14.58? All numbers are significant except 0 unless 0 is between two numbers
Answer:
216.31 (the work done by gravity is -216.31) positive for going up.
Explanation:
We look at this question first by getting the right equation for <em>work</em>.
Which should be... W = F x D.
From this, we can do everything, we need the Force (F) first - the question tells us that Joe is lying on his back and moves his arms upward to raise the barbell. This means that he is countering the force of graving on the object.
What is the formula for the force of gravity on an object near the earth?
Right here --- = mg
m = the mass and...
g = the acceleration due to gravity which is <em>9.81 m/s2</em>
Before we plug things in though, we need to convert everything to SI units,
the weight is in kg - so we're good to go there, but the length of Joe's arms are in "cm" we need m or meters. Converting 70 cm to m = .7 m.
Now, we just put it all together - (31.5kg)(9.81m/s2)(.7m) = 216.31 J or 216.31 N m.
Answer:
a)1mile=1609.344m
b)speed =om/s
c)om/s towards B
d)53.6448m/s
Explanation:
I viewed the question generally and that is what I thought of
The electric potential at the origin of the xy coordinate system is negative infinity
<h3>What is the electric field due to the 4.0 μC charge?</h3>
The electric field due to the 4.0 μC charge is E = kq/r² where
- k = electric constant = 9.0 × 10 Nm²/C²,
- q = 4.0 μC = 4.0 × 10 C and
- r = distance of charge from origin = x₁ - 0 = 2.0 m - 0 m = 2.0 m
<h3>What is the electric field due to the -4.0 μC charge?</h3>
The electric field due to the -4.0 μC charge is E = kq'/r² where
- k = electric constant = 9.0 × 10 Nm²/C²,
- q' = -4.0 μC = -4.0 × 10 C and
- r = distance of charge from origin = 0 - x₂ = 0 - (-2.0 m) = 0 m + 2.0 m = 2.0 m
Since both electric fields are equal in magnitude and directed along the negative x-axis, the net electric field at the origin is
E" = E + E'
= -2E
= -2kq/r²
<h3>What is the electric potential at the origin?</h3>
So, the electric potential at the origin is V = -∫₂⁰E".dr
= -∫₂⁰-2kq/r².dr
Since E and dr = dx are parallel and r = x, we have
= -∫₂⁰-2kqdxcos0/x²
= 2kq∫₂⁰dx/x²
= 2kq[-1/x]₂⁰
= -2kq[1/x]₂⁰
= -2kq[1/0 - 1/2]
= -2kq[∞ - 1/2]
= -2kq[∞]
= -∞
So, the electric potential at the origin of the xy coordinate system is negative infinity
Learn more about electric potential here:
brainly.com/question/26978411
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