Displacement is simply the change in position, or the difference in the final and initial positions:
Then
(a) ∆<em>d</em> = 5 m - 0 m = 5 m
(b) ∆<em>d</em> = 1 m - (-2 m) = 1 m + 2 m = 3 m
(c) ∆<em>d</em> = 2 m - (-2 m) = 2 m + 2 m = 4 m
(d) ∆<em>d</em> = 6 m - 2 m = 4 m
Chemical! physical would be a bruise or a cut.
The net force on the charge at the origin is -1.2×10-8
<u>Explanation:</u>
Solving the problem,
- Draw the x-axis and the locations of the given three charges.
- The forces applied on the charge at the origin and there are two of them, and since all the changes are positive, all the forces are repulsive.
- we have the formula, F = kq1Q/r².
- F1 = kq1Q/r²1 = (9.0*109Nm²/C²)(2.2*10^-9C)(3.5*10^-9C)/(1.5m)² = 31*10-9N = 3.1*10-8N. F1 points to the right (+x direction).
- F2 = kq2Q/r²2 = (9.0*109Nm²/C²)(5.4*10^-9C)(3.5*10^-9C)/(2.0m)² = 43*10^-9N = 4.3*10^-8N.
- F2 points to the left (-x direction).
- To find the net force we have to subtract the force F1 and force F2 .
- The net force is F(origin) = F1 - F2 = -1.2×10-8N.
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Answer:
she must increase the current by factor of 7
Explanation:
The magnetic field produced by a steady current flowing in a very long straight wire encircles the wire.In order to solve the question, we use this formula,
B= μo I/(2πr)
where,
'μo' represents permeability of free space i.e 4π*10-7 N/A2
B=magnetic field
I= current
r=radius
->When r= 1cm=> 0.01m
B1 = μo 
/(2π x 0.01)
->when r=7cm =>0.07m
B2 = μo 
/(2π x 0.07)
Now equating both of the magnetic fields, we have
B1= B2
μo /(2π x 0.01)= μo /(2π x 0.07)
/ = 0.01/0.07
/ = 1/ 7
Therefore, she must increase the current by factor of 7
Answer:
"A pendulum swinging back and forth" is an example of harmonic motion
X = Xo cos ω t
Explains the back and forth motion of the pendulum
