A block is attached to a spring, with spring constant k, which is attached to a wall. It is initially moved to the left a distan
1 answer:
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The power is 833.3 W
Explanation:
First of all, we need to calculate the work done in lifting the barbell, which is equal to the change in gravitational potential energy of the barbell:
where
mg = 1250 N is the weight of the barbell
h = 2 m is the change in height
Substituting,
Now we can calculate the power, which is equal to the work done per unit time:
where
W = 2500 J is the work done
t = 3 s is the time taken
Substituting,
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The total work done by the electric field on the charge is given by the scalar product between the electric force acting on the charge and the displacement of the charge:
where the force is F=qE, d=0.556 and
. Using the value of q and E given by the problem, we find
where the force is F=qE, d=0.556 and
Answer:
Explanation:
GIVEN
diameter = 15 fm =m
we use here energy conservation
there will be some initial kinetic energy but after collision kinetic energy will zero
on solving these equations we get kinetic energy initial
J ..............(i)
That is, the alpha particle must be fired with 35.33 MeV of kinetic energy. An alpha particle with charge q = 2 e
and gains kinetic energy K =e∆V ..........(ii)
by accelerating through a potential difference ∆V
Thus the alpha particle will
just reach the nucleus after being accelerated through a potential difference ∆V
equating (i) and second equation we get
e∆V = 35.33 Me V