Answer:
a) W = 6.75 J and b) v = 3.87 m / s
Explanation:
a) In the problem the force is nonlinear and they ask us for work, so we must use it's definition
W = ∫ F. dx
Bold indicates vectors. In a spring the force is applied in the direction of movement, whereby the scalar product is reduced to the ordinary product
W = ∫ F dx
We replace and integrate
W = ∫ (-60 x - 18 x²) dx
W = -60 x²/2 -18 x³/3
Let's evaluate between the integration limits, lower W = 0 for x = -0.50 m, to the upper limit W = W for x = 0 m
W = -30 [0- (-0.50) 2] -6 [0 - (- 0.50) 3]
W = 7.5 - 0.75
W = 6.75 J
b) Work is equal to the variation of kinetic energy
W = ΔK
W = ΔK = ½ m v² -0
v =√ 2W/m
v = √(2 6.75/ 0.90)
v = 3.87 m / s
Answer:
um, when you talk with other people about stuff. (I'm not trying to sound like a smarta s s I'm just giving a definition...)
Explanation:
SM/LAN which would demonstrate that there is a Cisco Internal Services Module or a remote LAN card. LAN interfaces the PC equipment in a confined territory, for example, an office or home. Normally, LANs utilize wired associations with connect the PCs to each other and to an assortment of fringe gadgets, for example, printers. LAN clients can speak with each other by talk or email.
Answer:
Explanation:
Let the plastic rod extends from - L to + L .
consider a small length of dx on the rod on the positive x axis at distance x . charge on it = λ dx where λ is linear charge density .
It will create a field at point P on y -axis . Distance of point P
= √ x² + .15²
electric field at P due to small charged length
dE = k λ dx x / (x² + .15² )
Its component along Y - axis
= dE cosθ where θ is angle between direction of field dE and y axis
= dE x .15 / √ x² + .15²
= k λ dx .15 / (x² + .15² )³/²
If we consider the same strip along the x axis at the same position on negative x axis , same result will be found . It is to be noted that the component of field in perpendicular to y axis will cancel out each other . Now for electric field due to whole rod at point p , we shall have to integrate the above expression from - L to + L
E = ∫ k λ .15 / (x² + .15² )³/² dx
= k λ x L / .15 √( L² / 4 + .15² )
Explanation:
It is given that,
Velocity of the electron, 
Magnetic field, 
Charge of electron, 
(a) Let 
is the force on the electron due to the magnetic field. The magnetic force acting on it is given by :
![F_e=1.6\times 10^{-19}\times [(2\times 10^6i+3\times 10^6j)\times (0.030i-0.15j)]](https://tex.z-dn.net/?f=F_e%3D1.6%5Ctimes%2010%5E%7B-19%7D%5Ctimes%20%5B%282%5Ctimes%2010%5E6i%2B3%5Ctimes%2010%5E6j%29%5Ctimes%20%280.030i-0.15j%29%5D)
(b) The charge of electron, 
The force acting on the proton is same as force on electron but in opposite direction i.e (-k). Hence, this is the required solution.
