Answer:
-6.6 km/h
Explanation:
In 7hr plane travelled 2020km;
For the first 4hr the average speed was 310km/h;
d=st, s=d/t;
Distance covered in first 4h is d = 310km/h×4h = 1240km;
See the image attached for further solution
Answer:
Explanation:
Analysis of structure gives
a=gsinθ−μkgcosθ
Notice that all the expression are right but we want to know of we can simplify the expression further.
We want to analyse if we can still further simplify the expression,
Inspecting the Right hand side of the equation, we notice that the acceleration due to gravity is common to both side, so we can bring it out i.e.
So option a is wrong because the expression can be simplified further to
a=g(sinθ−μkcosθ)
Option b is right and the best option.
Since we are given that, g=9.8m/s²
We can as well substitute that to option a
So we will have
a=9.8metre/second²(sinθ−μkcosθ)
Also option C is correct but it is not best inserting the values of g directly without simplifying the expression first
So it will have been the best option if it was written as
a=9.8metre/second²(sinθ−μkcosθ)
So the best option is B.
In collision that are categorized as elastic, the total kinetic energy of the system is preserved such that,
KE1 = KE2
The kinetic energy of the system before the collision is solved below.
KE1 = (0.5)(25)(20)² + (0.5)(10g)(15)²
KE1 = 6125 g cm²/s²
This value should also be equal to KE2, which can be calculated using the conditions after the collision.
KE2 = 6125 g cm²/s² = (0.5)(10)(22.1)² + (0.5)(25)(x²)
The value of x from the equation is 17.16 cm/s.
Hence, the answer is 17.16 cm/s.
Answer: Mathematical Model
Explanation:
Took the test
Answer:
P₁ = 2.3506 10⁵ Pa
Explanation:
For this exercise we use Bernoulli's equation and continuity, where point 1 is in the hose and point 2 in the nozzle
P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂
A₁ v₁ = A₂ v₂
Let's look for the areas
r₁ = d₁ / 2 = 2.25 / 2 = 1,125 cm
r₂ = d₂ / 2 = 0.2 / 2 = 0.100 cm
A₁ = π r₁²
A₁ = π 1.125²
A₁ = 3,976 cm²
A₂ = π r₂²
A₂ = π 0.1²
A₂ = 0.0452 cm²
Now with the continuity equation we can look for the speed of water inside the hose
v₁ = v₂ A₂ / A₁
v₁ = 11.2 0.0452 / 3.976
v₁ = 0.1273 m / s
Now we can use Bernoulli's equation, pa pressure at the nozzle is the air pressure (P₂ = Patm) the hose must be on the floor so the height is zero (y₁ = 0)
P₁ + ½ ρ v₁² = Patm + ½ ρ v₂² + ρ g y₂
P₁ = Patm + ½ ρ (v₂² - v₁²) + ρ g y₂
Let's calculate
P₁ = 1.013 10⁵ + ½ 1000 (11.2² - 0.1273²) + 1000 9.8 7.25
P₁ = 1.013 10⁵ + 6.271 10⁴ + 7.105 10⁴
P₁ = 2.3506 10⁵ Pa
