Answer:
B. Planets orbit a star, while moons orbit a planet
Explanation:
Every planet must orbit a star, like how the Earth orbits the Sun (the star of our solar system). Our moon orbits Earth. On other planets, there can be dozens of moons orbiting it.
Force is a vector quantity
so pulling from opposite side will be negative
so
750+(-500)= 250N
C is the right answer
becauseause the man on the right applies greater force.
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
Explanation:
As we know that
time = distance/speed
The time used for firs half of the trip was
(1 mi)(12 mi/hour) = 1/12 hours = 5 minutes
The last half of the trip will took 10 minutes, 1/6 hour.
Speed = distance/ time
(1 mil) = (1/6h) = 6 mil/h
so the speed for last half of the trip was = 6mph
the average speed was
(2mil)(1/4 hour) = 8 mil/hour
So the ling's average speed was 8mph.
Vocabulary should be, I think:
I. Hypothesis
II. Evidence, data
III. Experiment
What is your question exactly?