Answer:
The mass of the block is 1250g.
Explanation:
Given that the formula for density is ρ = mass/volume. Then you have to substitute the values into the formula :

Let density = 250,
Let volume = 5,



Explanation:
frequency =speed/wavelength
=5/0.5=10Hz
Answer:
E = k Q / [d(d+L)]
Explanation:
As the charge distribution is continuous we must use integrals to solve the problem, using the equation of the elective field
E = k ∫ dq/ r² r^
"k" is the Coulomb constant 8.9875 10 9 N / m2 C2, "r" is the distance from the load to the calculation point, "dq" is the charge element and "r^" is a unit ventor from the load element to the point.
Suppose the rod is along the x-axis, let's look for the charge density per unit length, which is constant
λ = Q / L
If we derive from the length we have
λ = dq/dx ⇒ dq = L dx
We have the variation of the cgarge per unit length, now let's calculate the magnitude of the electric field produced by this small segment of charge
dE = k dq / x²2
dE = k λ dx / x²
Let us write the integral limits, the lower is the distance from the point to the nearest end of the rod "d" and the upper is this value plus the length of the rod "del" since with these limits we have all the chosen charge consider
E = k 
We take out the constant magnitudes and perform the integral
E = k λ (-1/x)
Evaluating
E = k λ [ 1/d - 1/ (d+L)]
Using λ = Q/L
E = k Q/L [ 1/d - 1/ (d+L)]
let's use a bit of arithmetic to simplify the expression
[ 1/d - 1/ (d+L)] = L /[d(d+L)]
The final result is
E = k Q / [d(d+L)]
Answer:
All of the above are true.
Explanation:
(a). true
whenever charge particle move back and froth from its mean position then it will produce oscillating electric and magnetic fields, . so an em wave can be obtain by accelerating charge
(b). true
the electric field and the magnetic field have vibrations in the perpendicular direction along the motion of the wave so electromagnetic wave is a transverse wave. therefore, the EM wave is a Transverse wave
(c) true .
The Electromagnetic wave consists of the two mutually perpendicular electric and magnetic fields and also both fields are perpendicular to the direction of propagation of the wave.
(d) true .
An electromagnetic wave carry energy through vacuum with a speed of
so , all of the above are true.