1. Determine the moment of inertia (tensor) of a flat square of mass M, and side length A with respect to its center of mass.

The volume flow rate of the water through the hose is 4.58×10⁻⁴m³/sec.

Flow-restrictive lesions in the carotid, vertebral, and intracranial vessels have been assessed for severity and hemodynamic significance using MR volume flow rate measurements. The volume flow rate distal to the stenosis may significantly decrease as a result of a severe stenosis.

The volume flow rate () is the amount of fluid that moves through a cross-sectional area in a given amount of time. Volume flow rate and mass flow rate are related by the conservation of mass rule.

Given,

Radius = 0.0137 m

Area =  πr²

Area = 3.14×(0.0137)²

Area = 0.000589m².

Volume filled in 90sec = 25L

Rate = 25/90 = 0.27L/sec

Volume flow rate =  0.27× 1000/0.000589

Volume flow rate = 4.58×10⁻⁴m³/sec.

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Answer:

A Impulse = – 25 Ns

B. Time = 5 s

Explanation:

From the question given above, the following data were obtained:

Mass (M) = 5 Kg

Initial velocity (u) = 8 m/s

Final velocity (v) = 3 m/s

Impulse (I) =?

Time (t) =?

A. Determination of the Impulse.

Mass (M) = 5 Kg

Initial velocity (u) = 8 m/s

Final velocity (v) = 3 m/s

Impulse (I) =?

I = Ft = M(v – u)

I = M(v – u)

I = 5 (3 – 8)

I = 5 × – 5

I = – 25 Ns

NOTE: the negative sign indicates that the net force is acting in the negative direction.

B. Determination of the time.

Impulse (I) = 25 Ns

Force (F) = 5 N

Time (t) =?

I = Ft

25 = 5 × t

Divide both side by 5

t = 25 / 5

t = 5 s

Thus, it will take 5 s for the box to slide through the 15 m long ramp.

If you pull with a constant force of 400 N for 0.2 meters, then the work done will be equal to 80 J.

In physics, the word "work" involves the measurement of energy transfer that takes place when an item is moved over a range by an externally applied, at least a portion of which is applied within the direction of the displacement.

The length of the path is multiplied by the element of a force acting all along the path to calculate work if the force is constant. The work W is theoretically equivalent towards the force f times the length d, or W = fd, to portray this concept.

As per the given information in the question,

Force, f = 400 N

Displacement, d = 0.2 meters

\(Work done(W)=Force(f)*Displacement(d)\)

W = 400 × 0.2

W = 80 J.

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Answer:

1) convection.

2) thermometer.

3) Celsius scale.

4) Radiation.

5) Conduction.

6) Clinical thermometers.

Explanation:

so the questions have already given you a simple idea of their meanings.

Good luck.

The interval between the first and second echo is 7 seconds. This means that after the initial sound wave reaches the first cliff, it takes a total of 7 seconds for the sound to travel to the second cliff and then return to the student as the second echo.

To determine the interval between the first and second echo, we need to consider the time it takes for sound to travel from the student to the first cliff, and then from the first cliff to the second cliff, and finally back to the student.

Let's break down the distances and calculate the time for each part of the journey:

Distance from the student to the first cliff: 330 meters

Time taken: t₁ = distance / speed = 330 m / 330 m/s = 1 second

Distance from the first cliff to the second cliff: 990 meters

Time taken: t₂ = distance / speed = 990 m / 330 m/s = 3 seconds

Distance from the second cliff back to the student: 990 meters

Time taken: t₃ = distance / speed = 990 m / 330 m/s = 3 seconds

Now, we can calculate the total interval between the first and second echo by adding up the individual times:

Interval between first and second echo = t₁ + t₂ + t₃ = 1 s + 3 s + 3 s = 7 seconds

Therefore, the interval between the first and second echo is 7 seconds. This means that after the initial sound wave reaches the first cliff, it takes a total of 7 seconds for the sound to travel to the second cliff and then return to the student as the second echo.

