Answer:
The maximum data rate supported by this line is 39900 bps
Explanation:
The maximum data rate supported by this line can be obtained using the formula below
c = W*log2(S/N+1)
where;
c is the maximum data rate supported by the line
W is the bandwidth = 4kHz
S/N+1 is the signal to noise ratio = 1001
c = 4*log2(1001)
c = 39868.9 ≅ 39900 bps
Therefore, the maximum data rate supported by this line is 39900 bps
It’s not Cereal and Milk, carbonated drinks, and smoky air
Answer:
0.657 seconds
Explanation:
speed of wave= wavelength / time period
so
time period= wavelength / speed
= 4.6/7
=0.657 sec
Answer:D
Explanation:
Given
mass of object 
Distance traveled 
velocity acquired 
conserving Energy at the moment when object start falling and when it gains 12 m/s velocity
Initial Energy
Final Energy
where 
is friction work if any
Since Friction is Present therefore it is a case of Open system and net external Force is zero
An open system is a system where exchange of energy and mass is allowed and Friction is acting on the object shows that system is Open .
Explanation:
(a) Given:
Δx = 150 m
v₀ = 27 m/s
v = 54 m/s
Find: a
v² = v₀² + 2aΔx
(54 m/s)² = (27 m/s)² + 2a (150 m)
a = 7.29 m/s²
(b) Given:
Δx = 150 m
v₀ = 0 m/s
a = 7.29 m/s²
Find: t
Δx = v₀ t + ½ at²
150 m = (0 m/s) t + ½ (7.29 m/s²) t²
t = 6.42 s
(c) Given:
v₀ = 0 m/s
v = 27 m/s
a = 7.29 m/s²
Find: t
v = at + v₀
27 m/s = (7.29 m/s²) t + 0 m/s
t = 3.70 s
(d) Given:
v₀ = 0 m/s
v = 27 m/s
a = 7.29 m/s²
Find: Δx
v² = v₀² + 2aΔx
(27 m/s)² = (0 m/s)² + 2 (7.29 m/s²) Δx
Δx = 50 m
