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larisa86 [58]
3 years ago
9

A jet accelerates at a=3.92 m/s2 from rest until it reaches its takeoff velocity of 360 km/hr. It has to travel for 5 seconds at

its takeoff velocity before taking off. What's the minimum distance that the jet will travel on the runway?
Physics
1 answer:
Art [367]3 years ago
7 0

Answer:

\Delta s = 1775.510\,m

Explanation:

The minimum distance for takeoff is:

\Delta s = \Delta s_{1} + \Delta s_{2}

\Delta s = \frac{v_{f}^{2}-v_{o}^{2}}{2\cdot a} + v_{f}\cdot \Delta t

\Delta s = \frac{(100\,\frac{m}{s} )^{2}-(0\,\frac{m}{s} )^{2}}{2\cdot (3.92\,\frac{m}{s^{2}})}+(100\,\frac{m}{s} )\cdot (5\,s)

\Delta s = 1775.510\,m

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A maser is a laser-type device that produces electromagnetic waves with frequencies in the microwave and radio-wave bands of the
rusak2 [61]

Answers:

a) T=7.04(10)^{-10} s

b) 5.11(10)^{12} cycles

c) 2.06(10)^{26} cycles

d) 46000 s

Explanation:

<h2>a) Time for one cycle of the radio wave</h2>

We know the maser radiowave has a frequency f of 1,420,405,751.786 cycles/s

In addition we know there is an inverse relation between frequency and time T:

f=\frac{1}{T} (1)

Isolating  T: T=\frac{1}{f} (2)

T=\frac{1}{1,420,405,751.786 cycles/s} (3)

T=7.04(10)^{-10} s (4) This is the time for 1 cycle

<h2>b) Cycles that occur in 1 h</h2>

If 1h=3600s and we already know the amount of cycles per second 1,420,405,751.786 cycles/s, then:

1,420,405,751.786 \frac{cycles}{s}(3600s)=5.11(10)^{12} cycles This is the number of cycles in an hour

<h2>c) How many cycles would have occurred during the age of the earth, which is estimated to be 4.6(10)^{9} years?</h2>

Firstly, we have to convert this from years to seconds:

4.6(10)^{9} years \frac{365 days}{1 year} \frac{24 h}{1 day} \frac{3600 s}{1 h}=1.45(10)^{17} s

Now we have to multiply this value for the frequency of the maser radiowave:

1,420,405,751.786 cycles/s (1.45(10)^{17} s)=2.06(10)^{26} cycles This is the number of cycles in the age of the Earth

<h2>d) By how many seconds would a hydrogen maser clock be off after a time interval equal to the age of the earth?</h2>

If we have 1 second out for every 100,000 years, then:

4.6(10)^{9} years \frac{1 s}{100,000 years}=46000 s

This means the maser would be 46000 s off after a time interval equal to the age of the earth

7 0
3 years ago
True or false a moving wave source does not affect the frequancy of the wave encountered by the observer
Andru [333]
The answer is false your welcome
7 0
3 years ago
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A vacuum tube can be used to__. A. change alternating current into direct current B. increase the strength of a signal.. C. turn
vesna_86 [32]

The correct answer of this question is :  A) Change alternating current into direct current.

EXPLANATION :

As per the question, we are given vacuum tube. Vacuum tube can be of various types. Normally it contains two electrodes called cathode and anode which are enclosed in an evacuated glass chamber . There are also other types of vacuum tubes which contain extra electrodes like control grid .

The vacuum tube can be used as a rectifier. It means that it can be used as an electronic device which will convert alternating current into direct current. It may be a half wave rectifier or a full wave rectifier. Actually the direct current obtained during the rectification of alternating current is pulsating in nature.

Hence, the correct answer is that a vacuum tube can be used to change alternating current into direct current.


4 0
3 years ago
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A ball is thrown vertically downwards with a speed 7.3 m/s from the top of a 51 m tall building. With what speed will it hit the
ddd [48]

Answer:

32.46m/s

Explanation:

Hello,

To solve this exercise we must be clear that the ball moves with constant acceleration with the value of gravity = 9.81m / S ^ 2

A body that moves with constant acceleration means that it moves in "a uniformly accelerated motion", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.

When performing a mathematical demonstration, it is found that the equations that define this movement are the follow

\frac {Vf^{2}-Vo^2}{2.a} =X

Where

Vf = final speed

Vo = Initial speed =7.3m/S

A = g=acceleration =9.81m/s^2

X = displacement =51m}

solving for Vf

Vf=\sqrt{Vo^2+2gX}\\Vf=\sqrt{7.3^2+2(9.81)(51)}=32.46m/s

the speed with the ball hits the ground is 32.46m/s

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3 years ago
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"Influence" between bodies is reduced as the bodies are
azamat
Both particles and big objects will be influenced by each other less if they are moved further apart. 
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