It's important to note that this calculation assumes a straight path for the sound waves and neglects factors such as air temperature and wind that can affect the speed of sound. Additionally, it assumes perfect reflection of sound waves off the cliffs, which may not be the case in real-world scenarios.

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Note the complete questions is:

A student stands 330m from a tall cliff which is 990m from another tall cliff. If the speed of sound between the cliffs is 330m/s.What is the interval between the first and second echo?

Hi there!

a.

We can use the initial conditions to solve for w₀.

It is given that:

\(\frac{d\theta}{dt} = w_0e^{-\sigma t}\)

We are given that at t = 0, ω =  3.7 rad/sec. We can plug this into the equation:

\(\omega(0)= \omega_0e^{-\sigma (0)}\\\\3.7 = \omega_0 (1)\\\\\omega_0 = \boxed{3.7 rad/sec}\)

Now, we can solve for sigma using the other given condition:

\(2 = 3.7e^{-\sigma (8.6)}\\\\.541 = e^{-\sigma (8.6)}\\\\ln(.541) = -\sigma (8.6)\\\\\sigma = \frac{ln(.541)}{-8.6} = \boxed{0.0714s^{-1}}\)

b.

The angular acceleration is the DERIVATIVE of the angular velocity function, so:

\(\alpha(t) = \frac{d\omega}{dt} = -\sigma\omega_0e^{-\sigma t}\\\\\alpha(t) = -(0.0714)(3.7)e^{-(0.0714) (3)}\\\\\alpha(t) = \boxed{-0.213 rad\sec^2}\)

c.

The angular displacement is the INTEGRAL of the angular velocity function.

\(\theta (t) = \int\limits^{t_2}_{t_1} {\omega(t)} \, dt\\\\\theta(t) = \int\limits^{2.5}_{0} {\omega_0e^{-\sigma t}dt\\\\\)

\(\theta(t) = -\frac{\omega_0}{\sigma}e^{-\sigma t}\left \| {{t_2=2.5} \atop {t_1=0}} \right.\)

\(\theta = -\frac{3.7}{0.0714}e^{-0.0714 t}\left \| {{t_2=2.5} \atop {t_1=0}} \right. \\\\\theta= -\frac{3.7}{0.0714}e^{-0.0714 (2.5)} + \frac{3.7}{0.0714}e^{-0.0714 (0)}\)

\(\theta = 8.471 rad\)

Convert this to rev:

\(8.471 rad * \frac{1 rev}{2\pi rad} = \boxed{1.348 rev}\)

d.

We can begin by solving for the time necessary for the angular speed to reach 0 rad/sec.

\(0 = 3.7e^{-0.0714t}\\\\t = \infty\)

Evaluate the improper integral:

\(\theta = \int\limits^{\infty}_{0} {\omega_0e^{-\sigma t}dt\\\\\)

\(\lim_{a \to \infty} \theta = -\frac{\omega_0}{\sigma}e^{-\sigma t}\left \| {{t_2=a} \atop {t_1=0}} \right.\)

\(\lim_{a \to \infty} \theta = -\frac{3.7}{0.0714}e^{-0.0714a} + \frac{3.7}{0.0714}e^{-0.0714(0)}\\\\ \lim_{a \to \infty} \theta = \frac{3.7}{0.0714}(1) = 51.82 rad\)

Convert to rev:

\(51.82 rad * \frac{1rev}{2\pi rad} = \boxed{8.25 rev}\)

Answer:

C. is always perpendicular to the surface of the conductor

Explanation:

On a charged conductor , electric charge is uniformly distributed on its surface . The lines of forces are also uniformly  distributed on all directions . They repel each other so they emerge perpendicular to the surface so that they do nor cut each other and at the same time they remain at maximum distance from each other.

Answer:

B

Explanation:

First the battery uses chemical energy ===>    to make electrical energy  ===>    to make  mechanical energy

For any freely falling item, the acceleration in the kinematic equations is -9.8 m/s/s, whether or not this is explicitly stated.

The supplied problem informs us that a ball was dropped and is supposed to be falling freely. Since the problem does not specify whether any other forces were present to influence the ball's motion, we must assume that the initial speed is v0=0 m/s.

