10 POINTS
Which type of electromagnetic radiation is used in airport security screening?
O A. X-rays
• B. Ultraviolet radiation
• C. Gamma rays
O D. Infrared radiation
SUBMIT

10 POINTSWhich Type Of Electromagnetic Radiation Is Used In Airport Security Screening?O A. X-rays B.

Answers

Answer 1

Answer:

A) X-rays

Explanation:

Airport security screenings use X-rays because they can penetrate through bags to see objects that are located within that are obscured through normal view.

Answer 2

X- rays are used to  test  body in airports. Low energy waves are used to screen baggages and human body. Thus option A is correct.

What is electromagnetic radiation ?

Electromagnetic spectrum is the arrangement of radiations in the order of decreasing wavelength or increasing energy and frequency. The order is radio waves, microwaves, infrared, visible, ultraviolet, x-rays and gamma rays.

x -rays are highly energetic short waves which can be easily penetrate through a body. X-ray is used in different diagnostic techniques. X-rays easily penetrate through  tissues and provide the scanned images of internal parts.

In airport security screening x-rays are used to screen baggages and human bodies. High energetic waves are causing serious health risks. Hence, lower energy x-rays are used in testing.

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Related Questions

Spring
The pulling force acting on an object is 350.0 N. It also is being affected by a 50.0 N
kinetic friction force. As a result, its acceleration is 3.0 m/s?, what is its mass?
A. 67 kg
B. 83 kg
C. 100 kg
D. 600 kg
E. 0.0 kg
If the gravitational force of the Farth nulling you"down" in th

Answers

Answer:

100 kg

Explanation:

Net force = 350 - 50 = 300 N

F= ma

300 N  = m * 3 m/s^2

100 kg = mass

Mrs. dela Vega had 27 students in her class last year and this year she had 30 students in her math class. What is the percent of difference?

Answers

The class grew by (30-27) = 3 students.

3 is (3/27) = 1/9 of its original size.

The class grew by 1/9 or 11.1 % .

a boat travels 500 m down a strait river. if starts from the rest accelerates uniformly to a velocity of 5m/s. how long does this take?
PLSSSS HELPPPPP NOWWW!!!!!

Answers

Answer: 200 seconds

Explanation:
Initial Velocity = 0 m/s
Final Velocity = 5 m/s
Displacement = 500 m

Average Velocity = (Vo + Vf) / 2
(0 + 5) / 2 = 2.5

Time = Displacement / Average Velocity
Time = 500 / 2.5 = 200

A karate expert breaks a stack of bricks with his bare hand. If the force applied is 520 newtons and the impact time is 5.0 × 10 seconds, what is the value of impulse
A.
0.26 newton seconds
B.
0.58 newton seconds
C.
1.82 newton seconds
D.
2.20 newton seconds

Answers

Maybe the question is :

A karate expert breaks a stack of bricks with his bare hand. If the force applied is 520 newtons and the impact time is [tex] \sf{5.0 \times 10^{-4}} [/tex] seconds, what is the value of impulse ?

So, the value of impulse is (A). 0.26 newton seconds.

Introduction

Hi ! Here, I will help you about the impulse. Impulse is described as a quantity which expresses the integral of the amount force respect to time. The amount of impulse will be proportional to the amount of force or time (the greater  value of the force or the value of time, the value of impuls is greater too). The relationship between impulse, force, and time is expressed in this equation :

[tex] \boxed{sf{\bold{I = F \times \Delta t}}} [/tex]

With the following condition:

I = impulse that occurs (N.s)F = force that given (N)[tex] \sf{\Delta t} [/tex] = interval of the time (s)

Problem SolvingWe know that :F = force that given = 520 N[tex] \sf{\Delta t} [/tex] = interval of the time = [tex] \sf{5.0 \times 10^{-4} \: s}[/tex]

What was asked :

I = impulse that occurs = ... N.s

Step by step :

[tex] \sf{I = F \times \Delta t} [/tex]

[tex] \sf{I = 520 \times (5.0 \times 10^{-4}} [/tex]

[tex] \sf{I = 2,600 \times 10^{-4}} [/tex]

[tex] \sf{I = 2.60 \times 10^3 \times 10^{-4}} [/tex]

[tex] \sf{I = 2.60 \times 10^{3 + (-4)}} [/tex]

[tex] \sf{I = 2.60 \times 10^{-1} = \boxed{0.26 \: N.s}} [/tex]

Conclusion :

So, the value of impulse is (A). 0.26 newton seconds

theshold frequency ???​

Answers

Answer:

The threshold frequency is defined as the minimum frequency of the incident radiation below which photoelectric emission or emission of electrons is not possible. The threshold frequency refers to the frequency of light that will cause an electron to dislodge emit from the surface of the metal.

