Answer:
A fire hose must be able to shoot water to the top of a building 35.0 m tall ... Water enters this hose at a steady rate of 0.500 m3/s and shoots out of a round nozzle. ... I know that Flow rate=0.500 m3/s=A*V. I know the pressure needed to ... The first equation has no potential while the second has no kinetic.
Explanation:
Answer:
the correct affirmation is the 3
Explanation:
Let's analyze the problem with Newton's second law before looking at the claims.
X axis parallel to the plane, positive down
F -fr + Wₓ = ma
Y Axis perpendicular to the plane
N -Wy = 0
With trigonometry
Wₓ = W sin θ
Wy = w cos θ
Let's multiply by the displacement along the plane, to relate to the work, which has as expression W = F d
F d -fr d + Wx d = ma d
Push W₁ = Fd
frictional force W₂ = -fr d
gravity W₃ = Wx d
W₁ + W₃ -W₂ = m a d
Analysis affirmations:
R1) false. The work of gravity is the subtraction
R2) false. Each force contributes according to its magnitude
R3) true. In the equation we see that, if the acceleration is zero, W2 = W1 + W3
R4) False. It equals the difference
the correct affirmation is the 3
Nine
Basically you are using d = r * t
Frequency is 1 / t
d is more commonly represented by lambda (
). Lambda is the wavelength.
so the formula = c / f
More commonly it is written as *f = c.
Givens
c = 3*10^8 m/s
= 0.1 nm = 0.1 nm *[1 m /1 * 10^9 nm] = 0.1 * 10^ - 9 meters.
f = ??
f = 
f = 3 * 10^8 * 10^9 / 0.1 = 3 * 10^18
Answer: C
Ten
= c/f
= 3*10^8 / 1 * 10^16
= 3*10^-8 meters
= 30 * 10^-9 meters = 30 nm
Answer: D