Free-falling accelerates at a rate of 9.8 m/s2, which is referred to as the acceleration owing to gravity because it is being drawn towards the center of the earth. This indicates that a=9.8m/s2 is the value of the initial and final accelerations.

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Answer:

The correct answer is D C A

Answer:

parallel circuit, hopefully I wasn't late in sending the answer

Answer:

Proxima Centauri, the closest star to our own, is still 40,208,000,000,000 km away.

Explanation:

Proxima Centauri, → 40,208,000,000,000 km

An electron collides with an atom of element X and the atom emits a 1240 nm photon. The electron's speed after collision can be calculated by equating the energy of the photon to the kinetic energy of the electron, resulting in a speed of 1.872 x 10⁶ m/s.

The energy of the photon is equal to the difference between the initial and final kinetic energies of the electron. The initial kinetic energy of the electron can be calculated using the formula

KE = (1/2)mv²,

where m is the mass of the electron and v is its speed. The final kinetic energy of the electron can be calculated using the same formula, but with a different speed, which we will call vf. Since the energy is conserved, we can equate the initial and final energies

(1/2)mv² = (1/2)mvf² + E_photon

where E_photon is the energy of the photon emitted by the atom. We can convert the wavelength of the photon to its energy using the formula E_photon = hc/λ,

where h is Planck's constant and c is the speed of light. Plugging in the values given in the problem, we get

E_photon = (6.626 x 10⁻³⁴ J s)(3 x 10⁸ m/s)/(1240 x 10⁻⁹ m) = 1.006 x 10⁻¹⁹ J

Substituting this value back into the initial equation and solving for vf, we get

vf = √(2((1/2)mv² - E_photon)/m) = 1.872 x 10⁶ m/s

Therefore, the speed of the electron after the collision is 1.872 x 10⁶ m/s.

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Answer:

the sum of all force being applied to an object.

Explanation:

The earthquake would occur 13 minutes before 10:05 a.m. which will be at 9.52 am.

The p-waves travel with a constant velocity of 7 km/s

The time can be calculated by using the formula

t = d / v

where

T1 =  10:05 a.m

d is the distance they take to travel from the epicenter

v is the speed of the p-waves

On average, the speed of p-waves is

v = 7 km/s

d = 5600 km (given)

Substituting the values in the formula;

t = d / v

t = 5600 ÷ 7

t = 800 seconds

Converting into minutes,

t = 800 ÷ 60

t = 13.3

≈ 13 mins

T1 -  13 mins = T2

10:05 - 13 mins = 9.52 am

It means the earthquake occurred prior 13 minutes, that is at 9.52 am.

Therefore, the earthquake occurred at 9.52 am.

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Answer:

I know that 2 is called a foul I dont know about 1 though

The speed of the roller coaster at the bottom of the hill if the track was frictionless is 34.04 m/s.

Given that the weight of the roller coaster is 2000 kg and the initial vertical drop of the ride is 59.3 m. We are to find the speed of the roller coaster at the bottom of the hill if the track was frictionless.We know that the roller coaster will lose potential energy due to the vertical drop. Assuming there is no friction, the potential energy will be converted into kinetic energy at the bottom of the hill.Considering the conservation of energy between the potential and kinetic energy, we can set the initial potential energy equal to the final kinetic energy. We can use the formula to calculate potential energy, which is PE = mgh where m = 2000 kg, g = 9.8 m/s², and h = 59.3 m. Therefore,PE = 2000 kg × 9.8 m/s² × 59.3 m = 1,157,924 JWe can use the formula to calculate kinetic energy, which is KE = 1/2mv² where m = 2000 kg and v is the final velocity. Therefore,KE = 1/2 × 2000 kg × v².The total energy remains constant as we know there is no friction. Therefore the final kinetic energy will be equal to the initial potential energy,1,157,924 J = 1/2 × 2000 kg × v²v² = (2 × 1,157,924 J) / 2000 kgv² = 1157.924v = √1157.924v = 34.04 m/s.

