The answer is 60 mph.
The speed (v) is distance (d) per time (t): v = d/t
Car A:
v1 = ?
t1 = 2 h
d1 = ?
___
v1 = d1/t1
d1 = v1 * t1
Car B:
v2 = ?
t2 = 1.5 h
d2 = ?
___
v2 = d2/t2
d2 = v2 * t2
<span>Two cars traveled equal distances:
d1 = d2
</span>v1 * t1 = v2 * t2
<span>Car B traveled 15 mph faster than Car A:
v2 = v1 + 15
</span>v1 * t1 = v2 * t2
v2 = v1 + 15
________
v1 * 2 = (v1 + 15) * 1.5
2v1 = 1.5v1 + 22.5
2v1 - 1.5v1 = 22.5
0.5v1 = 22.5
v1 = 22.5/0.5
v1 = 45 mph
v2 = v1 + 15
v2 = 45 + 15
v2 = 60 mph
Answer:
So lift will be 30.19632 N
Explanation:
We have given area of the wing 
We know that density of air 
Speed at top surface
and speed at bottom surface 
According to Bernoulli's principle force is given by
Ngan's mass on earth is 85kg.
Ngan has a weight on Mars = 14.5 N
Ngan’s weight on Earth = 833.0 N
Ngan’s mass on Earth = ?
<span>Fg,earth = mg(earth)</span>
<span>M = Fg,earth </span><span>/ g(earth)</span>
<span>M = 833.0 N / 9.8 m/s2</span>
<span>M = 85 kg</span>
Answer:
The decibel of the remaining pigs is 51.5 dB.
Explanation:
Decibel (dB) is a unit of measure of the intensity of a given sound.
Number of pigs = 199, noise level = 74.3 dB.
Given that the intensity (I) of the sound from the pen is proportional to the number of pigs (N), thus:
I
N
I = kN
where k is the constant of proportionality.
⇒ k = 
= 
k = 0.3734
When 61 numbers of pigs were removed, the number of remaining pigs (N) squealing at their original level is 138.
Thus, the becibel level (I) of the remaining pigs can be determined by:
I = kN
= 0.3734 × 138
= 51.53 dB
The becibel level (I) of the remaining pigs is 51.53 dB.