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3 years ago

8

A plane moves at a certain velocity and then it accelerates at the rate of 52m/s for a distance of 2300m to reach a velocity of

620m/s .how fast the plane going before it began to accelerate

1 answer:

Illusion [34]3 years ago

6 0

52m/s is the answer because before it was gonna accelerate it was 52 m/s

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A 40-cm-long tube has a 40-cm-long insert that can be pulled in and out. A vibrating tuning fork is held next to the tube. As th

Umnica [9.8K]

Answer:

The frequency of the tuning is 1.065 kHz

Explanation:

Given that,

Length of tube = 40 cm

We need to calculate the difference between each of the lengths

Using formula for length

$\Delta L=L_{2}-L_{1}$

$\Delta L=74.7-58.6$

$\Delta L=16.1\ m$

For an open-open tube,

We need to calculate the fundamental wavelength

Using formula of wavelength

$\lambda=2\Delta L$

Put the value into the formula

$\lambda=2\times16.1$

$\lambda=32.2\ cm$

We need to calculate the frequency of the tuning

Using formula of frequency

$f=\dfrac{v}{\lambda}$

Put the value into the formula

$f=\dfrac{343}{32.2\times10^{-2}}$

$f=1065.2\ Hz$

$f=1.065\ kHz$

Hence, The frequency of the tuning is 1.065 kHz

3 0

3 years ago

A pendulum is formed by taking a 2.0 kg mass and hanging it from the ceiling using a steel wire with a diameter of 1.1 mm. it is

Lera25 [3.4K]

Answer: 1.39 s

Explanation:

We can solve this problem with the following equations:

$\frac{\Delta l}{l_{o}}=\frac{F}{AY}$ (1)

$T=2 \pi \sqrt{\frac{l_{o}}{g}}$ (2)

Where:

$\Delta l=0.05 mm=5(10)^{-5} m$ is the length the steel wire streches (taking into account 1mm=0.001 m)

$l_{o}$ is the length of the steel wire before being streched

$F=mg=(2 kg)(9.8 m/s^{2})=19.6 N$ is the force due gravity (the weight) acting on the pendulum with mass $m=2 kg$

$A$ is the transversal area of the wire

$Y=2(10)^{11} Pa$ is the Young modulus for steel

$T$ is the period of the pendulum

$g=9.8 m/s^{2}$ is the acceleration due gravity

Knowing this, let's begin by finding $A$ :

$A=\pi r^{2}=\pi (\frac{d}{2})^{2}=\pi \frac{d^{2}}{4}$ (3)

Where $d=1.1 mm=0.0011 m$ is the diameter of the wire

$A=\pi \frac{(0.0011 m)^{2}}{4}$ (4)

$A=9.5(10)^{-7}m^{2}$ (5)

Knowing this area we can isolate $l_{o}$ from (1):

$l_{o}=\frac{\Delta l AY}{F}$ (6)

And substitute $l_{o}$ in (2):

$T=2 \pi \sqrt{\frac{\frac{\Delta l AY}{F}}{g}}$ (7)

$T=2 \pi \sqrt{\frac{\frac{(5(10)^{-5} m)(9.5(10)^{-7}m^{2})(2(10)^{11} Pa)}{2(10)^{11} Pa}}{9.8 m/s^{2}}}$ (8)

Finally:

$T=1.39 s$

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Answer:

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It is called Arrhenius collision theory.

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