answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
OlgaM077 [116]
2 years ago
15

pitot tube on an airplane flying at a standard sea level reads 1.07 x 105 N/m2. What is the velocity of the airplane?

Physics
1 answer:
Allushta [10]2 years ago
8 0

Answer:

V_infinty=98.772 m/s

Explanation:

complete question is:

The following problem assume an inviscid, incompressible flow. Also, standard sea level density and pressure are 1.23kg/m3(0.002377slug/ft3) and 1.01imes105N/m2(2116lb/ft2), respectively. A Pitot tube on an airplane flying at standard sea level reads 1.07imes105N/m2. What is the velocity of the airplane?

<u>solution:</u>

<u>given:</u>

<em>p_o=1.07*10^5 N/m^2</em>

<em>ρ_infinity=1.23 kg/m^2</em>

<em>p_infinity=1.01*10^5 N/m^2</em>

p_o=p_infinity+(1/2)*(ρ_infinity)*V_infinty^2

V_infinty^2=9756.097

V_infinty=98.772 m/s

You might be interested in
A ball is dropped from the top of a building.After 2 seconds, it’s velocity is measured to be 19.6 m/s. Calculate the accelerati
zlopas [31]

Answer:

acceleration, a = 9.8 m/s²

Explanation:

'A ball is dropped from the top of a building' indicates that the initial velocity of the ball is zero.

u = 0 m/s

After 2 seconds, velocity of the ball is 19.6 m/s.

t = 2s, v = 19.6 m/s

Using

v = u + at

19.6 = 0 + 2a

a = 9.8 m/s²

6 0
2 years ago
Read 2 more answers
To practice Problem-Solving Strategy 17.1 for wave interference problems. Two loudspeakers are placed side by side a distance d
Nimfa-mama [501]

Complete Question

The compete question is shown on the first uploaded question

Answer:

The speed is  v  =  350 \  m/s  

Explanation:

From the question we are told that

   The  distance of separation is  d =  4.00 m  

  The distance of the listener to the center between the speakers is  I =  5.00 m

  The change in the distance of the speaker is by k  =  60 cm  =  0.6 \  m

    The frequency of both speakers is f =  700 \  Hz

Generally the distance of the listener to the first speaker is mathematically represented as

       L_1  =  \sqrt{l^2 + [\frac{d}{2} ]^2}

       L_1  =  \sqrt{5^2 + [\frac{4}{2} ]^2}

        L_1  =   5.39 \  m

Generally the distance of the listener to second speaker at its new position is  

          L_2  =  \sqrt{l^2 + [\frac{d}{2} ]^2 + k}

       L_2  =  \sqrt{5^2 + [\frac{4}{2} ]^2 + 0.6}

        L_2  =   5.64  \  m  

Generally the path difference between the speakers is mathematically represented as

        pD  = L_2 - L_1  =  \frac{n  *  \lambda}{2}

Here \lambda is the wavelength which is mathematically represented as

         \lambda =  \frac{v}{f}

=>    L_2 - L_1  =  \frac{n  *  \frac{v}{f}}{2}

=>    L_2 - L_1  =  \frac{n  *  v}{2f}  

=>    L_2 - L_1  =  \frac{n  *  v}{2f}  

Here n is the order of the maxima with  value of  n =  1  this because we are considering two adjacent waves

=>    5.64 - 5.39   =  \frac{1  *  v}{2*700}      

=>    v  =  350 \  m/s  

7 0
2 years ago
A future use of space stations may be to provide hospitals for severely burned persons. it is very painful for a badly burned pe
natta225 [31]
<span>1.5 minutes per rotation. The formula for centripetal force is A = v^2/r where A = acceleration v = velocity r = radius So let's substitute the known values and solve for v. So F = v^2/r 0.98 m/s^2 = v^2/200 m 196 m^2/s^2 = v^2 14 m/s = v So we need a velocity of 14 m/s. Let's calculate how fast the station needs to spin. Its circumference is 2*pi*r, so C = 2 * 3.14159 * 200 m C = 1256.636 m And we need a velocity of 14 m/s, so 1256.636 m / 14 m/s = 89.75971429 s Rounding to 2 significant digits gives us a rotational period of 90 seconds, or 1.5 minutes.</span>
5 0
2 years ago
An auto moves 10 meters in the first second of travel, 15 more meters in the next second, and 20 more meters during the third se
k0ka [10]

im guessing it's 5 m/s

6 0
2 years ago
The eyes of many older people have lost the ability to accommodate, and so an older person’s near point may be more than 25 cm f
tensa zangetsu [6.8K]

Answer:

Smaller refractive power

Explanation:

The refractive power of an eye is the extent to which it can converge or diverge the light rays.

Near point is the the closest point for an eye such that when an object is placed at that point the image it forms is sharp and clearly visible to the eye.

A the person ages, the ciliary muscles of the eyes weakens and as a result the lens contracts and the formation of the image takes place behind the retina instead of forming at the retina.

Thus the near point also increases and the refractive power becomes smaller.

4 0
2 years ago
Other questions:
  • An elevator is used to either raise or lower sacks of potatoes. In the diagram, a sack of potatoes of mass 10 kg is resting on a
    7·1 answer
  • Determine the cutting force f exerted on the rod s in terms of the forces p applied to the handles of the heavy-duty cutter.
    11·2 answers
  • Un electrón en un tubo de rayos catódicos acelera desde el reposo con una aceleración constante de 5.33x10¹²m/s² durante 0.150μs
    10·1 answer
  • The suns energy is classified by the
    15·2 answers
  • The bird is held in level flight due to the force exerted on it by the air as the bird beats its wings. What is the maximum valu
    5·1 answer
  • A particular string resonates in four loops at a frequency of 320 Hz . Name at least three other (smaller) frequencies at which
    6·1 answer
  • A diffusion couple composed of two silver– gold alloys is formed; these alloys have compositions of 98 wt% Ag–2 wt% Au and 95 wt
    10·1 answer
  • Magnus has reached the finals of a strength competition. In the first round, he has to pull a city bus as far as he can. One end
    9·1 answer
  • High-speed stroboscopic photographs show that the head of a 200 g golf club is traveling at 43.7 m/s just before it strikes a 45
    14·1 answer
  • A wire long and with mass is positioned horizontally near the earth's surface and perpendicular to a horizontal magnetic field o
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!