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Anastaziya [24]
3 years ago
10

Consider an aircraft powered by a turbojet engine that has a pressure ratio of 12. The aircraft is stationary on the ground, hel

d in position by its brakes. The ambient air is at 300 K and 95 kPa and enters the engine at a rate of 10 kg/s. The jet fuel has a heating value of 42,700 kJ/kg, and it is burned completely at a rate of 0.2 kg/s. Neglecting the effect of the diffuser and disregarding the slight increase in mass at the engine exit as well as the inefficiencies of engine components, determine the force that must be applied on the brakes to hold the aircraft stationary, in kN.
Physics
1 answer:
agasfer [191]3 years ago
3 0

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A Shaolin monk of mass 60 kg is able to do a ‘finger stand’: he supports his whole weight on his two index fingers, giving him a
ss7ja [257]

Answer:

P = 1471500 [Pa]

Explanation:

We must remember that pressure is defined as the relationship between Force over the area.

P=F/A

where:

P = pressure [Pa] (units of pascals)

F = force [N] (units of Newtons)

A = area of contact = 4 [cm²]

But first we must convert from cm² to m²

A = 4[cm^{2}]*\frac{1^{2} m^{2} }{100^{2} cm^{2} }

A = 0.0004 [m²]

Also, the weight should be calculated as follows:

w = m*g

where:

m = mass = 60 [kg]

g = gravity acceleration = 9.81 [m/s²]

Now replacing:

w = 60*9.81\\w = 588.6[N]

And the pressure:

P=588.6/0.0004\\P=1471500 [Pa]

Because 1 [Pa] = 1 [N/m²]

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3 years ago
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myrzilka [38]
Do you speak a little English cuz I can’t help you if a can’t understand you
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3 years ago
what is the force per meter on a straight wire carrying 5.0 a when it is placed in a magnetic field of 0.020 t
Dvinal [7]

The Force per meter on a straight wire carrying current in a magnetic field is<u>  0.045 N/m.</u>

<u>Calculation:-</u>

       F/ℓ = B I sin θ

  Where B – Magnetic field = 0.02 T I – Current = 5 A          

Substituting the values

F/ℓ = (0.02) (5) (sin 27 deg)

F/ℓ = <u>0.045 N/m</u>

A force is an influence that can alternate the motion of an item. A force can cause an item with mass to trade its pace, i.e., to boost up. force can also be described intuitively as a push or a pull. A pressure has both value and course, making it a vector quantity.

The push or pull on an item with mass causes it to change its velocity. force is an external agent capable of converting a frame's nation of relaxation or motion. It has significance and a path. A force is a push or pulls among gadgets. it is called an interplay because if one object acts on some other, its movement is matched with the aid of a reaction from the alternative object.

Learn more about force here:-brainly.com/question/12970081

#SPJ4

3 0
1 year ago
Sam is observing the velocity of a car at different times. After three hours, the velocity of the car is 53 km/h. After six hour
Serhud [2]
For the answer to the question above, first find out the gradient. 

<span>m = rise/run </span>
<span>=(y2-y1)/(x2-x1) </span>

<span>the x's and y's are the points given: "After three hours, the velocity of the car is 53 km/h. After six hours, the velocity of the car is 62 km/h" </span>
<span>(x1,y1) = (3,53) </span>
<span>(x2,y2) = (6,62) </span>

<span>sub values back into the equation </span>
<span>m = (62-53)/(6-3) </span>
<span>m = 9/3 </span>
<span>m = 3 </span>

<span>now we use a point-slope form to find the the standard form </span>
<span>y-y1 = m(x-x1) </span>
<span>where x1 and y1 are any set of point given </span>
<span>y-53 = 3(x-3) </span>
<span>y-53 = 3x - 9 </span>
<span>y = 3x - 9 + 53 </span>
<span>y = 3x + 44 </span>

<span>y is the velocity of the car, x is the time.
</span>I hope this helps.
4 0
3 years ago
2. If you want 0. 250 a (250 milliamps) to flow around a circuit with a resistance of 400 ohms, what voltage do you need?
grigory [225]

Answer:

0.000625 V

Explanation:

The formula linking current , resistance and voltage is :

V = I/R

Voltage = Current / Resistance

Now we substitute values given in question :

Voltage = 0.250 / 400

Voltage (V) = 0.000625

Our final answer is 0.000625 V

Hope this helped and have a good day

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