<span>umm it gives 1/v + 1/20 = -1/10 and v = -20/3 cm. Magnification = |v/u|=1/3.</span>
Everything but wormholes can be derived from the theory of relativity
They either push or pull and make the object magnetic.
Answer:
(a)
(b)
(c) 1 s
(d) 20 m
(e) 1 m
(f)
(g)
(h)
(i)
(j)
(k)
(l)
(m)
Explanation:
Since <em>x</em> is measured in meters and <em>t</em> in seconds, constants <em>a </em>and <em>b</em> must have units that gives meters when multiplied by square and cubic seconds respectivly, so that would mean for <em>a </em>and for <em>b</em>.
We can get the velocity <em>v </em>equation by deriving the position with respect to <em>t</em>, which gives:
And the acceleration <em>a</em> equation by deriving again:
Now for getting the maximun position between 0 and 4, we must find to points where the positions first derivate is equal to cero and evaluate those points. That is <em>v=0</em>, which gives
For <em>t = 0</em>,<em> x = 0</em> so the maximun position is archieved at 1 second, which gives <em>x = 1 meter</em>.
For obtaining it's displacement <em>r</em>, we can integrate the velocity from 0 seconds to 4 seconds, which gives the mean value of the position in that interval:
For the remaining questions, we just replace the values of <em>t</em> on the respective equations.
Answer:
h=0.425 L
Explanation:
Given that
θ = 54°
Coefficient of friction μ = 0.32
Mass of rod = m
Lets take mass of man = M = 4 m
C is the center of mass of the rod.
By balancing force in y and x direction
R= Fr
R = Fr= μ N
N = mg + Mg = mg + 4 m g ( M =4m)
N = 5 m g
Lets take distance cover by man is h along rod before sliding
Now taking moment about the lower end
M g h cosθ + m g cosθ L/2 = R L sinθ
2 M g h cosθ + m g cosθ L = 2 R L sinθ
Now by putting the value of R and M
8 m g h cosθ + m g cosθ L = 2 μ N L sinθ
8 m g h cosθ + m g cosθ L = 10 m g μ L sinθ
8 h cosθ + cosθ L = 10 μ L sinθ
8 h + L = 10 μ L tanθ
Now putting the value of θ and μ
8 h + L = 10 x 0.32 x tan54° x L
8 h + L = 4.4 L
8 h = 3.4 L
h=0.425 L