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Answer:

7.0 m/s

Explanation: I just did it

Answer:

x-component of velocity: 7.5 m/s

y-component of velocity: 13 m/s

Explanation:

This problem is pure trigonometry. Assuming you know trig, there are only a couple of steps to solving this problem. First, split the velocity into components; recall that any vector not directed along an axis has x and y components. Then, remember that sinΘ = opposite/hypotenuse. Applying this to your scenario, you get sin60° = vy/15. Multiplying this out gives you vy=15sin60. Put this into a calculator (make sure it's set to degree mode because the angle in this problem is in degrees) and you should get 12.99, which you can round up to 13 m/s. This is the velocity in the y-direction.

The procedure to find the x-velocity is very similar, but instead of using sine, we will use the cosine of theta. Recall that cosΘ=adjacent/hypotenuse. Once again plugging this scenario's numbers into that, you end up with cos60 = vₓ/15. Multiplying this out gives you vₓ = 15cos60. Once again, plug this into your calculator. 7.5 m/s should be your answer. This is the velocity in the x-direction.

By the way, a quick way to find the components of a vector, whether it's velocity, force, or whatever else, is to use these functions. Generally, if the vector points somewhere that's not along an axis, you can use this rule. The x-component of the vector is equal to hypotenuse*cosΘ and the y-component of the vector is equal to hypotenuse*sinΘ.

Answer:

5m

Explanation:

x=?

\(sin30^{0} =\frac{x}{10}\)

\(x=10sin30^{0} =10(0.5)=5\)

Hope this helps

Para sa aking Mahal na Magulang

Tula ni Eden Diao Apostol

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Pangalawa sa Diyos na s’yang lumikha

Utang ko sa inyo ang aking hininga

Minahal, hinubog ng inyong kalinga.

Mga sakripisyo’y sadyang hindi biro

Mula ng ako’y iniluwal sa mundo

Pag-ibig na iniukol sa ‘ki’y totoo

Pagmamahal ninyo’y nagsilbing lakas ko.

Ako’y tinuruan ng magandang asal

Sa gitna ng hirap ako’y pinag-aral

Upang ‘di mapariwara ang aking buhay

Diplomang natanggap sa inyo ini-alay.

Ngayon ang buhay ko ay sadyang kay-palad

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Sa aking pagmulat sa mundo ng kaligayahan

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Naging daan patungo sa kabutihan

Naging daan ng pag ibig sa tahanan

Kayo ang pundasyon, mabuting samahan

Maraming salamat sa pagmamahal.

Answer:

22/25 m/s^

Explanation:

The magnitude of the acceleration of the 20 kg box as it moves to the right is 7.84 m/s².

In this system, the force of gravity acts on both boxes, but the tension in the rope is the same on both sides of the pulley, so it cancels out. The frictional force acts only on the 20 kg box, opposing its motion to the right.

To find the acceleration of the 20 kg box, we need to use Newton's second law of motion, which states that the net force acting on an object is equal to its mass times its acceleration:

Net force = mass x acceleration

The net force on the 20 kg box is the force of gravity pulling it down minus the force of friction opposing its motion to the right:

net force = (20 kg)(9.8 m/s²) - (0.14)(20 kg)(9.8 m/s²) = 156.8 N

The net force on the 5 kg box is the force of gravity pulling it down plus the tension in the rope pulling it up:

Net force = (5 kg)(9.8 m/s²) + T

Since the tension is the same on both sides of the pulley, we can set these two equations equal to each other and solve for the tension:

T = (20 kg)(9.8 m/s²) - (0.14)(20 kg)(9.8 m/s²) - (5 kg)(9.8 m/s²) = 93.68 N

Now we can use the net force on the 20 kg box to find its acceleration:

Net force = (20 kg) x acceleration

156.8 N = (20 kg) x acceleration

Acceleration = 7.84 m/s²

Therefore, the magnitude of the acceleration of the 20 kg box as it moves to the right is 7.84 m/s².

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Answer:

they are quantities with magnitude without direction e.g weight,

Expansion of universe is not a hypothetical one but the universe is expanding in the sense increase distance over the space and dark matter.

The increase in distance over time between any two particular gravitationally unbound regions of the observable universe is known as the universe's expansion. 

As a result, the size of space itself expands intrinsically. It is not necessary for space to exist outside of the cosmos or for it to extend into anything.

The metric changes in scale rather than space or the things in space moving in the conventional sense. Objects move farther apart from one another at ever-increasing speeds as the spatial component of the universe's spacetime metric scales up.

Any viewer in the cosmos would see that all of space appears to be expanding, except for the galaxies receding with a speed proportional to the distance from the observer.

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The total equivalent resistance of the circuit which consists of 4 resistors ( 1202, 180, 320 and 3802 ) connected in series is 50.1 KΩ

Resistance of resistor,

Total equivalent resistance, R = R1 + R2 + R3 + R4

R = 12000 + 18 + 32 + 38000

R = 50.1 KΩ

Equivalent resistance is a another way of indicating total resistance. The equivalent resistance will be of a single resistor that replaces the total network without altering any effect on the system.

Therefore, the total equivalent resistance of a series circuit consists of 4 resistors connected in series: 1202, 180, 320, and a 3802 is 50.1 KΩ

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Approximate the instantaneous velocity by computing the average velocity of the car over the interval [1.95, 2.05] (i.e. the interval centered at t = 2 s with length 0.1 s).

By definition, average velocity is given by

v = ∆x / ∆t

So we have

v(2 s) ≈ (x(2.05 s) - x(1.95 s)) / (0.1 s) ≈ 13.94 m/s

The polynomial relationship between position and time for the bowling ball is linear.

A linear relationship between two variables is a term used to describe a straight-line relationship between the two variables.

Linear relationships can be expressed either in a graphical format or as a mathematical equation of the form y = mx + b.

From the equation of linear relationship between two variable, the highest power of x is one.

The given equation for position and time;

x(t) = vot + xo

From this given equation, the highest power of t is one, hence it is called linear relationship.

Thus, the polynomial relationship between position and time for the bowling ball is linear.

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The acceleration due to gravity near the surface of the hypothetical planet is 3.02 m/s².

The formula for acceleration due to gravity is:

g = GM/r² Where, g = acceleration due to gravity G = universal gravitational constant M = mass of the planet r = radius of the planet

In this case, since the mass of the hypothetical planet is the same as that of Earth, we can use the mass of Earth instead of M.

Therefore, g is proportional to 1/r².

So, using the ratio of radii given (1.8), we can write:

r = 1.8 x r Earth, where r Earth is the radius of Earth.

Substituting this value of r in the formula for acceleration due to gravity, we get:

g = GM/(1.8 x r Earth)² = GM/(3.24 x rEarth²) = (1/3.24)GM/rEarth²

We know that the acceleration due to gravity on Earth (g Earth) is 9.8 m/s².

Therefore, we can calculate the acceleration due to gravity on the hypothetical planet (gh) as follows:

gh = (1/3.24) x g Earth = 3.02 m/s²

Thus, the acceleration due to gravity near the surface of the hypothetical planet is 3.02 m/s².

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Taking into account the definition of wavelength, frecuency and propagation speed, the frequency of the waves in the tsunami is 2.59×10⁻⁴ Hz.

First of all, wavelength is the minimum distance between two successive points on the wave that are in the same state of vibration. It is expressed in units of length (m).

On the other side, frequency is the number of vibrations that occur in a unit of time. Its unit is s⁻¹ or hertz (Hz).

Finally, the propagation speed is the speed with which the wave propagates in the medium, that is, it is the magnitude that measures the speed at which the wave disturbance propagates along its displacement.

The propagation speed relate the wavelength (λ) and the frequency (f) inversely proportional using the following equation:

v = f×λ

In this case, you know:

Replacing:

0.194 m/s=f× 750 km

Solving:

f= 0.194 m/s ÷ 750 km

λ= 2.59×10⁻⁴ Hz

In summary, the frequency of the waves in the tsunami is 2.59×10⁻⁴ Hz.

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