How many centimeters is half of a 87.3 km road?
Your result must be in multiples of 108 cm. That means if, for example, you get a result of a
9.2200x108 just type 9.2200 in the answer box. Include four digit after the decimal point and
maximum of 1% of error is accepted in your answer.

The final velocity's angle with regard to the first asteroid's velocity is the same as the angle of the first asteroid's beginning velocity, which is 15° with respect to the x-axis.

To solve this problem, using the law of conservation of momentum, the two asteroids stick together, so consider them as a single system.

Let's start by finding the initial momentum of the system:

p₁i = m₁v₁i = (18 × 10³ kg)(780 m/s) = 1.404 × 10⁷ kg⋅m/s

p₂i = m₂v₂i = (23 × 10³ kg)(1050 m/s) = 2.415 × 10⁷ kg⋅m/s

The initial momentum of the system is the vector sum of these momenta:

pi ni = p₁i + p₂i = (1.404 + 2.415) × 10⁷ kg⋅m/s = 3.819 × 10⁷ kg⋅m/s

Find the final velocity and direction of the system. Since the two asteroids stick together, final mass is the sum of their initial masses:

mf = m₁ + m₂ = (18 × 10³ kg) + (23 × 10³ kg) = 41 × 10³ kg

To find the final velocity, using the law of conservation of momentum again:

pf in = mf vfin

Since momentum is conserved:

pfin = pini

Therefore:

mf vfin = pini

vfin = pini / mf = (3.819 × 10⁷ kg⋅m/s) / (41 × 10³ kg) = 930.49 m/s

To find the direction of the final velocity, using trigonometry, define the x-axis to be parallel to the initial velocity of the first asteroid, and the y-axis to be perpendicular to the x-axis. Initial velocity of first asteroid has components:

v₁ix = v₁i cos(15°) = 758.19 m/s

v₁iy = v₁i sin(15°) = 199.78 m/s

Similarly, the initial velocity of the second asteroid has components:

v₂ix = v₂i cos(-15°) = 1041.02 m/s

v₂iy = v₂i sin(-15°) = -269.41 m/s

(Note that we use -15° for the second asteroid since its velocity makes an angle of 165° with respect to the x-axis.)

The total momentum of the system has components:

pfinx = pini = 3.819 × 10⁷ kg⋅m/s

pfiny = 0

Therefore, the final velocity of the system makes an angle θ with respect to the x-axis, where:

tan θ = pfiny / pfinx = 0

Since the y-component of the final velocity is zero, the final velocity is parallel to the x-axis.

Therefore, the direction of the final velocity with respect to the velocity of the first asteroid is the same as the direction of the initial velocity of the first asteroid, which is 15° with respect to the x-axis.

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Answer:

D. red shift

Explaination: if the spectral lines of galaxy are shifted towards the red end of spectrum (red shift) it means the galaxy is going away from earth!

Amplitude is the maximum vertical displacement of the seismic waves on the Seismogram. As evident from the graph, it is 60 mm.

1. Time at which P waves arrive- 0 sec

Time at which S waves arrive- 18 sec

S-P interval= Time at which S wave arrive- Time at which P wave arrive

S-P interval= 18-0= 18 sec

2. Amplitude is the maximum vertical displacement of the seismic waves on the Seismogram. As evident from the graph, it is 60 mm.

3. The distance is around 150 km and is evident on the Distance/S-P bar chart.

4. The magnitude is about 4 and can be found by joining S-P interval (18 sec) with the Amplitude i.e., 6 mm. The point of intersection of the line to the magnitude line gives the magnitude, i.e., 4.

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Answer:

U Can check it out

Explanation:

the wire wound in a form of loop forms a coil or solenoid .when an electric current is passed through a coil or solenoid the resultant magnetic field is similar to that of a bar magnet .One end of the solenoid is North N-pole and other is S-pole like bar magnet .Since the field lines cannot cross each other, the Field lines are packed closer together inside. Therefore the field is stronger and uniform inside and weaker outside the coil.

The ring will have the greater moment of inertia.

Acceleration of a body rolling down an inclined plane without slipping is given by,

α = gsinθ/(1 + K²/R²)

Acceleration of the ring,

α = gsinθ/(1 + R²/R²)

α = 1/2 gsinθ

Acceleration of the sphere,

α = gsinθ(1 + 5/2)

α = 2/7 gsinθ

Acceleration of the solid cylinder,

α = g sinθ(1 + 1/2)

α = 2/3 gsinθ

The ring has the highest acceleration. Therefore, the ring will reach the bottom of the slope first.

The ring will have the greater moment of inertia among the three.

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The orbital speed of the satellite orbiting around a planet of radius 34 Km is found to be 2.59 km/s.

To find the orbital speed (v) of the satellite, we can use the formula,

v = √(GM/r), gravitational constant (6.674 x 10⁻¹¹ N(m/kg)²) is G, mass of the planet is M, and orbital radius of the satellite is r. To calculate M, we can use the formula,

g = GM/r², rearranging this formula, we get,

M = gr²/G

Substituting the values, we get,

M = 3.3(34,000)²/(6.674 x 10⁻¹¹)

M = 6.06 x 10²⁰ kg

Now, substituting the values of G, M, and r into the formula for orbital speed, we get,

v = √((6.674 x 10⁻¹¹)(6.06 x 10²⁰)/(34,000))

v = 2.59 x 10³ m/s

Therefore, the satellite's orbital speed is approximately 2.59 km/s.

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The time taken by the ball to fall through a distance of 30m is, 2 seconds

To calculate the time taken to fall through a distance of 30m, we have given data as,

mass of ball = 200g

g = 10 m/s^2

Initial height = 50 meter

Final height = 30 meter

So by the laws of motion, we have,

Δh = (ut) + (at²/2)

where, (u = initial velocity), and (t = time)

So,

50-30 = (0.t) + [10 x t²)/2]

20 = 5t²

Therefore, t = ± (2 seconds)

On eliminating the negative value as time can`t be negative here, we get

t = 2 seconds

Therefore, the time taken by the ball to fall through a distance of 30m is, 2 seconds

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To solve this problem, we can use the conservation of momentum and the conservation of kinetic energy.First, let's find the velocity of the 0.530-kg cart after the collision. We can use the conservation of momentum:m1v1 + m2v2 = m1v1' + m2v2'where m1 and v1 are the mass and velocity of the 0.530-kg cart before the collision, m2 and v2 are the mass and velocity of the 0.25-kg cart before the collision, and v1' and v2' are the velocities of the carts after the collision.Plugging in the numbers, we get:(0.530 kg)(0.572 m/s) + (0.25 kg)(0 m/s) = (0.530 kg)v1' + (0.25 kg)v2'Solving for v1', we get:v1' = [(0.530 kg)(0.572 m/s) + (0.25 kg)(0 m/s)] / (0.530 kg + 0.25 kg) = 0.378 m/s to the rightSo the 0.530-kg cart moves off to the right at 0.378 m/s after the collision.Next, let's find the maximum compression of the spring. We can use the conservation of kinetic energy:(1/2)m2v2^2 = (1/2)kx^2where k is the spring constant and x is the maximum compression of the spring.We know the mass and velocity of the 0.25-kg cart before the collision (v2 = 0 m/s), so we can solve for k:k = 2(1/2)m2v2^2 / x^2 = m2v2^2 / x^2Plugging in the numbers, we get:k = (0.25 kg)(0 m/s)^2 / x^2 = 0This means that the spring constant is 0, which is not physically possible. Therefore, there must be an error in the problem statement or some missing information that would allow us to calculate the maximum compression of the spring.

A car travels 200 km in the first 2.5 hour then stop for half hour then travels the final speed of 200 km in 2 hours. The average speed of the car is 80 km/hour.

To find the average speed of the car, we need to calculate the total distance traveled and the total time taken.

In the first 2.5 hours, the car travels 200 km.

Then, it stops for half an hour.

After that, the car travels another 200 km in 2 hours.

So the total distance traveled is 200 km + 200 km = 400 km.

The total time taken is 2.5 hours + 0.5 hours + 2 hours = 5 hours.

Therefore, the average speed of the car is:

Average speed = total distance / total time

= 400 km / 5 hours

= 80 km/hour.

So the average speed of the car is 80 km/hour.

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The breaking rate refers to the rate at which an object slows down due to braking or deceleration. In other words, it is the rate of change of velocity in the opposite direction of the object's motion.

Looking at the data provided, we can see that the velocity decreases from 50 m/s to 0 m/s over a period of 10 seconds, which means the object is decelerating at a constant rate. To calculate the breaking rate, we can use the formula:

breaking rate = (final velocity - initial velocity) / time taken

In this case, the breaking rate is:

breaking rate = (0 - 50) / 10 = -5 m/s^2

So, the object is decelerating at a rate of 5 m/s^2.

To compare this to the acceleration, we need to know the acceleration of the object before it starts breaking. If we assume that the object was accelerating at a constant rate of 5 m/s^2 before it started breaking, then the acceleration and breaking rates are equal in magnitude but opposite in direction. In other words, the acceleration and breaking rates are both 5 m/s^2, but the acceleration is positive while the breaking rate is negative.

It's worth noting that the breaking rate can vary depending on various factors such as the mass of the object, the friction between the object and the surface it is moving on, and the force applied to the brakes.

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The proportionality constant of the moving object experiencing a drag force is 0.01875 Ns²/m².

The work-energy principle states that the work done on an object is equal to its change in kinetic energy. So, the work done by the drag force can be found as follows:

W = (1/8)mv² - (1/2)mv₀²

where m = mass of the object, v = final speed, and v₀ = initial speed.

The work done by the drag force is also given by the formula:

W = ∫F(x)dx

where F(x) = force function and x = position of the object.

In this case, the force function is F(x) = -Cv², since the drag force is in the opposite direction of motion. So:

W = ∫-Cv²dx

Since the force is proportional to v², rewrite this as:

W = -C∫v²dx

Integrating both sides with respect to x:

W = -(1/3)Cv³

So, equating the two expressions for W:

(1/8)mv² - (1/2)mv₀² = -(1/3)Cv³

Substituting m = 0.01 kg, v₀ = 10 m/s, and solving for C:

C = -(3/8) × (m/v₀³) × (v² - v₀²) = -(3/8) × (0.01/10³) × (1/8 × 10² - 10²) = 0.01875 Ns²/m²

Therefore, the proportionality constant is C = 0.01875 Ns²/m².

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The electric field at the center of the cube is approximately 5.12 × 10⁴ N/C.

The electric field at the center of a face of the cube is approximately 4.54 × 10⁴ N/C.

The electric field at the center of the cube if one of the corner charges is removed is approximately 4.54 × 10⁴ N/C.

(a) To find the electric field at the center of the cube, we can use the principle of superposition, which states that the total electric field at a point in space is the vector sum of the electric fields due to each individual charge. Since all eight charges are identical and have the same distance to the center of the cube, the electric field due to each charge has the same magnitude and direction.

Using Coulomb's law, we can calculate the magnitude of the electric field due to one charge at the center of the cube as:

E = (kQ) / r²

where k is the Coulomb constant, Q is the charge on each point charge, and r is the distance from the charge to the center of the cube. Since the charges are at the corners of a cube with sides of length 10 cm, the distance from each charge to the center is sqrt√/2 times the length of the side, or 5√(3) cm.

Thus, the magnitude of the electric field due to one charge at the center of the cube is:

E = (kQ) / (5√(3) cm)² = 1.24 × 10⁴ N/C

Since there are eight charges, the total electric field at the center of the cube is:

E_total = 8E = 9.95 × 10⁴ N/C

(b) To find the electric field at the center of a face of the cube, we can again use the principle of superposition. Since the face of the cube is equidistant from four of the charges, the electric field due to those charges has the same magnitude and direction, while the electric field due to the other four charges cancels out.

So, the magnitude of the electric field at the center of a face of the cube is:

E_face = 4E = 4.96 × 10⁴ N/C

(c) If one of the corner charges is removed, the electric field at the center of the cube is no longer spherically symmetric. However, we can still use the principle of superposition to calculate the electric field due to the remaining seven charges. The electric field due to these charges at the center of the cube has the same magnitude as the electric field due to one charge at the center of a face of the cube.

Since the distance from the center to each of the remaining charges is √(2) times the length of the side of the cube.

Thus, the magnitude of the electric field due to the remaining charges is:

E_remaining = 7E = 3.18 × 10⁴ N/C

Therefore, the electric field at the center of the cube if one of the corner charges is removed is approximately 4.54 × 10⁴ N/C, which is the average of the electric fields at the centers of adjacent faces of the cube.

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The recoil speed of the rifle is 0.5 m/s.

Weight of the rifle, W = 30 N

Mass of the rifle, M = W/g = 30/10 = 3 kg

Mass of the bullet, m = 5 g = 5 x 10⁻³kg

Speed of the bullet, v = 300 m/s

a) The expression for the recoil speed of the rifle is given by,

v(r) = mv/M

v(r) = 5 x 10⁻³ x 300/3

v(r) = 0.5 m/s

b) Weight of the man, W' = 700 N

Mass of the man, M' = W'/g = 700/10 = 70 kg

So, the combined mass of the man and the rifle,

M₁ = M + M'

M₁ = 3 + 70

M₁ = 73 kg

Therefore, the recoil speed of man and rifle,

v(r)' = mv/M₁

v(r)' = 5 x 10⁻³ x 300/73

v(r)' = 0.0205 m/s

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Engineers use electrical insulators to prevent electricity from flowing to the wrong place.

Insulators are materials that do not conduct electricity easily and are used to separate electrical conductors to prevent current leakage or short circuits. Common insulating materials include rubber, plastic, glass, and ceramic. By using insulators, engineers can ensure that electrical energy is directed along the intended path and that electrical equipment operates safely and efficiently.

. Insulators are commonly used in a variety of applications, including electrical wiring, power transmission and distribution systems, electronic devices, and high-voltage equipment. Common insulating materials include rubber, plastic, glass, ceramic, and air. The choice of insulating material depends on various factors such as the required level of insulation, the operating temperature, and the environment in which the insulator will be used.

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The pressure exerted on the fluid by the force applied on the small piston can be calculated using the formula:

P = F/A

where P is the pressure, F is the force, and A is the area on which the force is applied. Since the force is applied on the smaller piston, we need to use its area:

A_small = πr_small^2

where r_small is the radius of the smaller piston. Thus,

A_small = π(0.02 m)^2 = 1.2566 x 10^-3 m^2

The force applied on the small piston is 250 N. Thus,

P = F/A_small = 250 N / 1.2566 x 10^-3 m^2 = 1.989 x 10^5 Pa

Therefore, the increase in pressure caused by the 250 N force on the small piston is 1.989 x 10^5 Pa, which is approximately equal to 2 x 10^5 Pa (to two significant figures).

b) We can use the principle of conservation of volume to determine how far the smaller piston moves when the larger piston moves 5 cm. The volume of the fluid in the hydraulic lift remains constant, so we have:

A_small × h_small = A_large × h_large

where h_small and h_large are the heights of the fluid columns above the smaller and larger pistons, respectively. Since the lift is filled with an incompressible fluid, the pressure is the same throughout the fluid. Thus,

P = F/A_small = F/A_large

Multiplying both sides of this equation by the areas of the pistons, we get:

F × A_small = F × A_large

Substituting the given values, we get:

250 N × (π(0.02 m)^2) = F × (π(0.20 m)^2)

Solving for F, we get:

F = 250 N × (0.02 m/0.20 m)^2 = 25 N

Now, we can use the force applied on the larger piston and the area of the smaller piston to calculate the force on the smaller piston:

F_small = F × (A_small/A_large) = 25 N × (1.2566 x 10^-3 m^2 / (π(0.20 m)^2)) = 0.1989 N

Using the formula for pressure, we can calculate the height of the fluid column above the smaller piston:

P = F_small/A_small = h_small × ρ × g

where ρ is the density of the fluid and g is the acceleration due to gravity. Since the density of the fluid and the acceleration due to gravity are constants, we can simplify this equation to:

h_small = F_small/(A_small × ρ × g)

Substituting the given values, we get:

h_small = 0.1989 N / (1.2566 x 10^-3 m^2 × 1000 kg/m^3 × 9.81 m/s^2) = 0.0159 m

Therefore, the smaller piston moves 0.0159 m (or approximately 1.6 cm) when the larger piston moves 5 cm.

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The given statement that Large speaker cones produce deeper frequencies than small speaker cones is true.

The frequency response of a speaker refers to its ability to reproduce sound across different frequencies. In general, larger speaker cones are capable of moving more air and producing lower frequencies than smaller cones. This is because the size of the speaker cone affects the amount of air it can displace and the amount of force it can generate.

Low-frequency sounds require more movement of air to be heard, and larger cones are better suited to move the necessary amount of air. However, it's worth noting that there are other factors that can affect a speaker's frequency response, such as the design of the speaker cabinet, the materials used in the speaker cone, and the quality of the electronics used to power the speaker.

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The magnitude of the acceleration of the ball  applied at the bottom of the circle can be expressed in the form  of FTension-FGravity/M.

Option D is correct.

Along the string's circular and vertical paths, the tension changes. As long as the total quantity of kinetic and potential energy is constant throughout, the ball's speed can change.

Centripetal force varies as a result of motion variations.

We can determine how tight a string that is traveling in a vertical circle is using the expression below:

FC = mv2 /r.

A moving item attached to a string experiences centripetal force, which is determined by the product of the object's mass (mg) and the string tension. (T).

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Complete question:  

A student swings ball of mass M on the end of a string in vertical circle of radius R,as shown in the figure below. Also shown is diagram representing all the forces exerted on the ball at the bottom of the circle where its speed is What is the magnitude of the acceleration of the ball at the bottom of the circle? FTension FGravity

A)Fi /M

B)Fc /M

C)Fr+Fg/M

D) Ft- Fg/M

Answer:

The answer is 13.5g/cm³

Explanation:

m1=20g

m3=70g

m2=695g

v=20cm³

m2-m1/m3-m1

R.d=695-20/70-20

R.d=675/50

R.d=13.5

R.d=density of liquid/density of water

density of liquid =R.d×density of water

D=13.5×1

D=13.5g/cm³

Answer:

Sound waves cause particles to vibrate perpendicular to the direction of the wave.

If we rank these magnets from the strongest to the weakest magnetic field the correct order is 4, 3, 2, 1.

The magnetic field and radius are related in the context of a charged particle moving in a circular path under the influence of a magnetic field. When a charged particle moves in a circular path under the influence of a magnetic field, the force on the particle is directed toward the center of the circle. In this force, the radius can be expressed as r = mv / Bq.

This equation shows that the radius of the circular path is directly proportional to the velocity of the particle, and inversely proportional to the magnetic field strength and the charge of the particle.

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Answer:

A screw gauge, also known as a micrometer screw gauge, is a precision measuring instrument used to measure small distances with high accuracy. It consists of a calibrated screw and a calibrated thimble with a spindle and an anvil face. The spindle is attached to the screw, and the anvil face is fixed to the body of the instrument.

To use a screw gauge, the object being measured is placed between the spindle and the anvil face, and the screw is turned until the spindle makes contact with the object. The thimble is then turned, which moves the spindle and the screw, until the spindle is firmly in contact with the object. The reading on the scale of the thimble is then taken, which gives the distance between the spindle and the anvil face, with a high degree of precision.

Screw gauges are commonly used in a variety of fields, including engineering, machining, and scientific research, where accurate measurements are required.

Screw gauge measures diameter with high accuracy using a screw and nut mechanism. It has a pitch and graduated thimble.

A screw check is an estimating instrument used to gauge the width of items, particularly wires and chambers, with high exactness. It deals with the rule of a screw and nut component.The screw measure comprises of a U-molded outline with a screw toward one side and a thimble at the other.

The screw has a pitch, which is the distance gone by the screw in one complete turn. The pitch is as a rule of the request for 0.5 mm. The thimble is graduated into 50 or 100 equivalent parts. The roundabout scale on the thimble must be changed to such an extent that the zero of the scale matches with the zero of the pitch scale.

To utilize the screw check, the item whose width is to be estimated is set between the iron block and the shaft. The screw is turned until the shaft simply contacts the item. The pitch scale perusing is noted.

The thimble is then turned until the zero of the thimble scale concurs with the middle line of the pitch scale. The thimble scale perusing is then noted. The breadth of the item is then determined by adding the pitch scale perusing and the thimble scale perusing.

Along these lines, a screw check can gauge widths to an exactness of 0.01 mm or better. It is broadly utilized in designing, fabricating, and logical applications.

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Answer: the quantity of the substance being measured and make sure to stir the substance before taking the readings, also make your eyesight perpendicular to the scale reading to avoid parallax error

Explanation:

Answer: She passed at 19 years old.

Explanation:

Answer:

a) The order in which the shapes reach the bottom of the slope will be the sphere, solid cylinder, and ring.

b) The time it takes for each shape to reach the bottom of the slope can be calculated using the following equation:

t = (2d / g)^(1/2)

Where t is the time, d is the height of the inclined plane (1m in this case), and g is the acceleration due to gravity (9.8 m/s^2).

For the sphere:

t = (2 x 1 / 9.8)^(1/2) = 0.45 seconds

For the solid cylinder:

t = (2 x 1 / 9.8)^(1/2) x (5/7) = 0.36 seconds

For the ring:

t = (2 x 1 / 9.8)^(1/2) x (2/5) = 0.28 seconds

c) The moment of inertia depends on the shape of the object and how the mass is distributed around its axis of rotation. For a solid sphere, the moment of inertia is given by I = (2/5)MR^2, for a solid cylinder it is I = (1/2)MR^2, and for a ring it is I = MR^2. Therefore, the order of increasing moment of inertia is the ring, the solid cylinder, and the sphere.

d) The linear acceleration of each shape can be calculated using the following equation:

a = gsinθ / (1 + I / MR^2)

Where a is the linear acceleration, g is the acceleration due to gravity (9.8 m/s^2), θ is the angle of the inclined plane (20° in this case), I is the moment of inertia, M is the mass, and R is the radius.

For the sphere:

a = (9.8 x sin20) / (1 + (2/5)) = 2.34 m/s^2

For the solid cylinder:

a = (9.8 x sin20) / (1 + (1/2)) = 3.29 m/s^2

For the ring:

a = (9.8 x sin20) / (1 + 1) = 4.16 m/s^2

e) The tangential (linear) velocity of each shape at the bottom of the slope can be calculated using the following equation:

v = ωR

Where v is the tangential velocity, ω is the angular velocity, and R is the radius.

The angular velocity can be calculated using the following equation:

ω = (2a / R)^(1/2)

For the sphere:

ω = (2 x 2.34 / 0.05)^(1/2) = 21.8 rad/s

v = 21.8 x 0.05 = 1.09 m/s

For the solid cylinder:

ω = (2 x 3.29 / 0.05)^(1/2) = 30.7 rad/s

v = 30.7 x 0.05 = 1.53 m/s

For the ring:

ω = (2 x 4.16 / 0.05)^(1/2) = 36.4 rad/s

v = 36.4 x 0.05 = 1.82 m/s

mark me brilliant

Answer:

c

Explanation:

The correct answer is using ultrasonic waves to create images of objects found inside other objects.

Infrasonic Wave: An infrasonic wave is a longitudinal elastic wave whose frequency is lower than that of sound, or 20Hz. For instance: It typically occurs during an earthquake. Ultrasonic Wave: An ultrasonic wave is a longitudinal wave with a frequency greater than the threshold of human hearing, or 20 kHz.

The audible range of frequencies below 20 Hz is known as infrasonic. Volcanoes, earthquakes, and thunder all emit sound in the infrasonic range. Human ears cannot hear this sound, but elephants and whales can. Ultrasonic frequencies are those that are above 20,000 Hz.

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The information about ultraviolet radiation supports or contradicts the safety of solar radiation exposure to astronauts on the international space station. They have wore suits that protects the astronaut from the UV light.

A space station is a sort of space habitat because it can sustain a human crew in orbit for a lengthy period of time. Major landing or propulsion systems are absent. An artificial satellite, also known as an orbital station or orbital space station, is a kind of orbital spaceflight. To allow other spacecraft to dock and transfer personnel and cargo, stations need to have docking ports. Depending on the programmed, a given orbiting outpost has a different role. Military launches have also taken place, although scientific launches of space stations have predominated. astronaut have wore suits that protects the astronaut from the UV light.  

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The heel of shoe D applies the greatest force to the carpet because it has the largest surface area in contact with it, and thus the largest force is distributed over that larger area.

The correct statement explaining how a megaphone amplifies sound is: "A megaphone amplifies sound by focusing sound energy into one specific direction."

By increasing the acoustic impedance perceived by the vocal chords and bringing them into closer proximity to the air, the loudhailer amplifies the sound and increases the amount of sound power that is emitted.

Many people are familiar with the distinctively distorted sound of a human voice amplified by a loudhailer thanks to its use in train and bus stations and sporting venues. It produces the sound of a vintage acoustic phonograph record player when used with music.

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