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3 years ago

find the potential energy of an aircraft weighing 10000 bs at 5000 ft true altitude and 125 kts true air speed

Physics

1 answer:

Cloud [144]3 years ago

6 0

Answer:

$U=5*10^7ft.Ib$

Explanation:

From the question we are told that

Weight $W= 10000bs$

Altitude $H=5000ft$

Speed $V=125kts\\1kts=0.514m/s\\V=125*0.514=>64.25m/s$

Generally the equation for Potential energy ids mathematically given as

$Potential\ Energy\ U=mgh$

$U=Wh$

$U=10000*5000$

$U=5*10^7ft.Ib$

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Tuning an Instrument. A musician tunes the C-string of her instrumeut to a fundamental frequency of 65.4 Hz. The vibrating porti

mylen [45]

(a) The tension the musician must stretch it is 147.82 N.

(b) The percent increase in tension is needed to increase the frequency is 26%.

<h3>Tension in the string</h3>

v = √T/μ

where;

v is speed of the wave
T is tension
μ is mass per unit length = 0.0144 kg / 0.6 m = 0.024 kg/m

v = Fλ

in fundamental mode, v = F(2L)

v = 2FL

v = 2 x 65.4 x 0.6 = 78.48 m/s

v = √T/μ

v² = T/μ

T = μv²

T = 0.024 x (78.48)²

T = 147.82 N

<h3>When the frequency is 73.4 Hz;</h3>

v = 2FL = 2 x 73.4 x 0.6 = 88.08 m/s

T = μv²

T = (0.02)(88.08)²

T = 186.19 N

<h3>Increase in the tension</h3>

= (186.19 - 147.82)/(147.82)

= 0.26

= 0.26 x 100%

= 26 %

Thus, the tension the musician must stretch it is 147.82 N.

The percent increase in tension is needed to increase the frequency is 26%.

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1 year ago

A remote controlled toy car starts from rest and begins to accelerate in a straight line. The figure below represents "snapshots

Nikolay [14]

(a). The car's average velocity between t = 1.0s to t = 1.5s will be - $1\;m/s$

(b). The car's acceleration at t = 1.5s will be - $0.4\;m/s^{2}$

(c). Car's speed is increasing with time.

We have a a remote controlled toy car that starts from rest and begins to accelerate in a straight line.

We have to determine -

The car's average velocity (in m/s) in the interval between -

t = 1.0 s to t = 1.5 s.

The car's acceleration at t = 1.5 s.
Determining whether car's speed increasing or decreasing with time.

<h3>What is Acceleration?</h3>

The rate of change of velocity with respect to time is called Acceleration. Mathematically -

$$a=\frac{dv}{dt}$

According to the question, we have the following data for the Car -

t = 0s → x = 0m

t = 0.5s → x = 0.1m

t = 1.0s → x = 0.4m

t = 1.5s → x = 0.9m

t = 2.0s → x = 1.6m

PART - A

The car's average velocity between t = 1.0s to t = 1.5s will be -

$$v_{avg} = \frac{0.9-0.4}{1.5-1}= 1 m/s$

PART - B

Velocity at t = 1.5 s will be -

$$v(1.5)=\frac{0.9}{1.5}= 0.6\;m/s$

The car's acceleration at t = 1.5s will be -

$$a(1.5) = \frac{v}{t} = \frac{0.6}{1.5} = 0.4\;m/s^{2}$

PART - C

Since, the acceleration of the car is positive, this means that the car is accelerating in the forward direction. Hence, its speed is increasing with time.

[ The following data was missing in your answer. The complete question would include this data also -

t = 0s → x = 0m

t = 0.5s → x = 0.1m

t = 1.0s → x = 0.4m

t = 1.5s → x = 0.9m

t = 2.0s → x = 1.6m ]

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7 0

1 year ago

Read 2 more answers

Lynn rubs a balloon with a piece of wool, which causes the balloon to pick up some of the electric charges from the wool. Lynn t

Hitman42 [59]

Answer:

electrostatics

gravitational forces

Explanation:

rubbing the ballon leads to it acquiring charges that are opposite to that in the wall.the electostatic forces attract the ballon to the wall and tyhe gravitational pull later acts on the ballon causing it to fall down

5 0

3 years ago

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What is defined by force per unit area

sleet_krkn [62]

Explanation:

Force per unit area is called Pressure.

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3 years ago

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A crude approximation for the x component of velocity in an incompressible laminar boundary layer is a linear variation from u =

slega [8]

Answer:

2.5 * 10^-3

Explanation:

solution:

The simplest solution is obtained if we assume that this is a two-dimensional steady flow, since in that case there are no dependencies upon the z coordinate or time t. Also, we will assume that there are no additional arbitrary purely x dependent functions f (x) in the velocity component v. The continuity equation for a two-dimensional in compressible flow states:

δu/δx+δv/δy=0

so that:

δv/δy= -δu/δx

Now, since u = Uy/δ, where δ = cx^1/2, we have that:

u=U*y/cx^1/2

and we obtain:

δv/δy=U*y/2cx^3/2

The last equation can be integrated to obtain (while also using the condition of simplest solution - no z or t dependence, and no additional arbitrary functions of x):

v=∫δv/δy(dy)=U*y/4cx^1/2

=y/x*(U*y/4cx^1/2)

=u*y/4x

which is exactly what we needed to demonstrate.

Also, using u = U*y/δ in the last equation we can obtain:

v/U=u*y/4*U*x

=y^2/4*δ*x

which obviously attains its maximum value for the which is y = δ (boundary-layer edge). So, finally:

(v/U)_max=δ^2/4δx

=δ/4x

=2.5 * 10^-3

7 0

3 years ago