Answer:
Current flowing in the cell will be equal to 0.1284 mA
Explanation:
We have given charge q = 3.70 C
And time through which charge is flowing = 8 hour
We know that 1 hour = 60 minutes, and 1 minute = 60 sec
So 1 hour = 60×60 = 3600 sec
So 8 hour = 8×3600 = 28800 sec
We know that current is rate time rate of flow of charge
So current 
So current flowing in the cell will be equal to 0.1284 mA
Answer:
It should be 1,029.105 m³
But I think you can ignore the 105 because it essentially doesn't make a difference
The number of significant digits to the answer of the following problem is four.
<h3>What are the significant digits?</h3>
The number of digits rounded to the approximate integer values are called the significant digits.
The following problem is
(2.49303 g) * (2.59 g) / (7.492 g) =
On solving we get
= 0.86184566204
The answer is approximated to 0.86185
Thus, the significant digits must be four.
Learn more about significant digits.
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Answer:
Explanation:
Sam mass=75kg
Height is 50m
20° frictionless slope
Horizontal force on Sam is 200N
According to the work energy theorem, the net work done on Sam will be equal to his change in kinetic energy.
Therefore
Wg - Ww =∆K.E
Note initial the body was at rest at top of the slope.
Then, ∆K.E is K.E(final) - K.E(initial)
K.E Is given as ½mv²
Since initial velocity is zero then, K.E(initial ) is zero
Therefore, ∆K.E=½mVf²
Wg is work done by gravity and it is given by using P.E formulas
Wg=mgh
Wg=75×9.8×50
Wg=36750J
Ww is work done by wind and it's is given by using formulae for work
Work=force × distance
Ww=horizontal force × horizontal distance
Using Trig.
TanX=opposite/adjacent
Tan20=h/x
x=h/tan20
x=50/tan20
x=137.37m
Then,
Ww=F×x
Ww=200×137.37
We=27474J
Now applying the formula
Wg - Ww =∆K.E
36750 - 27474 =½×75×Vf²
9276=37.5Vf²
Vf²=9275/37.5
Vf²= 247.36
Vf=√247.36
Vf=15.73m/s
The slope of the line on a velocity vs. time graph represents acceleration
