Answer:
E = k Q₁ / r²
Explanation:
For this exercise that asks us for the electric field between the sphere and the spherical shell, we can use Gauss's law
Ф = ∫ E .dA = 
/ ε₀
where Ф the electric flow, qint is the charge inside the surface
To solve these problems we must create a Gaussian surface that takes advantage of the symmetry of the problem, in this almost our surface is a sphere of radius r, that this is the sphere of and the shell, bone
R <r <R_a
for this surface the electric field lines are radial and the radius of the sphere are also, therefore the two are parallel, which reduces the scalar product to the algebraic product.
E A = q_{int} /ε₀
The charge inside the surface is Q₁, since the other charge Q₂ is outside the Gaussian surface, therefore it does not contribute to the electric field
q_{int} = Q₁
The surface area is
A = 4π r²
we substitute
E 4π r² = Q₁ /ε₀
E = 1 / 4πε₀ Q₁ / r²
k = 1/4πε₀
E = k Q₁ / r²
I believe the correct answer from the choices listed above is the first option. Decreasing a telescope's eyepiece focal length will increase magnification. <span>The magnification of the </span>telescope<span> image is (</span>focal length<span> of the objective) divided by (</span>focal length <span>of the </span>eyepiece<span>). Hope this answers the question.</span>
<h2>
Answer: 12 s</h2>
Explanation:
The situation described here is parabolic movement. However, as we are told <u>the instrument is thrown upward</u> from the surface, we will only use the equations related to the Y axis.
In this sense, the main movement equation in the Y axis is:
(1)
Where:
is the instrument's final position
is the instrument's initial position
is the instrument's initial velocity
is the time the parabolic movement lasts
is the acceleration due to gravity at the surface of planet X.
As we know and 
when the object hits the ground, equation (1) is rewritten as:
(2)
Finding :
(3)
(4)
(5)
Finally:
Answer:
Explanation:
Given:
Initial mass of isotope (m₀) = 20 g
Half life of the isotope 
= (ln 4) years
The general form for the radioactive decay of a radioactive isotope is given as:
Where,
So, the equation is: 
At half-life, the mass is reduced to half of the initial value.
So, at 
. Plug in these values and solve for 'k'. This gives,
Hence, the equation for the mass remaining is given as:
