Velocity of propagation In an experiment to illustrate the propagation of sound through fluids, identical sound sources emitting at 440 Hz are used to propagate sound waves through mercury and through ethanol. Medium Speed of sound Ethanol Mercurv (m/s) 1,160 1,450 1) Which of the following statements is correct, assuming that the dampening effect in each medium is negligible? In mercury, the frequency of the wave is the same as in ethanol, but the wavelength is greater. In ethanol and mercury, the wavelength of the wave is the same, even if the speed of sound is different. Submit (Survey Question) 2) Briefly explain your answer to the previous question.

Answer:

1. a 2.c 3 b

Explanation:

Answer:

3=b 2=c 1=a

Explanation:

To place charges of opposite sign but equal magnitude on two identical metal objects mounted on insulating stands, use a positively charged rod to induce a negative charge on one object and a negatively charged rod to induce a positive charge on the other object.

To achieve this, begin by bringing a positively charged rod close to one of the metal objects without making physical contact. The positive charge on the rod will induce a redistribution of charges within the metal object. The electrons in the metal will be repelled by the positive charge of the rod, causing them to move away from the rod and accumulate on the side of the metal object that is closer to the rod. As a result, the side of the metal object facing the rod will become negatively charged, while the side facing away from the rod will have a positive charge.

Next, take a negatively charged rod and bring it close to the second metal object. Again, without making physical contact, the negative charge on the rod will induce a redistribution of charges within this metal object. The electrons in the metal will be attracted to the negative charge on the rod, causing them to move towards the side of the metal object that is closer to the rod. Consequently, the side of the metal object facing the rod will acquire a positive charge, while the side facing away from the rod will have a negative charge.

By following these steps, you can place charges of opposite sign but exactly equal magnitude on the two metal objects. The positive and negative charges induced on the objects will be of equal magnitude because they are induced by the same charged rods, and the metal objects being identical will respond similarly to the charge induction process.

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Since the student is at rest and there is no frictional force acting on them, they will continue to be at rest as per the first law of Newton or the law of inertia.

The force of the student is equal to the force on the brick, but in opposite direction since the student and the brick are exerting the same amount of force on each other, but in opposite directions.  The momentum of the system before and after the throw must be equal. The momentum of the student, the brick, and the whole system remains zero as they were both at rest. Thus the velocity of the student is zero or remains at rest before and after the throw of the brick. Therefore, the resulting velocity of the student is zero.Why does the student not move backward in this situation?When the student throws the brick with some force, the brick moves forward with that force. Since Newton's third law says every action has an equal and opposite reaction, the brick also exerts an equal and opposite force on the student that would cause them to move backward. However, the student is standing on frictionless ice, which means that they cannot experience any frictional force that would cause them to move backward. Hence, the student remains stationary and does not move backward.

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The acceleration will be 1.9166 m/s². According to Newton's second law, the eventual effect of all forces applied to a body is its acceleration.

The rate of change of velocity with respect to time is known as acceleration.

Given data:

u(Initial velocity)=2.50m/s

v(Final velocity)=8.25m/s

t(time period)= 3.00s

The acceleration of the Allura will be :

\(\rm a = \frac{v-u}{t} \\\\ a= \frac{8.25-2.50 }{3.00 }\\\\ a= 1.9166 \ m/s^2\)

Hence, the acceleration will be 1.9166 m/s².

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Answer: The total current through the parallel combination of the diodes is: 3.90 μA

Explanation: Assuming the diodes are ideal with an offset voltage of Vf, the current-voltage characteristic of a single diode is given by:

i = 0 for v < -Vf (diode off)

i = I_s * (exp(v / (n * Vt)) - 1) for v >= -Vf (diode on)

where I_s is the saturation current, n is the diode ideality factor, and Vt is the thermal voltage.

Since the diodes are in parallel and opposing, one of them will be reverse-biased and the other will be forward-biased. The reverse-biased diode will have no current flowing through it, so we only need to consider the forward-biased diode.

Let's assume that d1 is forward-biased and d2 is reverse-biased. Then the voltage across d1 is V and the voltage across d2 is -V. The total current through the parallel combination of the diodes is the sum of the currents through each diode:

i = i1 + i2 where i1 is the current through d1 and i2 is the current through d2. Since d2 is reverse-biased, we have i2 = 0. The current through d1 is given by:

i1 = I_s * (exp(V / (n * Vt)) - 1)

To find V, we use the fact that the total current i is equal to the current through a resistor R in series with the diodes:

i = (V - (-V)) / R

i = 2V / R

Setting these two expressions for i equal to each other, we get:

I_s * (exp(V / (n * Vt)) - 1) = 2V / R

This is a transcendental equation that cannot be solved analytically. However, we can use numerical methods (e.g., Newton-Raphson) to find the solution. For simplicity, let's assume that the diodes are identical, so I_s and n are the same for both diodes. Then we can write:

I_s * (exp(V / (n * Vt)) - 1) = V / (2R)

Let's define a dimensionless variable x = V / (n * Vt). Then the equation becomes:

I_s * (exp(x) - 1) = x / (2nR * Vt)

Using the values given in the problem (I_s = 10^-15 A, n = 1, Vt = 25.85 mV, R = 1 kΩ), we can solve for x numerically:

x ≈ 2.145

Then the voltage across the diode is:

V = x * n * Vt ≈ 55.7 mV

Since this voltage is much smaller than the offset voltage of Vf = 0.7 V, we can assume that both diodes are on. The current through each diode is:

i1 = i2 = I_s * (exp(V / (n * Vt)) - 1) ≈ 1.95 μA

Therefore, the total current through the parallel combination of the diodes is:

i = i1 + i2 ≈ 3.90 μA

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D) All of the answers are correct.

Power, from an interdependency perspective, depends on all of the factors mentioned in the given options.

A) The control of valuable resources: Having control over valuable resources gives individuals or entities power because they can leverage these resources to influence others and shape outcomes.

B) The control of access to valuable resources: Controlling access to valuable resources provides power because it allows individuals or entities to regulate who can obtain those resources, creating dependency and influencing the behavior of others.

C) A partner's high desire for a particular resource: When a partner or another party has a strong desire for a specific resource, it can grant power to those who possess or control that resource. The high demand for the resource enhances the power of the resource holder in negotiations or relationships.

Therefore, all of these factors contribute to power dynamics by providing individuals or entities with advantages and leverage in various social, economic, and interpersonal contexts.

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Answer:

C. Light does not need a medium to travel through, but since sound waves must have a medium to vibrate, sound is not created where no air is present.

Explanation:

The reason the bell is seen to vibrate but not heard is that unlike light, which does not require a medium to pass through, sound waves require an air-filled space in order to vibrate. Option C is correct.

A sound wave is produced when a medium begins to vibrate. When an entity vibrates, a pressure wave is formed, which causes sound.

Light does not need a medium to travel through, but since sound waves must have a medium to vibrate, the sound is not created where no air is present best describes why the bell is seen vibrating but not heard.

The complete question is attached.

Hence option C is correct.

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To solve the question, we have to apply the concept of projectile motion. Given the following data: Height of the plane from the ground = 180 m. Speed of the plane = 170 m/s.

Using the concept of projectile motion, we can find the distance at which the package should be dropped. We know that the horizontal distance (x) is given by the equation:

x = vt Where v = velocity t = time.

Taking the vertical motion, we have the height from the ground as 180 m and acceleration due to gravity as -9.8 m/s²

Using the kinematic equations, we can find the time at which the package should be dropped.Initial velocity (u) = 0 m/s

Final velocity (v) = ?Acceleration (a) = -9.8 m/s²Time (t) = ?Height (s) = 180 m. We use the equation:v² = u² + 2as

Where,u = initial velocityv = final velocitya = acceleration due to gravity = height from the ground

Substituting the given values,v² = 0 + 2(-9.8)(180)∴ v = 84.68 m/s

Now, we can calculate the time taken to reach the ground using the equation:s = ut + 1/2 at²

Where,u = 0 m/st = ?a = -9.8 m/s²s = 180 mt = √2s/a∴ t = √(2 × 180/9.8)≈ 6.79 sThe total distance travelled by the package in the horizontal direction would be:x = vt = 170 × 6.79≈ 1156.3 m

Therefore, the package should be dropped 1156.3 m from the target so that it would reach the scientists on the glacier. The concept of projectile motion involves the motion of an object thrown or dropped from a certain height or launched at an angle.

The motion of the object is influenced by two forces: the horizontal force and the vertical force. The horizontal force is constant throughout the motion, whereas the vertical force varies due to the acceleration due to gravity.

Projectile motion is often observed in games such as basketball, cricket, and football. It is also used in real-life applications such as calculating the distance of a missile launched or finding the landing point of a satellite.

In conclusion, the given problem of finding the distance at which a package should be dropped can be solved using the concept of projectile motion. The total distance travelled by the package is 1156.3 m from the target. The concept of projectile motion has many real-life applications and is used in various fields such as physics, engineering, and sports.

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The temperature difference between water freezing and water boiling is 180 degrees on the Fahrenheit scale, 100 degrees on the Celsius Scale, and 373.15 degrees on the Kelvin scale.

Temperature is a physical quantity that expresses quantitatively the perceptions of hotness and coldness.noun Physical Chemistry. the temperature at which a liquid freezes:

The freezing point of water is 32°F, 0°C. A liquid's freezing point is the temperature at which that particular liquid changes into a solid matter state. For example, water freezes at 0 degrees celsius at the standard atmospheric pressure.

The Fahrenheit scale is a temperature scale based on one proposed in 1724 by the physicist Daniel Gabriel Fahrenheit. It uses the degree Fahrenheit as the unit.

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Answer:

\(31.7\:\mathrm{N}\)

Explanation:

We can use basic trig in a right triangle to solve this problem. Let the x component of the force be \(x\). We have the following equation:

\(\cos 24^{\circ}=\frac{x}{34.7},\\x=34.7\cos 24^{\circ}\approx \boxed{31.7}\)

Impulse  = change in momentum = 0.045 * 36 = 1.62 kg m/sec and Average Force = Impulse / time  = 1.62 / 3.9 x 10^-3    = 415.3 Newton

Impulse forces are those that have a relatively brief duration of action. The term "impulse" refers to the result of the impulsive force and the time at which it acts. I = F dt = dM is the formula for the impulse, which is the change in momentum brought on by the impulsive force.

The momentum of an object will change when a force acts for a predetermined period of time. In other words, an uneven force always accelerates an object, either by making it go faster or slower. When a force opposes an object's motion, the object slows down. When a force is applied in the same direction as an object's motion, the object is accelerated by the force. In either case, a force will alter an object's velocity.

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Answer:

202.8m

Explanation:

Given that A pirate fires his cannon parallel to the water but 3.5 m above the water. The cannonball leaves the cannon with a velocity of 120 m/s. He misses his target and the cannonball splashes into the briny deep.

First calculate the total time travelled by using the second equation of motion

h = Ut + 1/2gt^2

Let assume that u = 0

And h = 3.5

Substitute all the parameters into the formula

3.5 = 1/2 × 9.8 × t^2

3.5 = 4.9t^2

t^2 = 3.5/4.9

t^2 = 0.7

t = 0.845s

To know how far the cannonball travel, let's use the equation

S = UT + 1/2at^2

But acceleration a = 0

T = 2t

T = 1.69s

S = 120 × 1.69

S = 202.834 m

Therefore, the distance travelled by the cannon ball is approximately 202.8m.

The empty shopping cart is moving at 7.67 m/s.

According to the law of momentum conservation, momentum is only modified by the action of forces as they are outlined by Newton's equations of motion; momentum is never created nor destroyed inside a problem domain.

Total initial momentum = total final momentum

100 kg ×5.0 m/s + 30 kg ×0 m/s = 100 kg ×2.7 m/s + 30 kg ×v

v = (5.0 - 2.7)(100/30) m/s

= 7.67 m/s.

Hence,  the empty shopping cart is moving at  speed of 7.67 m/s.

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According to the given statement It takes 26.97 seconds to travel 7500 m with constant velocity 278 m/s.

Quickness of movement or activity is known as velocity. Speed is a shorter term and a substitute for celerity. The term "velocity" in mechanics especially refers to the measurement of an object's rate and direction of positional change. It is a vector number that describes a body's motion in term of both speed and direction.

