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2 years ago

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

Physics

1 answer:

adelina 88 [10]2 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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Match the following.

Eddi Din [679]

Answer:

I don't know

Explanation:

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3 years ago

For the airfoil and conditions in Problem 2.2, calculate the lift-to-drag ratio. Comment on its magnitude.

raketka [301]

Answer:

L/D= 112

Explanation:

Aerodynamics can be defined as the branch of dynamics which deals with the motion of air, their properties and the interaction between the air and solid bodies.

Aerodynamics law explains how an airplane is able to fly. There are four forces of flight, and they are; lift, weight, thrust and drag. The amount of lift generated by a wing divided by the aerodynamic drag is known as the lift to drag ratio.

Lift increases proportionally to the square of the speed.

The solutions to the question is the file attached to this explanation.

Lift,L= qC(l). S---------------------------(1).

and,

Drag,D = qC(d).S ----------------------(2).

Hence, Lift to drag ratio,L/D= C(l)/C(d).

Therefore, we have to compute various angle of attack.(check attached file)...

Then, (L/D) will then be equal to 112.

8 0

3 years ago

An automobile starter motor has an equivalent resistance of 0.0500Ω and is supplied by a 12.0-V battery with a 0.0100-Ω internal

alexandr1967 [171]

Answer

given,

resistance = 0.05 Ω

internal resistance of battery = 0.01 Ω

electromotive force = 12 V

a) ohm's law

V = IR

and volage

$V = \epsilon - Ir$

now,

$IR = \epsilon - Ir$

$I(R+r) = \epsilon$

$I= \dfrac{\epsilon}{R+r}$

inserting the values

$I= \dfrac{12}{0.05+0.01}$

I = 200 A

b) Voltage

V = I R

V = 200 x 0.05

V = 10 V

c) Power

P = I V

P = 200 x 10 = 2000 W

d) total resistance = 0.05 + 0.09 = 0.14 Ω

$I= \dfrac{\epsilon}{R+r}$

$I= \dfrac{12}{0.14+0.01}$

I = 80 A

V = 80 x 0.05 = 4 V

P = 4 x 80 = 320 W

6 0

3 years ago

Read 2 more answers

In an attempt to reduce the extraordinarily long travel times for voyaging to distant stars, some people have suggested travelin

alexandr402 [8]

Answer:

a) $v=0.999124c$

b) $E=7.566*10^{22}$

c) $E_a=760 times\ larger$

Explanation:

From the question we are told that

Distance to Betelgeuse $d_b=430ly$

Mass of Rocket $M_r=20000$

Total Time in years traveled $T_d=36years$

Total energy used by the United States in the year 2000 $E_{2000}=1.0*10^20$

Generally the equation of speed of rocket v mathematically given by

$v=\frac{2d}{\triangle t}$

$v=860ly/ \triangle t$

where

$\triangle t=\frac{\triangle t'}{(\sqrt{1-860/ \triangle t)^2}}$

$\triangle t=\frac{36}{(\sqrt{1-860/ \triangle t)^2}}$

$\triangle t=\sqrt{(860)^2+(36)^2}$

$\triangle t=860.7532$

Therefore

$v=\frac{860ly}{ 860.7532}$

$v=0.999124c$

Generally the equation of the energy E required to attain prior speed mathematically given by

$E=\frac{1}{\sqrt{1-(v/c)^2} }-1(20000kg)(3*10^8m/s)^2$

$E=7.566*10^{22}$

c)Generally the equation of the energy $E_a$ required to accelerate the rocket mathematically given by

$E_a=\frac{E}{E_{2000}}$

$E_a=\frac{7.566*10^{22}}{1.0*10^{20}}$

$E_a=760 times\ larger$

8 0

3 years ago

Explain the roles of piston valve of a water pump.

Shtirlitz [24]

Answer:

The upstroke of the piston draws water, through a valve, into the lower part of the cylinder. On the downstroke, water passes through valves set in the piston into the upper part of the cylinder. On the next upstroke, water is discharged from the upper part of the cylinder via a spout.

Explanation:

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2 years ago