The surface of a lake has an area of 15.5 km2. What is the area of the lake in m2?

Answers

Answer:

1.55 x 10⁷ m²

Explanation:

Unit conversion

km² → m² 1 km² = 10⁶ m²

Solving

15.5 km² = 10⁶ x 15.5 m²155 x 10⁵ m²1.55 x 10⁷ m²

15C of charge flow through the filament of a light bulb in 22 seconds. What is the strength of the current in the filament?

Answers

Answer:

Current(I)= charge(q)/time

I=15/22

I=0.680.68A

Explanation:

Current is flow of charge per unit time. Always make sure the time is in seconds. istthe

Why is there two π in the formula of a mathematical pendulum (T=2π √l/g)?

Answers

Answer:

The π is from the initial formula of period T=2π/omega

-What is the potential energy at point A?

-What is the kinetic energy at point A?

-What is the kinetic energy at point B?

-What is the potential energy at point D?

-What is the kinetic energy at point D?

Answers

Answer:

1560520156

Explanation:

Assuming there is no friction or other force involved, recall that energy is conserved in a system as long as no external force acts on the system.

Using the data from point C, we can find out that the total energy of the system is 156 because  [tex]E = K+Pe[/tex].

Since at point A the object doesn't move, it has Kinetic energy of 0, because [tex]K=\frac{1}{2} mv^2[/tex], therefore [tex]0=\frac{1}{2} m0^2[/tex]. However at point A it has maximum Potential energy, because [tex]Pe=mgh[/tex].

At point B, we can find the Kinetic energy by using [tex]E = K+Pe[/tex]. Substitute values:

[tex]156=104+K\\52=K[/tex]

At point D, the object has maximum kinetic energy and no potential energy, therefore it's the opposite of point A.

What is refractive index?​

Answers

[tex] \huge \mathbb \red{HEY \: THERE ♡}[/tex]

Refractive Index:

The ratio of the sine of angle of incidence to the sine of angle of refraction in case of lens for light of a given colour and given pair of media is constant.

Note** It is also called Snell's Law of Refraction. [tex] \mathsf \orange{\frac{sine \: i}{sine \: r} = constant}[/tex]

[tex] \huge \mathbb\pink{HOPE \: IT \: HELPS}[/tex]

A 75-W light bulb is turned on. It has an operating voltage of 120 V. (A)How much current flows through the bulb? (B)What is the resistance of the bulb? (C)How much energy is used each second?

Answers

Required Answer:

Given:

Power (P) = 75 W Voltage (V) = 120 V

(A)

As we know that,

I = P/V

↠ Current = 75/120

↠ Current = 0.625 A

(B)

R = V/I

Here,

R is resistanceV is Voltage, and I is current

↠ Resistance = 120/0.625

↠ Resistance = 192 Ω

(C)

Energy = P × t

Here,

P is power T is time

↠ Energy = 75 × 1

↠ Energy = 75 J

As water is cooled, its density increases until it reaches about
O A. 25° C
B. -2° C
O c. 4°C
O D.O°C

Answers

Answer:

4 degrees C

Explanation:

this is just a 'known'

a true statement of
kinetic theory

Answers

Answer:

real kinetic theory means that kinetic energy

A scientist decreases the wavelength of the light used in a double-slit experiment and keeps every other aspect the same. What will be true
about the new interference pattern seen on the screen compared to the original interference pattern? (point)

O The spacing between the bright fringes will increase.

O The spacing between the dark fringes will remain the sam
same.

O The spacing between the bright fringes will decrease.

O The spacing between the dark fringes will increase.

Answers

The answer to your question is “The spacing between the bright fringes will increase”. Have a great day

A scientist decreases the wavelength of the light used in a double-slit experiment and keeps every other aspect the same. The new interference pattern seen on the screen compared to the original interference pattern is the spacing between the bright fringes will increase. Thus, the correct option is A.

What is Interference pattern?

Interference is the net effect of the combination of two or more wave forms or trains which are moving on intersecting or coincident paths in a medium. The effect of this pattern is that of the addition of the amplitudes of the individual waves at each point which are affected by more than one wave in the medium.

A scientist when decreases the wavelength of the light used in a double-slit experiment and it keeps every other aspect of the wave the same. The new interference pattern seen on the screen as compared to the original interference pattern is that the spacing between the bright fringes will increase.

Therefore, the correct option is A.

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A uniformly charged insulating rod is bent into the shape of a semicircle of radius R = 5 cm. If the rod has a total charge of Q = 3.10-9C, find the magnitude and direction of the electric field at O, the center of the circle.

Answers

Hi there!

We can begin by using Coulomb's Law:

[tex]E = \frac{kq}{r^2}[/tex]

k = Coulomb's Constant (8.99 × 10⁹ Nm²/C²)

E = Electric field strength (N/C)
r = distance from point (m)

q = charge (C)

Since this is a continuous charge, we must use calculus.

We can express this as the following:
[tex]q = \lambda L[/tex]

λ = Linear charge density (C/m)

L = Length of rod (m)

Now, since this is an arc, L = s (arc length). Additionally, we must find the differential elements of each:
[tex]dq = \lambda ds\\\\dq = \lambda rd\theta[/tex]

Our new equation is:
[tex]dE = \frac{kdq}{r^2}\\\\dE = \frac{k\lambda rd\theta}{r^2}[/tex]

However, we will only take the cosine component of the electric field since the vertical components will cancel out. (Electric fields are a vector). Therefore:
[tex]dE = \frac{k\lambda rd\theta}{r^2}cos\theta\\\\dE = \frac{k\lambda}{r}cos\theta d\theta[/tex]

Integrate. For a semicircle, the bounds will be from -π/2 to π/2.

[tex]E = \frac{k\lambda}{r}\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}} {cos\theta} \, d\theta\\\\E = \frac{k\lambda}{r}sin\theta\left \|{\frac{\pi}{2}} \atop {-\frac{\pi}{2}}} \right. \\\\E = \frac{k\lambda}{r}(1 - (-1)) = \frac{2k\lambda}{r}[/tex]

We need to solve for λ, which is Q/ L:
[tex]\lambda = \frac{3.10 \times 10^{-9} C}{\pi (0.05)} = 1.9735 \times 10^{-8} \frac{C}{m}[/tex]

Now, plug and solve for the electric field strength:
[tex]E = \frac{2(8.99\times 10^9)(1.9735\times 10^{-8})}{0.05} = \boxed{7096.783 \frac{N}{C}}[/tex]

**A diagram was not provided, but if the hemisphere's focus was to the right, the electric field would be to the right, and etcetera.

List the chemicals and apparatus

Answers

Answer:

Beaker

Graduated Cylinders

Volumetric flask

Test tube

Funnel


Burns are a particular safety hazard when handling

Answers

Answer:

flammable  liquids

Explanation:when it catches on fire your going to get burned.hope this helps.

A wave is sent back and forth along a rope 4 m long with a mass of 0.6 kg by exerting a force a force of 30 N. Calculate the linear mass density of the rope (in kg/m).

Answers

The linear mass density of the rope in the given motion of the wave is 0.15 kg/m.

Linear mass density of the rope

The linear mass density of the rope in the given motion of the wave is determined by dividing the mass of the rope with the length of the entire rope.

The linear mass density of the rope is calculated as follows;

μ = m/L

μ = 0.6 kg / 4 m

μ = 0.15 kg/m

Thus, the linear mass density of the rope in the given motion of the wave is 0.15 kg/m.

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Difine Ripple Factor
-,- ty!!​

Answers

Answer:

The answer is in the picture above.

A rocket designed to place small payloads into orbit is carried to an altitude of 12.0 km above sea level by a converted airliner. When the airliner is flying in a straight line at a constant speed of 850 km/h, the rocket is dropped. After the drop, the airliner maintains the same altitude and speed and continues to fly in a straight line. The rocket falls for a brief time, after which its rocket motor turns on. Once its rocket motor is on, the combined effects of thrust and gravity give the rocket a constant acceleration of magnitude 3.00g directed at an angle of 30.0° above the horizontal. For reasons of safety, the rocket should be at least 1.00 km in front of the airliner when it climbs through the airliner’s altitude. Your job is to determine the minimum time that the rocket must fall before its engine starts. You can ignore air resistance. Your answer should include (i) a diagram showing the flight paths of both the rocket and the airliner, labeled at several points with vectors for their velocities and accelerations; (ii) an x-t graph showing the motions of both the rocket and the airliner; and (iii) a y-t graph showing the motions of both the rocket and the airliner. In the diagram and the graphs, indicate when the rocket is dropped, when the rocket motor turns on, and when the rocket climbs through the altitude of the airliner.

Answers

Answer:

Explanation:

A rocket designed to place small payloads into orbit is carried to an altitude of 12.0 km above sea level by a converted airliner. When the airliner is flying in a straight line at a constant speed of 850 km/h, the rocket is dropped. After the drop, the airliner maintains the same altitude and speed and continues to fly in a straight line. The rocket falls for a brief time, after which its rocket motor turns on. Once its rocket motor is on, the combined effects of thrust and gravity give the rocket a constant acceleration of magnitude 3.00g directed at an angle of 30.0° above the horizontal. For reasons of safety, the rocket should be at least 1.00 km in front of the airliner when it climbs through the airliner’s altitude. Your job is to determine the minimum time that the rocket must fall before its engine starts. You can ignore air resistance. Your answer should include (i) a diagram showing the flight paths of both the rocket and the airliner, labeled at several points with vectors for their velocities and accelerations; (ii) an x-t graph showing the motions of both the rocket and the airliner; and (iii) a y-t graph showing the motions of both the rocket and the airliner. In the diagram and the graphs, indicate when the rocket is dropped, when the rocket motor turns on, and when the rocket climbs through the altitude of the airliner.

1. A drag racer accelerates from rest at 18ft/sec^2. How long does it take to acquire a speed of 60mph? What is required?
2. A contestant ran a 100-m dash in 10.6sec. What was his speed a) In feet per second and b) in miles per hr?

Answers

(1) The time taken for the drag racer to accelerate is 4.89 s

(2) The speed of the contestant (a) in feet per second is 30.89 ft/s (b) in miles per hr is 21.12 miles per hr.

(1) To calculate the time required to accelerate 18 ft/sec² from rest to a velocity of 60 mph, we use the formula below.

Formula:

t = (v-u)/a........... Equation 1

Where:

t = timev = Final velocityu = initial velocitya = acceleration.

From the question,

Given:

a = 18 ft/sec² = (18×0.3048) = 5.4864 m/s²v = 60 mph = (60×0.44704) = 26.82 m/su = 0 m/s ( from rest)

Substitute these values into equation 1

t = (26.82-0)/5.4864t = 4.89 seconds

(2) To calculate the speed of the contestant, we use the formula below

Formula:

s = d/t............ Equation 1

Where:

s = speed of the contestantd = distancet = time.

From the question,

Given:

d = 100 mt = 10.6 s

Substitute these values into equation 1

s = 100/10.6s = 9.43 m/s

(a) In feet = (9.43/0.3048) = 30.94 ft/s

(b)  in miles per hr = (9.43×2.24) = 21.12 miles per hr

Hence, (1) The time taken for the drag racer to accelerate is 4.89 s (2) The speed of the contestant (a) in feet per second is 30.89 ft/s (b) in miles per hr is 21.12 miles per hr.

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Two horses pull horizontally on ropes attached to a stump. The two forces Fu and F2 that they apply to the stump are such that the net (resultant) force R has a magnitude equal to that of Fu and makes an angle of 90° with Fu. Let F1 = 1300 N and R = 1300 N also. Find the magnitude of F and its direction (relative to Fi)

Answers

For two horses pull horizontally on ropes attached to a stump, the magnitude of F and its direction is mathematically given as

F=1711 N

dF= 135 degrees

What is the magnitude of F and its direction?

Generally, the equation for the components of F1   is mathematically given as

For The x axis

1210 - F2x = 0

F2x = 1210

For the y axis

F2y = R

F2y = 1210

In conclusion

F = sqrt(2)*1210

F=1711 N

Fopt the direction of F2

dF= 90 + 45

dF= 135 degrees

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A positive charge +Q is moving to the right and experiences a vertical (upward) magnetic force. In which direction is the magnetic field?
O into the screen
O upward
O out of the screen
O to the left
O to the right

Answers

Answer:

out of the screen

Explanation:

because charge +Q is already moving upward

A 2 kg solid disk with a radius of 0.22 m has a tangential force of 300N applied to it.
a) What is the torque acting on the disk?
b) What is the moment of inertia of the disk?
c) What angular acceleration is produced by the torque?
d) If the disk starts from rest and the acceleration is constant for 3.0s, what is the angular velocity of the disk at the end of 3.0s?
e) Through what angle in radians has the disk rotated during this time?

Answers

(a) The torque acting on the disk is 66 Nm.

(b) The moment of inertia of the disk is 0.05 kgm².

(c) The angular acceleration is produced by the torque is 1,320 rad/s².

(d) The final angular velocity of the disk is  3,960 rad/s.

(e) The angle of rotation of the disk is 5,940 rad.

Torque acting on the disk

The torque acting on the disk is calculated as follows;

τ = Fr

τ = 300 x 0.22

τ = 66 Nm

Moment of inertia

The moment of inertia of a solid disk is calculated as follows;

I = ¹/₂MR²

I = ¹/₂ x 2 x (0.22)²

I = 0.05 kgm²

Angular acceleration of the disk

The angular acceleration of the disk is calculated as follows;

τ = Iα

[tex]\alpha = \frac{\tau }{I} \\\\\alpha = \frac{66}{0.05} \\\\\alpha = 1,320 \ rad/s^2[/tex]

Angular velocity of the disk after 3 s

ωf = ωi + αt

ωf = 0 + (1320 x 3)

ωf = 3,960 rad/s

Angle of rotation of the disk

ωf² = ωi²+ 2αθ

(3,960)² = 0 + 2(1320)θ

θ = (3,960²) / (2 x 1320)

θ = 5,940 rad

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Calculate the frequency of a sound wave produced when a tennis racquet string is plucked. The tension of the string is 274 N, the mass of the string is 28 kg and the length of the string is 0.74 m.

Answers

Answer:

Explanation:

The velocity of sound from the plucked string in the tennis racquet is:

[tex]v=\sqrt{\frac{FL}{m}}=\sqrt{\frac{274(0.74)}{28}}\approx 2.69 ms^{-1}[/tex]

Then the frequency will be:

[tex]f=\frac{v}{2L}=\frac{2.69}{2(0.74)}\approx 1.81 Hz[/tex]

Two children are playing tetherball, in which a ball at the end of a cord spins around a pole. After a really good hit, the ball makes three complete revolutions in 2.0 s. What is the angular speed of the ball?

Answers

Answer:

3 pi R / s     or   9.42 R/s    or  540 degrees/sec

Explanation:

Ball covers 2 pi radians each rotation

2 pi * 3 R / 2  s =   3 pi R / s

the angular speed of the ball is 3 pi R / s     or   9.42 R/s    or  540 degrees/sec

What is angular speed ?

Angular speed can be defined as the measures of speed of how fast the central angle of a rotating body changes with respect to time

Angular speed mainly implies how quickly the rotation of an object occur means it is described as the change in the angle of the object per unit of time.

To calculate the speed of a rotational motion the angular speed value need to be known, hence  The angular speeds formula is used to calculate the distance traveled by an object in terms of rotation and revolutions per unit of time.

The unit of angular speed of an object is radian per second and Both angular speed, angular velocity are determined by using the same formula, angular velocity is  vector quantity which describes both magnitude and direction.

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• (-/1 Points) DETAILS OSCOLPHYS2016 9.5.WA.040.
MY NOTES ASK YOUR TEACHER PRACTICE ANOTHER
A supertanker uses a windlass (a type of winch) like the one shown in the figure below to hoist its 21,200-kg anchor. Determine the force that must be exerted on the
outside wheel to lift the anchor at constant speed, neglecting friction and assuming the anchor is out of the water.
IN
doorg
-0.45 m
- 1.5m

Answers

The force that must be exerted on the outside wheel to lift the anchor at constant speed is 6.925 x 10⁵ N.

Force exerted outside the wheel

The force exerted on the outside of the wheel can be determined by applying the principle of conservation of angular momentum as shown below.

∑τ = 0

Let the distance traveled by the load = 1.5 mLet the radius of the wheel or position of the force = 0.45 m

∑τ = R(mg) - r(F)

rF = R(mg)

0.45F = 1.5(21,200 x 9.8)

F = 6.925 x 10⁵ N.

Thus, the force that must be exerted on the outside wheel to lift the anchor at constant speed is 6.925 x 10⁵ N.

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What season would Texas be having at point D?

Answers

Answer:

Almost winter so fall

Explanation:

Hope this helps! Please let me know if you want me to elaborate more or think my answer is incorrect. Brainliest would be MUCH appreciated. Have a wonderful day!

What are the essential components of successfully competing in the field events:


Power, technique, and timing


Power, speed, and strength


Technique, flexibility, and speed


None of the above

Answers

Answer:

Power, technique, and timing

Explanation:

I just did the quiz and it was correct!! Hope this helps!

A 0.5 kg turntable with a radius of 15 cm is rotating 78 times per minute.

A. What is the angular momentum of the turntable?

B. A 100 gram bar which is as long as the diameter of the turntable is dropped on the
turntable such that the bar is rotating about its center. What is the new angular velocity of the turntable-bar system?

C. In order to stop the turntable and bar in just one rotation, how much force must be applied to the edge of the turntable (assuming the force is applied perpendicular to the radius)?

Answers

Answer:

c

Explanation:

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