Answer:
3.1 m/s
Explanation:
First, find the time it takes for the cat to land. Take down to be positive.
Given:
Δy = 0.61 m
v₀ = 0 m/s
a = 9.81 m/s²
Find: t
Δy = v₀ t + ½ at²
(0.61 m) = (0 m/s) t + ½ (9.81 m/s²) t²
t = 0.353 s
Now find the horizontal velocity needed to travel 1.1 m in that time.
Given:
Δx = 1.1 m
a = 0 m/s²
t = 0.353 s
Find: v₀
Δx = v₀ t + ½ at²
(1.1 m) = v₀ (0.353 s) + ½ (0 m/s²) (0.353 s)²
v₀ = 3.1 m/s
The sound gets louder as it gets closer and when it passes is gets softer
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
Answer:
What is the pressure transmitted in the liquid on a hydraulic pump where an elephant with a weight of 40 000 N is placed on top of the large piston with an area of 40 m2. The small piston area is 4 m2.
Answer:
Part a)
Part b)
Explanation:
Part a)
As we know that both charge particles will exert equal and opposite force on each other
so here the force on both the charges will be equal in magnitude
so we will have
here we have
now we have
Part b)
Now for the force between two charges we can say
now we have
now we have
