The projectile has a height <em>h</em> at time <em>t</em> given by
<em>h</em> = (14.0 m/s) <em>t</em> - 1/2 <em>g t</em> ²
where <em>g</em> = 9.80 m/s² is the magnitude of the acceleration due to gravity. Solve for <em>t</em> when <em>h</em> = 0 :
0 = (14.0 m/s) <em>t</em> - 1/2 <em>g t</em> ²
0 = 1/2 <em>t</em> (28.0 m/s - <em>g t</em> )
1/2 <em>t</em> = 0 <u>or</u> 28.0 m/s - <em>g</em> <em>t</em> = 0
The first equation says <em>t</em> = 0, which refers to the moment the gun is first fired, so we ignore that solution. We're left with
28.0 m/s - <em>g t</em> = 0
<em>t</em> = (28.0 m/s) / <em>g</em>
<em>t</em> = (28.0 m/s) / (9.80 m/s²)
<em>t</em> ≈ 2.86 s
Answer
Gravitational force is between two object caused by energy in mass. Here when you are in a mine the radius of the earth in case of calculating gravity is reduced to the length between you and the center.
This means the mass of earth is decreasing even though the distance is decreasing. Relative to the decrease in distance squared the decrease in the mass of the second object(earth) is more. As a result your weight decreases.
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:
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Answer: a = 1.32m/s2
Therefore, the average acceleration is 1.32m/s2
Explanation:
Acceleration is the rate of change in the velocity per time
a = change in velocity/time
a = ∆v/t
average acceleration a = (v2 -v1)/t. ....1
Given;
Final velocity v2 = 1.63m/s
Initial velocity v1 = -1.15ms
time taken t = 2.11s
Substituting into eqn 1
a = [1.63 - (-1.15)]/2.11
a = (1.63+1.15)/2.11
a = 2.78/2.11
a = 1.32m/s2
Therefore, the average acceleration is 1.32m/s2
