = 5.59x10⁴ y

Explanation:

To calculate the time for the ¹⁴C drops to 1.02 decays/h, we need to use the next equation:

A_{t} = A_{0}\cdot e^{- \lambda t}    (1)

<em>where A_{t}: is the number of decays with time, A₀: is the initial activity, λ: is the decay constant and t: is the time.</em>

To find A₀ we can use the following equation:  

A_{0} = N_{0} \lambda   (2)

<em>where N₀: is the initial number of particles of ¹⁴C in the 1.03g of the trees carbon </em>

From equation (2), the N₀ of the ¹⁴C in the trees carbon can be calculated as follows:        

N_{0} = \frac{m_{T} \cdot N_{A} \cdot abundance}{m_{^{12}C}}

<em>where m_{T}: is the tree's carbon mass, N_{A}: is the Avogadro's number and m_{^{12}C}: is the ¹²C mass.  </em>

N_{0} = \frac{1.03g \cdot 6.022\cdot 10^{23} \cdot 1.3\cdot 10^{-12}}{12} = 6.72 \cdot 10^{10} atoms ^{14}C    

Similarly, from equation (2) λ is:

\lambda = \frac{Ln(2)}{t_{1/2}}

<em>where t 1/2: is the half-life of ¹⁴C= 5700 years </em>

\lambda = \frac{Ln(2)}{5700y} = 1.22 \cdot 10^{-4} y^{-1}

So, the initial activity A₀ is:  

A_{0} = 6.72 \cdot 10^{10} \cdot 1.22 \cdot 10^{-4} = 8.20 \cdot 10^{6} decays/y    

Finally, we can calculate the time from equation (1):

t = - \frac{Ln(A_{t}/A_{0})}{\lambda} = - \frac {Ln(\frac{1.02decays \cdot 24h \cdot 365d}{1h\cdot 1d \cdot 1y \cdot 8.20 \cdot 10^{6} decays/y})}{1.22 \cdot 10^{-4} y^{-1}} = 5.59 \cdot 10^{4} y              

I hope it helps you!

4 0
3 years ago
An electric motor with a 1200 hp output drives a machine with an efficiency of 60%. What is the energy output of the machine per
kirill [66]

Answer:

536,904 J/s

Explanation:

The energy output from motor is the input energy in the machine.

We know that efficiency is percentage energy ouput to energy input, and expressed as

n=\frac {E_o}{E_i}

Where n and E represent efficiency and energy respectively, subscripts o and i represent output and input respectively. Since for the machine we have the input energy then the output will be the product of efficiency and input energy

Energy output=0.6*1200 hp=720 hp

Converting hp to J/s we multiply by 745.7

Energy is 720*745.7=536,904 J/s

4 0
3 years ago
What is the only thing that can pull a beam of light towards itself ? ​
Elan Coil [88]

Answer:

<h2><em>The </em><em>only </em><em>thing</em><em> </em><em>that</em><em> </em><em>can</em><em> </em><em>pull</em><em> </em><em>a </em><em>beam </em><em>of</em><em> light</em><em> </em><em>towards</em><em> </em><em>it</em><em> </em><em>self</em><em> </em><em>is </em><em>a </em><em>black </em><em>hole.</em></h2>

4 0
3 years ago
Read 2 more answers
If the volume on your TV is low, turning the volume up one click of the remote control will make the TV seem louder than if the
marusya05 [52]

From what we know, we can confirm that this ratio (turning up the volume by one click relative to the TV's overall volume) can be quantified as the Weber fraction.

<h3>What is the Weber fraction?</h3>

This fraction describes the ratio needed for change to a stimulus in which the change is just barely noticeable. This question is a prime example in that it seeks to find out just how low of a difference is needed in TV volume in order for the difference to be noticeable.

Therefore, we can confirm that this ratio (turning up the volume by one click relative to the TV's overall volume) can be quantified as the Weber fraction.

To learn more about Weber visit:

brainly.com/question/5004433?referrer=searchResults

7 0
2 years ago
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