I'm not sure what "60 degree horizontal" means.
I'm going to assume that it means a direction aimed 60 degrees
above the horizon and 30 degrees below the zenith.
Now, I'll answer the question that I have invented.
When the shot is fired with speed of 'S' in that direction,
the horizontal component of its velocity is S cos(60) = 0.5 S ,
and the vertical component is S sin(60) = S√3/2 = 0.866 S . (rounded)
-- 0.75 of its kinetic energy is due to its vertical velocity.
That much of its KE gets used up by climbing against gravity.
-- 0.25 of its kinetic energy is due to its horizontal velocity.
That doesn't change.
-- So at the top of its trajectory, its KE is 0.25 of what it had originally.
That's E/4 .
Answer:
How does sperm cells get out of the genitals of the male's? What produces the sperms? Why does women have to hope on the genitals?
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Explanation:
<h2>K.E/P.E = m/k tan²φ x ω²</h2>
Explanation:
The given position of block x = x₀ cos(ωt + φ)
The velocity of block v = dx/dt = - x₀ sin(ωt + φ) x ω
The kinetic energy = 1/2 mv² = 1/2 m x₀² sin²(ωt + φ) x ω²
The potential energy of spring = 1/2 k x² , where k is the spring constant
Thus P.E = 1/2 x k x x₀² cos²(ωt + φ)
When t = 0
K.E = 1/2 m x₀²sin²φ x ω²
P.E = 1/2 k x₀² cos²φ
Dividing these , we have
K.E/P.E = m/k tan²φ x ω²
To develop the problem we will start by finding the energy taken by each cycle through the efficiency of the motor and the exhausted energy. Later the work will be found for the conservation of energy in which this is equivalent to the difference between the two calculated energy values. Finally the estimated time will be calculated with the work and the power given,
PART A)
Work done by the heat engine in each cycle = W
According to the value given we have that,
Power is defined as the variation of energy as a function of time therefore,
Therefore the interval for each cycle is 0.75s
