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ad-work [718]
2 years ago
8

The table above lists the wavelength range, frequency range, possible uses, and hazards of the seven types of electromagnetic ra

diation. Which type of radiation helps doctors image the interior of the human body but also causes cell damage?
A.
microwave
B.
infrared
C.
radio wave
D.
X-ray
Physics
1 answer:
m_a_m_a [10]2 years ago
6 0
I think it's D caz X-ray scans can diagnose possibly life-threatening conditions such as blocked blood vessels, bone cancer, and infections. However, x-rays produce ionizing radiation—a form of radiation that has the potential to harm living tissue. However, the risk of developing cancer from radiation exposure is generally small.
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A person drives a car around a circular cloverleaf with a radius of 77 m at a uniform speed of 10 m/s. What is the acceleration
umka2103 [35]

Answer:

770m/s

Explanation:

caculation using one of the newton law of motion

6 0
3 years ago
Energy is conventionally measured in Calories as well as in joules. One Calorie in nutrition is one kilocalorie, defined as 1 kc
Sergeeva-Olga [200]

Answer:

a) The student must run flight of stairs to lose 1.00 kg of fat 709.5 times.

b) Average power

P(w)= 1062.07 [w]

P(hp)=1.42 [hp]

c) This activity is highly unpractical, because the high amount of repetitions he has to due in order to lose, just 1 Kg of fat.

Explanation:

First, lets consider the required amount of work to move the mass of the student. (considering running stairs just as a vertical movement)

Work:

W= F*d= m*g*d

Where m is the mass of the student, g is gravity (9.8 m/s) and d is the total distance going up the stairs (0.15m *85steps= 12.75m )

W= F*d= m*g*d=85* 9.8*12.75=10620.75 [J]

Converting from Joules to Kcals:

\frac{10620.75}{4186} =2.537 Kcal

Now lets take into account the efficiency of the human body (20%)

2.537 ---> 20%

 x       ---> 100%

x=\frac{2.537*100}{20} =12.685

So the student is consuming 12.685 KCals each time he runs up the stairs.

Now,

1 g --> 9 Kcals

1000 g --> 9000KCals

Burning 1 g of fat, requieres 9 KCals, 1000g burns 9000KCals. So in order to burn a 1Kg of fat:

\frac{9000Kcals}{12.685Kcals} =709.5 times

He must run up the stairs 709.5 times, to burn 1 Kg of fat.

********************

For b) just converting units, taking into account the time lapse. (53103.75 is the 100% of the energy in joules, from converting 12.685Kcals to joules)

Power=\frac{Joules}{Seconds} =\frac{53103.75}{50} =1062.075 [W]\\

P(hp)=\frac{P(w)}{745.7} =\frac{1062.075}{745.7} =1.42[hp]

*****

4 0
3 years ago
Definition of artficial and natural radioactivity​
Genrish500 [490]
Search Results
Featured snippet from the web
The spontaneous emission of radiations from an unstable nuclei is known as natural radioactivity. on the other hand, The process of emission of radiations from naturally occurring isotopes when they are bombarded with sub-atomic particles or high levels of X-rays or gamma rays called artificial radioactivity.
8 0
2 years ago
Read 2 more answers
HELP PLEASE
4vir4ik [10]

Recall this kinematic equation:

a = \frac{Vi+Vf}{Δt}

This equation gives the acceleration of the object assuming it IS constant (the velocity changes at a uniform rate).

a is the acceleration.

Vi is the initial velocity.

Vf is the final velocity.

Δt is the amount of elapsed time.


Given values:

Vi = 0 m/s (the car starts at rest).

Vf = 25 m/s.

Δt = 10 s


Substitute the terms in the equation with the given values and solve for a:

a = \frac{0+25}{10}

<h3>a = 2.5 m/s²</h3>
7 0
3 years ago
Read 2 more answers
You are a member of an alpine rescue team and must get a box of supplies, with mass 2.20 kg , up an incline of constant slope an
a_sh-v [17]

Answer:

The minimum speed of the box bottom of the incline so that it will reach the skier is 8.19 m/s.

Explanation:

It is given that,

Mass of the box, m = 2.2 kg

The box is inclined at an angle of 30 degrees

Vertical distance, d = 3.1 m

The coefficient of friction, \mu=6\times 10^{-2}

Using the work energy theorem, the loss of kinetic energy is equal to the sum of gain in potential energy and the work done against friction.

KE=PE+W

\dfrac{1}{2}mv^2=mgh+W

W is the work done by the friction.

W=f\times d

f=\mu mg\ cos\theta

W=\mu mg\ cos\theta\times \dfrac{d}{sin\theta}

W=\dfrac{\mu mgh}{tan\theta}

\dfrac{1}{2}mv^2=mgh+\dfrac{\mu mgh}{tan\theta}

\dfrac{1}{2}v^2=gh+\dfrac{\mu gh}{tan\theta}

\dfrac{1}{2}v^2=9.8\times 3.1+\dfrac{6\times 10^{-2}\times 9.8\times 3.1}{tan(30)}

v = 8.19 m/s

So, the speed of the box is 8.19 m/s. Hence, this is the required solution.

8 0
3 years ago
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