Answer:
23.4 m/s
Explanation:
f = actual frequency of the wave = 6.2 x 10⁹ Hz
= frequency observed as the ball approach the radar
= frequency observed as the ball recede away from the radar
V = speed of light
= speed of ball
B = beat frequency = 969 Hz
frequency observed as the ball approach the radar is given as
eq-1
frequency observed as the ball recede the radar is given as
eq-2
Beat frequency is given as
Using eq-2 and eq-1
inserting the values
= 23.4 m/s
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
Answer:
Since opposite charges attract each other, the negatively charged electrons are attracted to the positively charged protons. ... Together, all of the electrons of an atom create a negative chargethat balances the positive charge of the protons in the atomic nucleus.
Explanation:
<h3>I hope it is helpful for you ...</h3>
Answer: The first one
Explanation:
Because it’s the higher frequency
