Answer:
30m/s
Explanation:
Suppose there's no friction between the office chair and the floor. The only thing that impact the motion of the boy is the ejected carbon dioxide. We can apply the law of momentum conservation to find the his (the boy, the chair and the empty extinguisher) after the extinguisher is emptied.
Before the ejects, the system momentum is 0, due to at rest and no speed
After the ejects, momentum of the carbon dioxide is
9*20 = 180 kg m/s
The momentum of his is then
P + 180 = 0
P = -180
MV = -180
(26 + 16 + 18)V = - 180
V = -180/(26 + 16 + 18) = -30 m/s
So he would have a speed of 30m/s in the opposite of the ejected carbon dioxide direction
Answer:
Thus simply tells us that The nut has no net charge and so therefore, There will be a negative charge on the left side, and an equal positive charge on the right side
Breaking down food molecules
Answer:
A = 150 cm at 120 degc
Where A + B = C find B
B = C - A add -A to both sides
Ax = 150 cos 60 = -75
Ay = 150 sin 60 = 129.9
Likewise
Cx = 114.7
Cy = 80.3
Bx = Cx - Ax = 114.7 + 75 = 189.7
By = Cy - Ay = 80.3 -129.9 = -49.6
B = (189.7^2 + 49.6^2)^1/2 = 196.7 length of B vector
tan B = By / Bx = -49.6 / 189.7 B = -14.65 deg
Also
sin B = By / B = -49.6 / 196.1 = -14.65 deg
So B is 196.1 cm at -14.65 deg
Answer:
![a. W_w=235\ J\\b. W_g=-343.54\ J\\c. F_N=463.100\ N\\d. W_t=235\ J](https://tex.z-dn.net/?f=a.%20%20W_w%3D235%5C%20J%5C%5Cb.%20W_g%3D-343.54%5C%20J%5C%5Cc.%20F_N%3D463.100%5C%20N%5C%5Cd.%20%20W_t%3D235%5C%20J)
Explanation:
Given: that,
Angle of inclination of the surface, ![\theta=34^{\circ}](https://tex.z-dn.net/?f=%5Ctheta%3D34%5E%7B%5Ccirc%7D)
mass of the crate, ![m=57\ kg](https://tex.z-dn.net/?f=m%3D57%5C%20kg)
Force applied along the surface, ![F=230\ N](https://tex.z-dn.net/?f=F%3D230%5C%20N)
distance the crate moves after the application of force, ![s=1.1\ m](https://tex.z-dn.net/?f=s%3D1.1%5C%20m)
a) work done = F× s
work done = 230 × 1.1
work done = 253 J
b) Work done by the gravitational force:
![W_g=m.g\times h](https://tex.z-dn.net/?f=W_g%3Dm.g%5Ctimes%20h)
where:
g = acceleration due to gravity
h = the vertically downward displacement
Now, we find the height:
![h=s\times sin\ \thetah=1.1\times sin\ 34^{\circ}h=0.615\ m](https://tex.z-dn.net/?f=h%3Ds%5Ctimes%20sin%5C%20%5Cthetah%3D1.1%5Ctimes%20sin%5C%2034%5E%7B%5Ccirc%7Dh%3D0.615%5C%20m)
So, the work done by the gravity:
![W_g=57\times 9.8\times (-0.615) \\= - 343.54 J](https://tex.z-dn.net/?f=W_g%3D57%5Ctimes%209.8%5Ctimes%20%28-0.615%29%20%5C%5C%3D%20-%20343.54%20J)
∵direction of force and displacement are opposite.
= - 343.54J
c)
The normal reaction force on the crate by the inclined surface:
![F_N=m.g.cos\ \thetaF_N=57\times 9.8\times cos\ 34F_N=463.100\ N](https://tex.z-dn.net/?f=F_N%3Dm.g.cos%5C%20%5CthetaF_N%3D57%5Ctimes%209.8%5Ctimes%20cos%5C%2034F_N%3D463.100%5C%20N)
d)
Total work done on crate is with respect to the worker:
![W_t=235\ J](https://tex.z-dn.net/?f=W_t%3D235%5C%20J)