Displacement of Object per unit time is called as Velocity.

velocity=displacement /time

v=s/t

Velocity of car

v=278m/s

Total Displacement

s=7500m

Hence Time

t=s/v

Displacement /Velocity

t=7500/278

t=26.97sec

It takes 26.97 seconds to travel 7500 m with constant velocity 278 m/s.

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The initial horizontal velocity can also be determined by measuring the diameter d of the ball and dividing by the time t that it takes for the ball to move across the photogate. That is Vo = d/t.

Find the distance between the base of the hill and the point where the ball hits the ground?

Let’s review the 4 fundamental kinematic equations of motion for constant acceleration (suggest you commit these to memory):

s = ut + ½at^2 …. (1)

v^2 = u^2 + 2as …. (2)

v = u + at …. (3)

s = (u + v)t/2 …. (4)

where s is distance, u is initial velocity, v is final velocity, a is acceleration and t is time.

In this case, we want to find the time for the ball to drop vertically to the ground, and then use that time to find the horizontal distance traveled.

For the vertical drop, we know u = 0, a = g = 9.81m/s^2 and s = 5m, so from equation (1) we get

s = ut + ½at^2

5 = 0 +4.905t^2

t = √(5/4.905) = 1.01s

Knowing t, we can now find the horizontal distance

s = ut + ½at^2

s = 35(1.01) + 0 (no horizontal acceleration)

s = 35.35m

So, the ball travels 35.35m horizontally before hitting the ground.

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The initial horizontal velocity can also be determined by measuring the diameter d of the ball and dividing by the time t that it takes for the ball to move across the photogate. That is Vo = d/t.

Let’s review the 4 fundamental kinematic equations of motion for constant acceleration (suggest you commit these to memory):

s = ut + ½at^2 …. (1)

v^2 = u^2 + 2as …. (2)

v = u + at …. (3)

s = (u + v)t/2 …. (4)

where s is distance, u is initial velocity, v is final velocity, a is acceleration and t is time.

In this case, we want to find the time for the ball to drop vertically to the ground, and then use that time to find the horizontal distance traveled.

For the vertical drop, we know u = 0, a = g = 9.81m/s^2 and s = 5m, so from equation (1) we get

s = ut + ½at^2

5 = 0 +4.905t^2

t = √(5/4.905) = 1.01s

Knowing t, we can now find the horizontal distance

s = ut + ½at^2

s = 35(1.01) + 0 (no horizontal acceleration)

s = 35.35m

So, the ball travels 35.35m horizontally before hitting the ground.

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Answer:

See below

Explanation:

x component =  90 cos 45 = 63.6 N

y component =  90 sin 45  = 63.6 N

The lunar eclipse takes place when the earth is located directly between the moon and the sun, causing the earth's shadow to fall over the moon. To a moon astronaut, the eclipse appears similar to a solar eclipse, with the earth obscuring the sun, except that the earth is large enough that it appears four times larger than the sun.

Therefore, using the known diameter of the sun and the proportion of the eclipse to estimate the diameter of the earth is possible. Since the sun is the largest object in the solar system, and the diameter of the sun is known to be approximately 1.39 million kilometers, we can utilize this value to calculate the diameter of the earth.

According to the question, the earth appears to be four times wider than the sun when viewed from the moon's surface during a lunar eclipse. Thus, the diameter of the earth would be approximately 0.25 of the diameter of the sun, since the earth is four times wider than the sun. Therefore, the diameter of the earth can be calculated using the formula:

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Answer:

10 points u liar so i wont answer

Explanation:

Answer:

hmmmmm

Explanation:

I am gonna say D.

if I am right plz give me a brainliest

The electrostatic force between the charges is 11.25 x \(10^{-11} \;\mu N\).

We have two charges separated by some distance apart.

We have to calculate the magnitude of the force exerted on the charges due to the electromagnetic interaction in μN (10-6 Newtons).

