Is a glacier nothing more than a huge mineral flowing downhill?
1 answer:
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Answer:
a) The work done by the force is 720 joules, b) The power supplied by the force is 138 watts.
Explanation:
a) Since force is uniform and parallel to the direction of motion, the work (), in joules, done by the force (
), in newtons, is defined by this formula:
(1)
Where is the travelled distance, in meters.
If we know that and
, then the work done by the force is:
The work done by the force is 720 joules.
b) And an expression for the power supplied by the force (), in watts, is concieved by differentiating (1) in time:
Where is the speed of the wagon, in meters per second.
If we know that and
, then the power supplied by the force is:
The power supplied by the force is 138 watts.
Answer:
The total distance is 24 m.
The tiger's resultant displacement is 8 m.
Explanation:
Given that,
Width of tiger= 8 m
Tiger starts from the right side of the cage,
Paces to the left side
Then, back to the right side,
finally, back to the left
We need to calculate the total distance covered
Using formula of distance
Put the value into the formula
We need to calculate the tiger's resultant displacement
Using formula of displacement
Put the value into the formula
Hence, The total distance is 24 m.
The tiger's resultant displacement is 8 m.
Answer:
a) q_inner = -Q
, q_outer = -2Q
b) E₁ = k Q / r² r<a
E₂ = 0 a<r<b
E₃ = - k 2Q/r² r>b
d) the charge continues inside the spherical shell, the results do not change
Explanation:
a) The point load in the center induces a load on the inner surface of the shell with constant opposite sign
q_inner = -Q
the outer shell of the shell the load is
q_outer = -3 Q + Q
q_outer = -2Q
b) To find the electric field again, use Gauss's law,
We define as a Gaussian surface a sphere
Ф = E. dA = /ε₀
in this case the electric field lines and the radii of the sphere are parallel, so the sclar product is reduced to the algegraic product
E A = q_{int}/ε₀
the area of a sphere is
A = 4 π r²
E = 1 / 4πε₀ Q/ r²
k = 1 / 4πε₀
let's apply this expression to the different radii
i) r <a
in this case the load inside is the point load
= + Q
E₁ = k Q / r²
ii) the field inside the shell
a <r <b
As the sphere is conductive, so that it is in electrostatic equilibrium, there can be no field within it.
E₂ = 0
iii) r>b
q_{int} = Q- 3Q = -2Q
E₃ = k (-2Q/r²)
E₃ = - k 2Q/r²
c) see attached
d) as the charge continues inside the spherical shell, the results do not change, since the lcharge inside remains the same and it does not matter its precise location, but remains within the Gaussian surface
