Answer:
F₁ = 1500 N
F₂ = 750 N
= 500 N
Explanation:
Given :
Power transmission, P = 7.5 kW
= 7.5 x 1000 W
= 7500 W
Belt velocity, V = 10 m/s
F₁ = 2 F₂
Now we know from power transmission equation
P = ( F₁ - F₂ ) x V
7500 = ( F₁ - F₂ ) x 10
750 = F₁ - F₂
750 = 2 F₂ - F₂ ( ∵F₁ = 2 F₂ )
∴F₂ = 750 N
Now F₁ = 2 F₂
F₁ = 2 x F₂
F₁ = 2 x 750
F₁ = 1500 N , this is the maximum force.
Therefore we know,
= 3 x 
where
is centrifugal force
=
/ 3
= 1500 / 3
= 500 N
Answer:
0.304 L of Freon is needed
Explanation:
Q = mCT
Q is quantity of energy that must be removed = 47 BTU = 47×1055.06 = 49587.82 J
C is specific heat of Freon = 74 J/mol.K = 74 J/mol.K × 1 mol/120 g = 0.617 J/g.K
T is temperature in the area of Mars = 189 K
m = Q/CT = 49587.82/(0.617×189) = 452.23 g = 452.24/1000 = 0.45223 kg
Density of Freon = specific gravity of Freon × density of water = 1.49 × 1000 kg/m^3 = 1490 kg/m^3
Volume of Freon = mass/density = 0.45223/1490 = 0.000304 m^3 = 0.000304×1000 = 0.304 L
Answer:
G = $37,805.65
Explanation:
I found this on another site:
475,000 = 25,000(P/A,10%,6) + G(P/G,10%,6)
475,000 = 25,000(4.3553) + G(9.6842)
9.6842G = 366,117.50
G = $37,805.65
When a person is turning onto a two-lane road divided by a broken yellow line, you know immediately that you are on a two-way road.
<h3>What is the road about?</h3>
Note that a Yellow centerlines can be seen in roads and it is one that is often used to separate traffic moving in different directions.
Note also that Broken lines can be crossed to allow slower-moving traffic and as such, When a person is turning onto a two-lane road divided by a broken yellow line, you know immediately that you are on a two-way road.
See full question below
You are turning onto a two-lane road divided by a broken yellow line. You know immediately that:
Answers
You are on a two-way road.
You are on a one-way road.
The road is under repair.
You must stay to the left of the broken yellow lines.
Learn more about two-way road from
brainly.com/question/13123201
#SPJ2
Answer:
≅ 111 KN
Explanation:
Given that;
A medium-sized jet has a 3.8-mm-diameter i.e diameter (d) = 3.8
mass = 85,000 kg
drag co-efficient (C) = 0.37
(velocity (v)= 230 m/s
density (ρ) = 1.0 kg/m³
To calculate the thrust; we need to determine the relation of the drag force; which is given as:
=
× CρAv²
where;
ρ = density of air wind.
C = drag co-efficient
A = Area of the jet
v = velocity of the jet
From the question, we can deduce that the jet is in motion with a constant speed; as such: the net force acting on the jet in the air = 0
SO, 
We can as well say:

We can now replace
in the above equation.
Therefore,
=
× CρAv²
The A which stands as the area of the jet is given by the formula:

We can now have a new equation after substituting our A into the previous equation as:
=
× Cρ 
Substituting our data from above; we have:
=
× 
= 
= 110,990N
in N (newton) to KN (kilo-newton) will be:
= 
= 110.990 KN
≅ 111 KN
In conclusion, the jet engine needed to provide 111 KN thrust in order to cruise at 230 m/s at an altitude where the air density is 1.0 kg/m³.