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aivan3 [116]
3 years ago
13

You are testing a vehicle fuel gauge float. The float pulls 40mA when the tank is 100% full and 2mA when the tank is empty. Assu

ming that the float's resistance scales linearly between 0% and 100% full, what current draw in milliamps should we see if we set the float to 49% full? Give your answer to the tenth of a milliamp.
Physics
1 answer:
Ad libitum [116K]3 years ago
3 0

Answer:

I=20.6mA

Explanation:

the relation between the current and the tank percentage is linear:

\frac{40mA-2mA}{100-0} = \frac{I-2mA}{49-0}

then:

I=49*38/100+2=20.6mA

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3 years ago
Lena is playing with a remote-controlled car in her backyard. She knows that the car uses a wheel and axle to move. What is the
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C. chemical energy from the batteries

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7 0
2 years ago
1. A friend measures the length of the school
shusha [124]

Answer:

option A

Explanation:

A soccer field is about 100 meters of length, with the other instruments is possible to make the measure, but it will be very complex and it will take time. Therefore with the 50-m tape, the student will need only two measures.

6 0
3 years ago
What is the relationship between force and motion described by Newton's first law
prisoha [69]
Newtons First Law of Motion:
An object at rest stays at rest and an object in motion<span> stays in </span>motion <span>with the same speed and in the same direction unless acted upon by an unbalanced force.</span>

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4 0
3 years ago
A car accelerates from rest at 1.0 m/s2 for 20.0 second along a straight road. It then moves at a constant speed for half an hou
Whitepunk [10]

Total distance = 36500 m

The average velocity = 19.73 m/s

<h3>Further explanation</h3>

Given

vo=initial velocity=0(from rest)

a=acceleration= 1 m/s²

t₁ = 20 s

t₂ = 0.5 hr = 1800 s

t₃= 30 s

Required

Total distance

Solution

State 1 : acceleration

\tt d=vo.t+\dfrac{1}{2}at^2\\\\d=\dfrac{1}{2}\times 1\times 20^2\rightarrow vo=0\\\\d=200~m

\tt vt=vo+at\\\\vt=at\rightarrow vo=0\\\\vt=1\times 20\\\\vt=20~m/s

State 2 : constant speed

\tt d=v\times t\\\\d=20\times 1800\\\\d=36000~m

State 3 : deceleration

\tt vt=vo+at\rightarrow vt=0(stop)\\\\vo=-at\\\\20=-a.30~s\\\\a=-\dfrac{2}{3}m/s^2(negative=deceleration)

\tt d=vot+\dfrac{1}{2}at^2\\\\d=20.30-\dfrac{1}{2}.\dfrac{2}{3}.30^2\\\\d=300~m

Total distance : state 1+ state 2+state 3

\tt 200 + 36000 + 300=36500~m

the average velocity = total distance : total time

\tt avg~velocity=\dfrac{36500}{20~s+1800~s+30~s}=19.73~m/s

4 0
2 years ago
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