If the horse lifts a 50 kilogram mass a vertical distance of 3 meters in 2 seconds, what is the horse’s minimum power output?
1 answer:
You might be interested in
The time required by the particles are as follows:
a. t = 1.5 seconds
b. t = 3 seconds
c. t = 0.4 seconds
<h3>What is the time required?</h3>The time required for the particles to be at several distances apart is calculated using the equation of motion given below:
a) Time required to be 30 m apart:
Assuming the distance covered by P is S1 and distance covered by Q is S2.
S1 + S2 = 51 - 30
S1 + S2 = 21
Substituting the values of velocity and acceleration in the equation of motion above:
5t + 1/2t^2 + 6t + 3t^2 = 21
2t^2 + 11t - 21 = 0
Solving for time, t by factorization, t = 1.5 seconds
b) Time required to meet:
Assuming the distance covered by P is S1 and distance covered by Q is S2.
S1 + S2 = 51
Substituting the values of velocity and acceleration in the equation of motion above:
5t + 1/2t^2 + 6t + 3t^2 = 51
2t^2 + 11t - 51 = 0
Solving for time, t by factorization, t = 3 seconds
c) Time required for velocity of P is ¾ of the velocity of Q:
Using the equation of motion: V = u + at
Vp = 3/4 Vq
4Vp= 3Vq
Substituting the values:
4(5 + t) = 3(6 + 3t)
5t = 2
t = 0.4 seconds
Learn more about distance, velocity and acceleration at: brainly.com/question/14344386
#SPJ1