Answer:
Option D is correct: 170 µW/m²
Explanation:
Given that,
Frequency f = 800kHz
Distance d = 2.7km = 2700m
Electric field Eo = 0.36V/m
Intensity of radio signal
The intensity of radial signal is given as
I = c•εo•Eo²/2
Where c is speed of light
c = 3×10^8m/s
εo = 8.85 × 10^-12 C²/Nm²
I = 3×10^8 × 8.85×10^-12 × 0.36²/2
I = 1.72 × 10^-4W/m²
I = 172 × 10^-6 W/m²
I = 172 µW/m²
Then, the intensity of the radio wave at that point is approximately 170 µW/m²
Answer:
KE = 2.03 J
Explanation:
After impact, the kinetic energy of the bullet+block will convert to potential energy
½mv² = mgh
v = √(2gh) = √(2(9.81)(0.00500) = 0.0981 m/s
conservation of momentum during the collision
0.015u + 2.50(0) = (2.50 + 0.015)(0.0981)
u = 16.4481 m/s
KE = ½mv² = ½(0.015)16.4481² = 2.0290499...
KE = 2.03 J
Answer:
due to its distance ..............
Radiocarbon saying assume the half life for radioistopes change over time. The half life of C -14 is 5730 years
