Answer:
1)![t=2.26\: s](https://tex.z-dn.net/?f=t%3D2.26%5C%3A%20s)
2)![S=33.9\: m](https://tex.z-dn.net/?f=S%3D33.9%5C%3A%20m)
3)![v=26.77\: m/s](https://tex.z-dn.net/?f=v%3D26.77%5C%3A%20m%2Fs)
4)
Explanation:
1)
We can use the following equation:
![y_{f}=y_{0}+v_{iy}t-0.5*g*t^{2}](https://tex.z-dn.net/?f=y_%7Bf%7D%3Dy_%7B0%7D%2Bv_%7Biy%7Dt-0.5%2Ag%2At%5E%7B2%7D)
Here, the initial velocity in the y-direction is zero, the final y position is zero and the initial y position is 25 m.
![0=25-0.5*9.81*t^{2}](https://tex.z-dn.net/?f=0%3D25-0.5%2A9.81%2At%5E%7B2%7D)
![t=2.26\: s](https://tex.z-dn.net/?f=t%3D2.26%5C%3A%20s)
2)
The equation of the motion in the x-direction is:
![v_{ix}=\frac{S}{t}](https://tex.z-dn.net/?f=v_%7Bix%7D%3D%5Cfrac%7BS%7D%7Bt%7D)
![15=\frac{S}{2.26}](https://tex.z-dn.net/?f=15%3D%5Cfrac%7BS%7D%7B2.26%7D)
![S=33.9\: m](https://tex.z-dn.net/?f=S%3D33.9%5C%3A%20m)
3)
The velocity in the y-direction of the stone will be:
![v_{fy}=v_{iy}-gt](https://tex.z-dn.net/?f=v_%7Bfy%7D%3Dv_%7Biy%7D-gt)
![v_{fy}=0-(9.81*2.26)](https://tex.z-dn.net/?f=v_%7Bfy%7D%3D0-%289.81%2A2.26%29)
![v_{fy}=-22.17\: m/s](https://tex.z-dn.net/?f=v_%7Bfy%7D%3D-22.17%5C%3A%20m%2Fs)
Now, the velocity in the x-direction is 15 m/s then the velocity will be:
![v=26.77\: m/s](https://tex.z-dn.net/?f=v%3D26.77%5C%3A%20m%2Fs)
4)
The angle of this velocity is:
Then α=55.92° negative from the x-direction.
I hope it helps you!
Answer:
Input force of pulley system = 200 N
Explanation:
Given:
Mechanical advantage of pulley system = 5
Output force from pulley system = 1,000 N
Find;
Input force of pulley system
Computation:
Mechanical advantage = Output force / Input force
Mechanical advantage of pulley system = Output force from pulley system / Input force of pulley system
5 = 1,000 / Input force of pulley system
Input force of pulley system = 1,000 / 5
Input force of pulley system = 200 N
Answer:
M.A = load/ effort
1200N/400N
= 3
velocity ratio= radius of wheel/radius of Axle
40cm/10cm
=4
efficiency= 3/4*100
75%
Answer:
with teamwork
Explanation:
you need to use team work so the right answer is C
Answer:
A) μ = A.m²
B) z = 0.46m
Explanation:
A) Magnetic dipole moment of a coil is given by; μ = NIA
Where;
N is number of turns of coil
I is current in wire
A is area
We are given
N = 300 turns; I = 4A ; d =5cm = 0.05m
Area = πd²/4 = π(0.05)²/4 = 0.001963
So,
μ = 300 x 4 x 0.001963 = 2.36 A.m².
B) The magnetic field at a distance z along the coils perpendicular central axis is parallel to the axis and is given by;
B = (μ_o•μ)/(2π•z³)
Let's make z the subject ;
z = [(μ_o•μ)/(2π•B)] ^(⅓)
Where u_o is vacuum permiability with a value of 4π x 10^(-7) H
Also, B = 5 mT = 5 x 10^(-6) T
Thus,
z = [ (4π x 10^(-7)•2.36)/(2π•5 x 10^(-6))]^(⅓)
Solving this gives; z = 0.46m =