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3 years ago

32. Which type of electromagnetic wave is used for nuclear power and medical treatment?

Physics

1 answer:

adelina 88 [10]3 years ago

3 0

Answer:

Answers at the bottom!!

Explanation:

For the first answer: Gamma ray

Gamma rays and x-rays consist of high-energy waves that can travel great distances at the speed of light and generally have a great ability to penetrate other materials. For that reason, gamma rays (such as from cobalt-60) are often used in medical applications to treat cancer and sterilize medical instruments.

For the second answer: They both have technological uses.

In my opinion I'm pretty sure it's A because we do use microwaves and x-rays as technological uses.

Hope this helps!!

Happy Holidays and Season Greeting!!

ii Feliz Navidad a todos !!

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How do you disseminate a car if you rub it with rubber

natta225 [31]

I have honestly never read anything about a car being disseminated,
or any instructions on how to do it, or any of the theory behind it.

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3 years ago

If a migrating loon flies at an average speed of 19 m/s, what is its average speed in mi/hr?

zlopas [31]

Unit conversion is a way of converting some common units into another without changing their real value. The average speed of the migrating loon flies is 45.5019 miles/hr.

<h3>What is Unit conversion?</h3>

Unit conversion is a way of converting some common units into another without changing their real value. for, example, 1 centimeter is equal to 10 mm, though the real measurement is still the same the units and numerical values have been changed.

Given that migrating loon flies at an average speed of 19 m/s. Now, since 1 meter is equal to (1/1609.34) miles and 1 second is equal to (1/3600) seconds. Therefore, we can write the speed as,

$\rm Speed = 19\ \dfrac{meter}{second}\\$

$=19 \times \dfrac{\frac{1}{1609.34}\rm\ miles}{\frac{1}{3600} \rm\ hours}$

= 19 × (3600/1609.34) miles/hr

= 42.5019 miles/ hr

Hence, the average speed of the migrating loon flies is 45.5019 miles/hr.

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brainly.com/question/4736731

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1 year ago

1. You released a pendulum of mass 1kg from a height of 0.05m

photoshop1234 [79]

a. The speed of the pendulum when it reaches the bottom is 0.9 m/s.

b. The height reached by the pendulum is 0.038 m.

c. When the pendulum no longer swing at all, all the kinetic energy of the pendulum has been used to overcome frictional force.

<h3>Kinetic energy of the pendulum when it reaches bottom</h3>

K.E = 100%P.E - 18%P.E

where;

P.E is potential; energy

K.E(bottom) = 0.82P.E

K.E(bottom) = 0.82(mgh)

K.E(bottom) = 0.82(1 x 9.8 x 0.05) = 0.402 J

<h3>Speed of the pendulum</h3>

K.E = ¹/₂mv²

2K.E = mv²

v² = (2K.E)/m

v² = (2 x 0.402)/1

v² = 0.804

v = √0.804

v = 0.9 m/s

<h3>Final potential energy </h3>

P.E = 100%K.E - 7%K.E

P.E = 93%K.E

P.E = 0.93(0.402 J)

P.E = 0.374 J

<h3>Height reached by the pendulum</h3>

P.E = mgh

h = P.E/mg

h = (0.374)/(1 x 9.8)

h = 0.038 m

<h3>when the pendulum stops</h3>

When the pendulum no longer swing at all, all the kinetic energy of the pendulum has been used to overcome frictional force.

Thus, the speed of the pendulum when it reaches the bottom is 0.9 m/s.

The height reached by the pendulum is 0.038 m.

When the pendulum no longer swing at all, all the kinetic energy of the pendulum has been used to overcome frictional force.

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2 years ago

A series LRC circuit consists of a 12.0-mH inductor, a 15.0-µF capacitor, a resistor, and a 110-V (rms) ac voltage source. If th

finlep [7]

Answer:

61.85 ohm

Explanation:

L = 12 m H = 12 x 10^-3 H, C = 15 x 10^-6 F, Vrms = 110 V, R = 45 ohm

Let ω0 be the resonant frequency.

$\omega _{0}=\frac{1}{\sqrt{LC}}$

$\omega _{0}=\frac{1}{\sqrt{12\times 10^{-3}\times 15\times 10^{-6}}}$

ω0 = 2357 rad/s

ω = 2 x 2357 = 4714 rad/s

XL = ω L = 4714 x 12 x 10^-3 = 56.57 ohm

Xc = 1 / ω C = 1 / (4714 x 15 x 10^-6) = 14.14 ohm

Impedance, Z = $\sqrt{R^{2}+\left ( XL - Xc \right )^{2}}$

Z = \sqrt{45^{2}+\left ( 56.57-14.14 )^{2}} = 61.85 ohm

Thus, the impedance at double the resonant frequency is 61.85 ohm.

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