Answer:
W = 637 N
Explanation:
For this exercise we will use Newton's second law
F -W = m a
where F is the force, m the mass and the acceleration
in this case the force is the weight of the body and as the elevator goes at constant speed the acceleration is zero
F = W
a = 0
F = W
W = m g
W = 65 9.8
W = 637 N
Answer:
Question one is b, Question two is b, and question three is b, im pretty positive that what it is
Answer:
Time = 1.75[s]; Distance traveled = 21.5 [m]; Max height = 15 [m]
Explanation:
First, we have to break down the velocity vector into the X & y components.
![(v_{x})_{0} = 15 * cos( 35)= 12.28[m/s]\\(v_{y})_{0} = 15 * sin( 35)= 8.6[m/s]\\\\](https://tex.z-dn.net/?f=%28v_%7Bx%7D%29_%7B0%7D%20%3D%2015%20%2A%20cos%28%2035%29%3D%2012.28%5Bm%2Fs%5D%5C%5C%28v_%7By%7D%29_%7B0%7D%20%3D%2015%20%2A%20sin%28%2035%29%3D%208.6%5Bm%2Fs%5D%5C%5C%5C%5C)
To find the time t that lasts the ball of cannon in the air we must use the following equation of kinematics, in this equation the value of y is equal to zero because it will be proposed that the ball lands at the same level that was fired.
![y=(v_{y} )_{0}-\frac{1}{2}*g*t^{2} \\where:\\g=9.81[m/s^2]\\t = time[s]\\y=0[m]](https://tex.z-dn.net/?f=y%3D%28v_%7By%7D%20%29_%7B0%7D-%5Cfrac%7B1%7D%7B2%7D%2Ag%2At%5E%7B2%7D%20%20%20%5C%5Cwhere%3A%5C%5Cg%3D9.81%5Bm%2Fs%5E2%5D%5C%5Ct%20%3D%20time%5Bs%5D%5C%5Cy%3D0%5Bm%5D)
![0=8.6*t-\frac{1}{2}*9.81*t^{2} \\4.905*t^{2}=8.6*t\\ t=1.75[s]](https://tex.z-dn.net/?f=0%3D8.6%2At-%5Cfrac%7B1%7D%7B2%7D%2A9.81%2At%5E%7B2%7D%20%20%5C%5C4.905%2At%5E%7B2%7D%3D8.6%2At%5C%5C%20t%3D1.75%5Bs%5D)
In order to find the distance traveled horizontally from the cannonball, we must use the speed kinematics equation in the X coordinate.
![x = (v_{x})_{0} *t\\x=12.28*1.75\\x=21.5 [m]](https://tex.z-dn.net/?f=x%20%3D%20%28v_%7Bx%7D%29_%7B0%7D%20%20%2At%5C%5Cx%3D12.28%2A1.75%5C%5Cx%3D21.5%20%5Bm%5D)
In order to find the last value, we must bear in mind that when the cannonball reaches the maximum height, the velocity in the component y is equal to zero, and we can find the value of and with the following kinematic equation
![y = (v_{y})_{0} *t+\frac{1}{2} *g*(t)^{2} \\y = 0*t+\frac{1}{2} *9.81*(1.75)^{2}\\ y=15 [m]](https://tex.z-dn.net/?f=y%20%3D%20%28v_%7By%7D%29_%7B0%7D%20%2At%2B%5Cfrac%7B1%7D%7B2%7D%20%2Ag%2A%28t%29%5E%7B2%7D%20%5C%5Cy%20%3D%200%2At%2B%5Cfrac%7B1%7D%7B2%7D%20%2A9.81%2A%281.75%29%5E%7B2%7D%5C%5C%20y%3D15%20%5Bm%5D)
The range of force exerted at the end of the rope is 285.7 N to 1,000 N.
<h3>Net horizontal force of the cylinder</h3>
The net horizontal force of the cylinder when it is at equilibrium position is determined by applying Newton's second law of motion.
∑F = 0
F - μFn = 0
F - 0.2(5,000) = 0
F - 1,000 = 0
F = 1,000 N
The strength of the applied force increases as the number of turns of the rope increases.
minimum force = total force/number of turns of rope
minimum force = 1,000/3.5
minimum force = 285.7 N
Thus, the range of force exerted at the end of the rope is 285.7 N to 1,000 N.
Learn more about Newton's second law of motion here: brainly.com/question/3999427
15 lb is nearly 6 kg amd 50 n is nearly 5 kg,so first puppy weighs more