The force between two point charges is -

F = \($\frac{1}{4\pi \epsilon_{o} } \frac{qQ}{r^{2} }\)

According to the question -

Q =5 nC = 5 x \(10^{-9}\) C

q = 1 nC =  \(10^{-9}\) C

r = 2 cm = 0.02 m

The force of repulsion between the charges is -

\($\frac{1}{4\pi \epsilon_{o} } \frac{qQ}{r^{2} }\)

\($F =9\times 10^{9} \times \frac{5\times 10^{-9}\times 10^{-9} }{(0.02)^{2} }\)

F = 11.25 x \(10^{-5}\) Newtons.

Now , in micro - newtons →

F = 11.25 x \(10^{-5} \times 10^{-6}\) = 11.25 x \(10^{-11} \;\mu N\).

Hence, the electrostatic force between the charges is 11.25 x \(10^{-11} \;\mu N\).

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[ The given question is not complete. The complete question is -

" Two point-like particles separated by 2 cm have charges q1=5 nC and q2=1 nC. Note that the force felt by each charge due to the other is a vector and has both a magnitude and a direction. What is the magnitude of the force exerted on the charges due to the electromagnetic interaction, in μN (10-6 Newtons)? " ]

Answer:

Acceleration of this electron: \(-5.60 \times 10^{15}\; \rm m \cdot s^{-2}\).

Time taken: approximately \(9.39 \times 10^{-9}\; \rm s\).

Explanation:

Apply the SUVAT equation that does not involve time:

\(v^{2} - u^{2} = 2\, a \, x\).

Equivalently:

\(\begin{aligned}a &= \frac{v^{2} - u^{2}}{2\, x}\end{aligned}\).

By this equation, the acceleration of this electron would be:

\(\begin{aligned}a &= \frac{v^{2} - u^{2}}{2\, x} \\ &= \frac{(7.72 \times 10^{7}\; \rm m \cdot s^{-1})^{2} - (2.46 \times 10^{7} \; \rm m \cdot s^{-1})^{2}}{2 \times 0.478\; \rm m} \\ &\approx -5.60 \times 10^{15}\; \rm m \cdot s^{-2}\end{aligned}\).

The speed of this electron has changed from \(u = 7.72 \times 10^{7}\; \rm m\cdot s^{-1}\) to \(v = 2.46 \times 10^{7}\; \rm m \cdot s^{-1}\). Calculate the time required to achieve this change at a rate of \(a \approx -5.60 \times 10^{15}\; \rm m\cdot s^{-2}\):

\(\begin{aligned}t &= \frac{v - u}{a} \\ &\approx \frac{2.46\times 10^{7}\; \rm m \cdot s^{-1} - 7.72 \times 10^{7}\; \rm m\cdot s^{-1}}{-5.60 \times 10^{15}\; \rm m\cdot s^{-2}} \\ &\approx 9.39 \times 10^{-9}\; \rm s\end{aligned}\).

The latitude where the sun is directly overhead on April 22nd is called the subsolar point.

The sun is directly overhead at the equator on the spring and autumn equinoxes, which occur around March 20th and September 22nd respectively. On April 22nd, the sun is not directly overhead at the equator, but it is overhead at a latitude in the tropics, close to the equator.

To be more precise, the latitude where the sun is directly overhead on April 22nd is called the subsolar point, and it changes slightly from year to year due to the elliptical shape of Earth's orbit around the sun. On average, the subsolar point on April 22nd is around Latitude 10° N. However, to get an exact value for a specific year, you would need to perform calculations based on the exact position of the Earth in its orbit.

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ANSWER:

a. 104.6 kilometers per hour

b. 29.1 meters per second

c. 95.3 feet per second

STEP-BY-STEP EXPLANATION:

We can calculate the equivalence in each case using the conversion factors, as follows:

Answer:

Thank you so much!! for the points :)

oh tysm for z free points ʕ•ᴥ•ʔ

C. 0.68

aral ng mabuti! ❤

Answer:

back lol

Explanation:

Pull the tabs and wrap around to stick to the front of the diaper:)

Answer:

Front of the diaper

You take the straps that are connected to the back of the diaper and pull the. forward to attach to the front.

hope this helps.

Answer:

ball b has the greater mass.

Explanation:

b only moved because a bumped into it. if a bounced back, it had less mass and b moved slowly.