Answer:
C) only part of the bandwidth of the AM signal is amplified, causing some of the sideband information to be lost and distortion results.
Explanation:
Selectivity is the ability of a receiver to respond only to a specific signal on a wanted frequency and reject other signals nearby in frequency.
If a receiver is overly selective, only part of the bandwidth of the AM signal is amplified, causing some of the sideband information to be lost and distortion results. Whereas, if a receiver is underselective, the receiver can pick different signals on different frequencies at the same time.
Distance is 90km,time 1 hr
distance 2 is 82km,time 5 hrs
average speed=total distance travelled/total time taken.
90+82=172/5+1=6
average speed=28.7km over hour
V₁ = (1/g)₁ = Way₁ = 20(9.81)(0) = 0
V₂ (Vg)₂ = -WAy₂ = -20(9.81)(0.5) = -98.1J
The kinetic energy because the pool rotates about a fixed axis
W = VA/rA = VA/0.2 5VA
Mass momen of inertila about fixed axis which passes through point 0
I₀ = mko² = 50(0.280)² = 3.92 kg. m²
∴ The kinetic energy of the system is
T = 1/2 I₀w² + 1/2mAVA²
= 1/2(3.92)(5VA)² + 1/2 (20) VA² = 59VA²
Now that the system is at rest then T₁ = 0
Energy conservation is
T₁ +V₁ = T₂ + V₂
0+ 0 = 59VA² + (-98.1)
VA = 1.289 m/s
= 1.29 m/s
Answer:
Explanation:
given
T = 3months = 7.9 × 10⁶s
orbital speed = 88 × 10³m/s
V= 2πr÷T
∴ r = (V×T) ÷ 2π
r = (88km × 7.9 × 10⁶s) ÷ 2π
r = 1.10 × 10⁸km
using kepler's 3rd law
mass of both stars = (seperation diatance)³/(orbital speed)²
M₁ + M₂ = (2r)³/(
year)²
= (1.06 × 10²⁵)/(6.2×10¹³)
1.71×10¹²kg
since M₁ = M₂ =1.71×10¹²kg ÷ 2
M₁ = M₂ = 8.55×10¹¹kg
Answer: The moon would continue to move on its orbital path.
Explanation:
