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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?

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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While flying over the grand canyon the pilot

Veronika [31]

We cannot answer the question if we do not have the answers - sorry!!

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3 years ago

In designing circular rides for amusement parks, mechanical engineers must consider how small variations in certain parameters c

Bess [88]

Answer:

Part a)

$dF = -\frac{mv^2}{r^2} dr$

Part b)

$dF = \frac{2mvdv}{r}$

Part c)

$dT = - \frac{2\pi r}{v^2} dv$

Explanation:

Part a)

As we know that force on the passenger while moving in circle is given as

$F = \frac{mv^2}{r}$

now variation in force is given as

$dF = -\frac{mv^2}{r^2} dr$

here speed is constant

Part b)

Now if the variation in force is required such that r is constant then we will have

$F = \frac{mv^2}{r}$

so we have

$dF = \frac{2mvdv}{r}$

Part c)

As we know that time period of the circular motion is given as

$T = \frac{2\pi r}{v}$

so here if radius is constant then variation in time period is given as

$dT = - \frac{2\pi r}{v^2} dv$

8 0

2 years ago

If a stone with an original velocity of 0 is falling from a ledge and takes 8 seconds to hit the ground, what is the final veloc

sergey [27]

I believe the answer is C - 78.4 m/s.

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2 years ago

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There are free body diagrams in the picture, number (#). Which of the diagram(s) shows an object with a net force right?

ryzh [129]

4 only diagram net Force

6 0

3 years ago

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The masses are m1 = m, with initial velocity 2v0, and m2 = 7.4m, with initial velocity v0. Due to the collision, they stick toge

lesya [120]

Answer:

Loss, $\Delta E=-10.63\ J$

Explanation:

Given that,

Mass of particle 1, $m_1=m =0.66\ kg$

Mass of particle 2, $m_2=7.4m =4.884\ kg$

Speed of particle 1, $v_1=2v_o=2\times 6=12\ m/s$

Speed of particle 2, $v_2=v_o=6\ m/s$

To find,

The magnitude of the loss in kinetic energy after the collision.

Solve,

Two particles stick together in case of inelastic collision. Due to this, some of the kinetic energy gets lost.

Applying the conservation of momentum to find the speed of two particles after the collision.

$m_1v_1+m_2v_2=(m_1+m_2)V$

$V=\dfrac{m_1v_1+m_2v_2}{(m_1+m_2)}$

$V=\dfrac{0.66\times 12+4.884\times 6}{(0.66+4.884)}$

V = 6.71 m/s

Initial kinetic energy before the collision,

$K_i=\dfrac{1}{2}(m_1v_1^2+m_2v_2^2)$

$K_i=\dfrac{1}{2}(0.66\times 12^2+4.884\times 6^2)$

$K_i=135.43\ J$

Final kinetic energy after the collision,

$K_f=\dfrac{1}{2}(m_1+m_2)V^2$

$K_f=\dfrac{1}{2}(0.66+4.884)\times 6.71^2$

$K_f=124.80\ J$

Lost in kinetic energy,

$\Delta K=K_f-K_i$

$\Delta K=124.80-135.43$

$\Delta E=-10.63\ J$

Therefore, the magnitude of the loss in kinetic energy after the collision is 10.63 Joules.

7 0

3 years ago