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2 years ago

9

a) CALCULATE the acceleration of a 300,000 kg jumbo jet, just before takeoff when the thrust of each of its 4 engines is 36,000

N. If the jumbo jet take off speed is 184 mph, b) CONVERT 184 mph to m/s and c) DETERMINE how many seconds would it take for the plane to reach this speed? Show your work and round your answer to the nearest hundredth

1 answer:

Zanzabum2 years ago

5 0

Answer:

a) 0.12m/s^2

b)0.0511m/s

c)0.4258 seconds

Explanation:

a)

$acceleration= \frac{36000}{300000}$

= 0.12m/s^2

b) 184m/h

184/3600 = 0.0511m/s

c)

$time = \frac{0.0511}{0.12}$

Time = 0.4258 seconds

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3 years ago

A 10 g particle undergoes SHM with an amplitude of 2.0 mm and a maximum acceleration of magnitude 8.0 multiplied by 103 m/s2, an

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Answer:

a) $T=0.0031416s$

b) $v_{max}=6.283\frac{m}{s}$

c) $E=0.1974J$

d) $F=80N$

e) $F=40N$

Explanation:

1) Important concepts

Simple harmonic motion is defined as "the motion of a mass on a spring when it is subject to the linear elastic restoring force given by Hooke's Law (F=-Kx). The motion experimented by the particle is sinusoidal in time and demonstrates a single resonant frequency".

2) Part a

The equation that describes the simple armonic motion is given by $X=Acos(\omega t +\phi)$ (1)

And taking the first and second derivate of the equation (1) we obtain the velocity and acceleration function respectively.

For the velocity:

$\frac{dX}{dt}=v(t)=-A\omega sin(\omega t +\phi)$ (2)

For the acceleration

$\frac{d^2 X}{dt}=a(t)=-A\omega^2 cos(\omega t+\phi)$ (3)

As we can see in equation (3) the acceleration would be maximum when the cosine term would be -1 and on this case:

$A\omega^2=8x10^{3}\frac{m}{s^2}$

Since we know the amplitude A=0.002m we can solve for $\omega$ like this:

$\omega =\sqrt{\frac{8000\frac{m}{s^2}}{0.002m}}=2000\frac{rad}{s}$

And we with this value we can find the period with the following formula

$T=\frac{2\pi}{\omega}=\frac{2 \pi}{2000\frac{rad}{s}}=0.0031416s$

3) Part b

From equation (2) we see that the maximum velocity occurs when the sine function is euqal to -1 and on this case we have that:

$v_{max}=A\omega =0.002mx2000\frac{rad}{s}=4\frac{m rad}{s}=4\frac{m}{s}$

4) Part c

In order to find the total mechanical energy of the oscillator we can use this formula:

$E=\frac{1}{2}mv^2_{max}=\frac{1}{2}(0.01kg)(6.283\frac{m}{s})^2=0.1974J$

5) Part d

When we want to find the force from the 2nd Law of Newton we know that F=ma.

At the maximum displacement we know that X=A, and in order to that happens $cos(\omega t +\phi)=1$ , and we also know that the maximum acceleration is given by::

$|\frac{d^2X}{dt^2}|=A\omega^2$

So then we have that:

$F=ma=mA\omega^2$

And since we have everything we can find the force

$F=ma=0.01Kg(0.002m)(2000\frac{rad}{s})^2 =80N$

6) Part e

When the mass it's at the half of it's maximum displacement the term $cos(\omega t +\phi)=1/2$ and on this case the acceleration would be given by;

$|\frac{d^2X}{dt^2}|=A\omega^2 cos(\omega t +\phi)=A\omega^2 \frac{1}{2}$

And the force would be given by:

$F=ma=\frac{1}{2}mA\omega^2$

And replacing we have:

$F=\frac{1}{2}(0.01Kg)(0.002m)(2000\frac{rad}{s})^2 =40N$

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3 years ago

8. A rectangle is measured to be 6.4 +0.2 cm by 8.3 $0.2 cm.

mamaluj [8]

Answer:

a) The perimeter of the rectangle is 29.4 centimeters.

b) The uncertainty in its perimeter is 0.8 centimeters.

Explanation:

a) From Geometry we remember that the perimeter of the rectangle ( $p$ ), measured in centimeters, is represented by the following formula:

$p = 2\cdot (w+l)$ (1)

Where:

$w$ - Width, measured in centimeters.

$l$ - Length, measured in centimeters.

If we know that $w = 6.4\,cm$ and $l = 8.3\,cm$ , then the perimeter of the rectangle is:

$p = 2\cdot (6.4\,cm+8.3\,cm)$

$p = 29.4\,cm$

The perimeter of the rectangle is 29.4 centimeters.

b) The uncertainty of the perimeter ( $\Delta p$ ), measured in centimeters, is estimated by differences. That is:

$\Delta p = 2\cdot (\Delta w + \Delta l)$ (2)

Where:

$\Delta w$ - Uncertainty in width, measured in centimeters.

$\Delta l$ - Uncertainty in length, measured in centimeters.

If we know that $\Delta w = 0.2\,cm$ and $\Delta l = 0.2\,cm$ , then the uncertainty in perimeter is:

$\Delta p = 2\cdot (0.2\,cm+0.2\,cm)$

$\Delta p = 0.8\,cm$

The uncertainty in its perimeter is 0.8 centimeters.

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2 years ago