Jody has a mass of 50 kilograms what is his weight on earth
2 answers:
Answer:
Weight of body = 490 Newton
Explanation:
Mass of the body is the total amount of matter contained in the body but Weight of the body is the force exerted by gravity on the body. Weight of the body depends upon acceleration due to gravity of the planet.
Thus,
Weight = mass × acceleration due to gravity
W = mg
W = weight of the body
m = mass of the body
g = acceleration due to gravity
W = 50 × 9.8
W = 490
W = 490 Newton
Hence, weight of the body is 490 Newton.
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Answer:
It would take approximately 289 hours for the population to double
Explanation:
Recall the expression for the continuous exponential growth of a population:
where N(t) measures the number of individuals, No is the original population, "k" is the percent rate of growth, and "t" is the time elapsed.
In our case, we don't know No (original population, but know that we want it to double in a certain elapsed "t". We also have in mind that the percent rate "k" would be expressed in mathematical form as: 0.0024 (mathematical form of the given percent growth rate).
So we need to solve for "t" in the following equation:
Which can be rounded to about 289 hours
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
One charge is enough in order to have an electric field.
In fact, every charge generates an electric field: for example, the electric field generated by a single point positive charge is radial, as shown in the attached figure. More complicate electric field configurations can be obtained adding charges or using more complicate charge distributions, but one charge is enough to have an electric field.
In fact, every charge generates an electric field: for example, the electric field generated by a single point positive charge is radial, as shown in the attached figure. More complicate electric field configurations can be obtained adding charges or using more complicate charge distributions, but one charge is enough to have an electric field.