The formula for kinetic energy = ½m·v<span>2
1/2 * 55 kg x 5,87 m/s ^2 = 27.5 x </span>34.4569 = <span>947.56475 Joule </span>≈ 948 J
I think we will use the law of conservation of linear momentum;
M1V1 = M2V2
M1 = 4 kg (mass of the water balloon launcher)
V1=?
M2= 0.5 kg ( mass of the balloon)
V2 = 3 m/s
Therefore; 4 V1 = 0.5 × 3
4V1= 1.5
V1= 1.5/4
= 0.375 m/s
Answer:
η = 0.882 = 88.2 %
Explanation:
The efficiency of the pulley system can be given as follows:
![\eta = \frac{W_{out}}{W_{In}}\\\\](https://tex.z-dn.net/?f=%5Ceta%20%3D%20%5Cfrac%7BW_%7Bout%7D%7D%7BW_%7BIn%7D%7D%5C%5C%5C%5C)
where,
η = efficiency of pulley system = ?
W_out = Output Work = (600 N)(0.6 m) = 360 J
W_in = Input Work = (35.7 N)(11.43 m) = 408.051 J
Therefore,
![\eta = \frac{360\ J}{408.051\ J}](https://tex.z-dn.net/?f=%5Ceta%20%3D%20%5Cfrac%7B360%5C%20J%7D%7B408.051%5C%20J%7D)
<u>η = 0.882 = 88.2 %</u>
It does take on new set of proerties