Answer:
![T_1=6655.295917 \approx 6655.3N](https://tex.z-dn.net/?f=T_1%3D6655.295917%20%5Capprox%206655.3N)
Explanation:
From the question we are told that
Angle of cable 2 ![\theta=37.0\textdegree](https://tex.z-dn.net/?f=%5Ctheta%3D37.0%5Ctextdegree)
Weight of sculpture ![W=5000 N](https://tex.z-dn.net/?f=W%3D5000%20N)
Generally the Tension from cable 2
is mathematically given by
![T_2sin37\textdegree=5000N](https://tex.z-dn.net/?f=T_2sin37%5Ctextdegree%3D5000N)
![T_2=5000N/sin37\textdegree](https://tex.z-dn.net/?f=T_2%3D5000N%2Fsin37%5Ctextdegree)
![T_2=8308.2N](https://tex.z-dn.net/?f=T_2%3D8308.2N)
Generally the Tension from Cable 1
is mathematically given by
![T_1=T_2 cos37\textdegree](https://tex.z-dn.net/?f=T_1%3DT_2%20cos37%5Ctextdegree)
![T_1=8308.2* cos 37\textdegree](https://tex.z-dn.net/?f=T_1%3D8308.2%2A%20cos%2037%5Ctextdegree)
![T_1=6655.295917 \approx 6655.3N](https://tex.z-dn.net/?f=T_1%3D6655.295917%20%5Capprox%206655.3N)
Answer:
a) 0.462 m/s^2
b) 31.5 rad/s
c) 381 rad
d) 135m
Explanation:
the linear acceleration is given by:
![a=\alpha *r\\a=1.30rad/s^2*(35.5*10^{-2}m)\\a=0.462m/s^2](https://tex.z-dn.net/?f=a%3D%5Calpha%20%2Ar%5C%5Ca%3D1.30rad%2Fs%5E2%2A%2835.5%2A10%5E%7B-2%7Dm%29%5C%5Ca%3D0.462m%2Fs%5E2)
the angular speed is given by:
![\omega=\frac{v}{r}\\\\\omega=\frac{11.2m/s}{35.5*10^{-2}m}\\\\\omega=31.5rad/s](https://tex.z-dn.net/?f=%5Comega%3D%5Cfrac%7Bv%7D%7Br%7D%5C%5C%5C%5C%5Comega%3D%5Cfrac%7B11.2m%2Fs%7D%7B35.5%2A10%5E%7B-2%7Dm%7D%5C%5C%5C%5C%5Comega%3D31.5rad%2Fs)
to calculate how many radians have the wheel turned we need the apply the following formula:
![\theta=\frac{1}{2}\alpha*t^2\\\\t=\frac{\omega}{\alpha}\\\\t=\frac{31.5rad/s}{1.30rad/s^2}\\\\t=24.2s\\\\\theta=\frac{1}{2}*1.30rad/s^2*(24.2s)^2\\\\\theta=381rad](https://tex.z-dn.net/?f=%5Ctheta%3D%5Cfrac%7B1%7D%7B2%7D%5Calpha%2At%5E2%5C%5C%5C%5Ct%3D%5Cfrac%7B%5Comega%7D%7B%5Calpha%7D%5C%5C%5C%5Ct%3D%5Cfrac%7B31.5rad%2Fs%7D%7B1.30rad%2Fs%5E2%7D%5C%5C%5C%5Ct%3D24.2s%5C%5C%5C%5C%5Ctheta%3D%5Cfrac%7B1%7D%7B2%7D%2A1.30rad%2Fs%5E2%2A%2824.2s%29%5E2%5C%5C%5C%5C%5Ctheta%3D381rad)
the distance is given by:
![d=\theta*r](https://tex.z-dn.net/?f=d%3D%5Ctheta%2Ar)
![d=381rad*(35.5*10^{-2}m)\\d=135m](https://tex.z-dn.net/?f=d%3D381rad%2A%2835.5%2A10%5E%7B-2%7Dm%29%5C%5Cd%3D135m)
Answer:
It is not correct because the amplitude of the waves can be bigger than others and the graph can be going up and down
Explanation: I got the question right
Could be easy for some people and hard for some people.
Answer:
All statement are correct.
Explanation:
1. Electric field lines are the same thing as electric field vectors, electric field are mathematically vectors quantity. These vectors point in the direction in which a positive test charge would move.
2. Electric field line drawings allow you to determine the approximate direction of the electric field at a point in space. Yes it is correct tangent drawn at any point on these lines gives the direction of electric filed at that point.
3. The number of electric field lines that start or end at a charged particle is proportional to the magnitude of charge on the particle, is a correct statement.
4.The electric field is strongest where the electric field lines are close together, again a correct statement as relative closeness of field lines indicate a stronger strength of electric field.
Hence we can say that all the statement are correct.