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PolarNik [594]
3 years ago
14

you’ve just been involved in a traffic accident that leaves you stranded on the side of the road. Which part of the EM spectrum

would be of the most use to you and why?
Physics
1 answer:
Akimi4 [234]3 years ago
4 0

You’ve just been involved in a traffic accident that leaves you stranded on the side of the road. <u>Microwaves from the EM spectrum would be of the most use.</u>

<u></u>

Explanation:

  • Microwaves are a form of electromagnetic radiation with wavelengths ranging from about one meter to one millimeter, with frequencies between 300 MHz and 300 GHz.
  • Microwaves have a range of applications, including communications, radar and, perhaps best known by most people, cooking
  • Microwave is a line-of-sight wireless communication technology that uses high frequency beams of radio waves to provide high speed wireless connections that can send and receive voice, video, and data information.
  • Microwaves and radio waves are used to communicate with satellites.
  • Microwaves pass straight through the atmosphere and are suitable for communicating with distant geostationary satellites, while radio waves are suitable for communicating with satellites in low orbit.
  • Microwaves are widely used in modern technology, <u>for example</u> : in point-to-point communication links, wireless networks, microwave radio relay networks, radar, satellite and spacecraft communication, medical diathermy and cancer treatment, etc
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A ball of mass m is thrown straight upward from ground level at speed v0. At the same instant, at a distance D above the ground,
n200080 [17]

Answer:

a. t = \frac{v_{0}  +/- \sqrt{v_{0} ^{2} - gD} }{g}  b. D = v₀²/2g

Explanation:

Here is the complete question

A ball is thrown straight up from the ground with speed v₀ . At the same instant, a second ball is dropped from rest from a height D , directly above the point where the first ball was thrown upward. There is no air resistance

Find the time at which the two balls collide.

Express your answer in terms of the variables D ,v₀ , and appropriate constants..

t = ?!

Part B

Find the value of D in terms of v₀ and g so that at the instant when the balls collide, the first ball is at the highest point of its motion.

Express your answer in terms of the variables v₀ and g .

D =?!

Solution

The distance moved by the ball dropped from distance,D with velocity v₀, H₁ = D - (v₀t - gt²/2) = D + v₀t + gt²/2.

The distance moved by the ball thrown straight upward with velocity v₀ is H₂ = v₀t - gt²/2.

The two balls collide when their vertical distances are equal. That is H₁ = H₂

So, D - v₀t + gt²/2 = v₀t - gt²/2

Collecting like terms

D + gt²/2 + gt²/2 = v₀t + v₀t

D +gt² = 2v₀t

gt² - 2v₀t + D = 0.

Using the quadratic formula,

t = \frac{-(-2v_{0} ) +/- \sqrt{(-2v_{0} )^{2} - 4 X g XD} }{2g} = \frac{2v_{0}  +/- \sqrt{4v_{0} ^{2} - 4gD} }{2g} = \frac{v_{0}  +/- \sqrt{v_{0} ^{2} - gD} }{g}

B. At its highest point, the velocity of the first ball, v = 0. Using v² = u² - 2gs where s = highest point of first ball when they collide and u = v₀.

0 = v₀² - 2gs

s = v₀²/2g.

Also, the time it takes the first ball to reach its highest point is gotten from v = u - gt. At highest point, v = 0 and u = v₀. So,

 0 = v₀ - gt₀

t₀ = v₀/g

Also H = s₁ + s where s₁  = distance moved by second ball in time t₀ for collision = v₀t₀ - gt₀²/2.

So, H = v₀t₀ - gt₀²/2 + v₀²/2g = v₀(v₀/g) - g(v₀/g)²/2 + v₀²/2g = v₀²/2g - v₀²/2g + v₀²/2g = v₀²/2g

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A stone is thrown horizontally with an initial speed of 10 m/s from the edge of a cliff. A stopwatch measures the stone's trajec
olga2289 [7]

Answer:

The height of the cliff is 90.60 meters.

Explanation:

It is given that,

Initial horizontal speed of the stone, u = 10 m/s

Initial vertical speed of the stone, u' = 0 (as there is no motion in vertical direction)

The time taken by the stone from the top of the cliff to the bottom to be 4.3 s, t = 4.3 s

Let h is the height of the cliff. Using the second equation of motion in vertical direction to find it. It is given by :

h=u't+\dfrac{1}{2}gt^2

h=\dfrac{1}{2}gt^2

h=\dfrac{1}{2}\times 9.8\times (4.3)^2

h = 90.60 meters

So, the height of the cliff is 90.60 meters. Hence, this is the required solution.

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