The momentum does not change because he is going the same speed just a different way.
Answer:
so the height will be increased by 9 times
Explanation:
The truck is in its pontential energy at rest
The potential energy is converted to kinetic energy when in motion
Given that,
h = height of the hill
m = mass of the truck
g = acceleration due to gravity
v = velocity of the truck at the bottom of the hill.
At the top of the hill,
Potential energy(PE) = mgh
Kinetic energy(KE) = 0
Total energy, E = PE + KE = mgh + 0 = mgh____(1)
At the bottom of the hill :
Potential energy, PE = 0
Kinetic energy,KE = (1/2)mv²
Total energy, E = PE + KE
= 0 + (1/2)mv²_____(2)
In law of conservation of energy,
(1) = (2)
mgh = (1/2)mv²
h = (1/2g)v²
given,
v = 2 m/s
g = 9.81 m/s²
h = (1/2(9.81)) × 2²
h = 0.20 m
so,
if v = 6 m/s
h = (1/2(9.81)) × 6²
h = 1.83 m
from the first h = 0.20m and the second h = 1.83m
1.83m / 0.20m
= 9.15
so the height will be increased by 9 times
Answer:
P₂ = 392720.38 Pa = 392.72 kPa
Explanation:
Given
D₁ = 5 cm = 0.05 m
D₂ = 10 cm = 0.10 m
v₁ = 8 m/s
P₁ = 380 kPa = 380000 Pa
α = 1.06
ρ = 1000 kg/m³
g = 9.8 m/s²
We can use the following formula
(P₁ / (ρg)) + α*(V₁² / (2g)) + z₁ = (P₂ / (ρg)) + α*(V₂² / (2g)) + z₂ + +hL
knowing that z₁ = z₂ we have
(P₁ / (ρg)) + α*(V₁² / (2g)) = (P₂ / (ρg)) + α*(V₂² / (2g)) + +hL <em> (I)
</em>
Where
V₂ can be obtained as follows
V₁*A₁ = V₂*A₂ ⇒ V₁*( π* D₁² / 4) = V₂*( π* D₂² / 4)
⇒ V₂ = V₁*(D₁² / D₂²) = (8 m/s)* ((0.05 m)² / (0.10 m)²)
⇒ V₂ = 2 m/s
and
hL is a head loss factor: hL = α*(1 - (D₁² / D₂²))²*v₁² / (2*g)
⇒ hL = (1.06)*(1 – ((0.05 m) ² / (0.10 m)²))²*(8 m/s)² / (2*9.8 m/s²)
⇒ hL = 1.9469 m
Finally we get P₂ using the equation <em>(I)
</em>
⇒ P₂ = P₁ - ((V₂² - V₁²)* α*ρ / 2) – (ρ*g* hL)
⇒ P₂ = 380000 Pa - (((2 m/s)² - (8 m/s)²)*(1.06)*(1000 kg/m³) / 2) – (1000 kg/m³*9.8 m/s²*1.9469 m)
⇒ P₂ = 392720.38 Pa = 392.72 kPa
Answer: she did no work while she posed because she did not move the torch or the book
Explanation: just did it on edge