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sdas [7]
3 years ago
8

How many meters in 2.50 miles? (Use these two conversions: 1000 m = 1 km and 1.00 km = .621 mi )

Physics
1 answer:
Artyom0805 [142]3 years ago
7 0

2.50 miles is equal to 4026 m.

<u>Explanation:</u>

As it is known that 1000 m =1 km and 1 km = 0.621 miles. So first we have to convert miles to km and then to metre as follows.

As 1 km = 0.621 miles, then

             \text { 1 miles }=\frac{1}{0.621} \mathrm{km}

So, 2.50 miles will be equal to

            2.50 \text { miles }=\frac{2.50}{0.621} \mathrm{km}=4.026 \mathrm{km}

Then, in order to get the answer in meters, we have to convert this km to meter by the conversion of 1000 m =1 km.  So,

           1 \mathrm{km}=1000 \mathrm{m}

Thereby,

          4.026 \mathrm{km}=4026 \mathrm{m}

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2 years ago
How many joules of kinetic energy does a pendulum have when it has 100 joules of potential energy
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Answer:

The maximum kinetic energy is 100 j.    

Explanation:

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3 years ago
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A rocket engine has a chamber pressure 4 MPa and a chamber temperature of 2000 K. Assuming isentropic expansion through the nozz
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This question is incomplete, the complete question is;

A rocket engine has a chamber pressure 4 MPa and a chamber temperature of 2000 K. Assuming isentropic expansion through the nozzle, and an exit Mach number of 3.2, what are the stagnation pressure and temperature in the exit plane of the nozzle?  Assume the specific heat ratio is 1.2.

Answer:

- stagnation pressure is 274.993 Mpa

- the stagnation temperature Tt is 4048 K

Explanation:

Given the data in the question;

To determine the stagnation pressure and temperature in the exit plane of the nozzle;

we us the expression;

Pt/P = (1 + (γ-1 / 2) M²)^(γ/γ -1) = ( Tt/T )^(γ/γ -1)

where Pt is stagnant pressure = ?

P is static pressure = 4 MPa = 4 × 10⁶ Pa  

Tt is stagnation temperature = ?

T is the static temperature  = 2000 K

γ is ratio of specific heats = 1.2

M is Mach number M = 3.2

we substitute

Pt/P = (1 + (γ-1 / 2) M²)^(γ/γ -1)

Pt = P(1 + (γ-1 / 2) M²)^(γ/γ -1)

Pt = 4 × 10⁶(1 + (1.2-1 / 2) 3.2²)^(1.2/1.2 -1)

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Pt = 274.993 × 10⁶ Pa

Pt = 274.993 Mpa

Therefore stagnation pressure is 274.993 Mpa

Now, to get our stagnation Temperature

Pt/P = ( Tt/T )^(γ/γ -1)

we substitute

274.993 × 10⁶ Pa / 4 × 10⁶ Pa =  ( Tt / 2000 )^(1.2/1.2 -1)

68.7484 =  Tt⁶ / 6.4 × 10¹⁹

Tt⁶ = 68.7484 × 6.4 × 10¹⁹

Tt⁶ = 4.3998976 × 10²¹

Tt = ⁶√(4.3998976 × 10²¹)

Tt = 4047.999 ≈ 4048 K

Therefore, the stagnation temperature Tt is 4048 K

6 0
3 years ago
If each of the three rotor helicopter blades is 3.50 m long and has a mass of 120 kg , calculate the moment of inertia of the th
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Answer:

1470kgm²

Explanation:

The formula for expressing the moment of inertial is expressed as;

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Since there are three rotor blades, the moment of inertia will be;

I = 3(1/3mr²)

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Required

Moment of inertia

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