Answer: I believe his brother, Mike, who is 6'2, will have the longer shadow.
Answer:
p2 = 9.8×10^4 Pa
Explanation:
Total pressure is constant and PT = P = 1/2×ρ×v^2
So p1 + 1/2×ρ×(v1)^2 = p2 + 1/2×ρ×(v2)^2
from continuity we have ρ×A1×v1 = ρ×A2×v2
v2 = v1×A1/A2
and
r2 = 2×r1
then:
A2 = 4×A1
so,
v2 = (v1)/4
then:
p2 = p1 + 1/2×ρ×(v1)^2 - 1/2×ρ×(v2)^2 = p1 + 1/2×ρ×(v1)^2 - 1/2×ρ×(v1/4)^2
p2 = 3.0×10^4 Pa + 1/2×(1000 kg/m^3)×(12m/s)^2 - 1/2×(1000kg/m^3)×(12^2/16)
= 9.75×10^4 Pa
= 9.8×10^4 Pa
Therefore, the pressure in the wider section is 9.8×10^4 Pa
Answer:
C. f = 440 Hz
Explanation:
- In any wave, there is a fixed relationship between the wavelength (distance between two sucesive crests), the frequency (number of cycles per unit time) and the perturbation speed, as follows:
where v = speed of the wave
λ = wavelength
f = frequency
- Replacing by the givens in (1) and solving for f, we have:
Answer:
(c) 0.20 m/s²
Explanation:
Using The equation of motion,
V² = u² + 2gs...................... equation 1
Where V = final velocity, u = initial velocity, g = gravitational acceleration,
s = distance.
Making g the subject of the equation in equation 1 above,
g =( V² - u²)/2s............................ equation 2
Where V= 8 m/s, u = 5 m/s, s= 100 m.
g = (8²-5²)/(2×100)
g = (64-25)/200
g = 39/200
g =0.195 m/s²
g ≈ 0.20 m/s².
The gravitational acceleration = 0.20 m/s²
