A car starting from rest is accelerated at a constant rate of 3.0 meters per second^2. What
is its speed at the end of 8.0 seconds?

Gravitational force between two masses m, and m, is represented as F = Gm₂ m₂ / r^2 where r = xi+yj + zkG, m, m₂ are nonzero constants and let's assume that I = 0

a) Calculation:For F = Gm₂ m₂ / r^2.

Using r = xi+yj + zk and let r^2 = x^2 + y^2 + z^2∴ F = Gm₂ m₂ / (x^2 + y^2 + z^2), Where G, m, m₂ are nonzero constants. Divergence of F = ∇ · F= 1/r^2(d/dx(r^2Fx) + d/dy(r^2Fy) + d/dz(r^2Fz))= 1/r^2(d/dx(r^2Gm₂ m₂ x/(x^2+y^2+z^2)^(3/2)) + d/dy(r^2Gm₂ m₂ y/(x^2+y^2+z^2)^(3/2)) + d/dz(r^2Gm₂ m₂ z/(x^2+y^2+z^2)^(3/2)))= 1/r^2(d/dx(r^2Gm₂ m₂ x/(x^2+y^2+z^2)) * (x^2+y^2+z^2)^(3/2) + d/dy(r^2Gm₂ m₂ y/(x^2+y^2+z^2)) * (x^2+y^2+z^2)^(3/2) + d/dz(r^2Gm₂ m₂ z/(x^2+y^2+z^2)) * (x^2+y^2+z^2)^(3/2))= 1/r^2(Gm₂ m₂ [2x(x^2+y^2+z^2)-3x^2]/(x^2+y^2+z^2)^(5/2) + Gm₂ m₂ [2y(x^2+y^2+z^2)-3y^2]/(x^2+y^2+z^2)^(5/2) + Gm₂ m₂ [2z(x^2+y^2+z^2)-3z^2]/(x^2+y^2+z^2)^(5/2))= 1/r^2(Gm₂ m₂ [(2x^2+2y^2+2z^2-3x^2)/(x^2+y^2+z^2)^(3/2)] + [2x^2+2y^2+2z^2-3y^2]/(x^2+y^2+z^2)^(3/2)] + [2x^2+2y^2+2z^2-3z^2]/(x^2+y^2+z^2)^(3/2)])= 1/r^2(Gm₂ m₂ [x^2+y^2+z^2]/(x^2+y^2+z^2)^(3/2))= 0.

Curl of F = ∇ × F= i(d/dy(Fz) - d/dz(Fy)) - j(d/dx(Fz) - d/dz(Fx)) + k(d/dx(Fy) - d/dy(Fx))= i(d/dy(Gm₂ m₂ z/(x^2+y^2+z^2)) - d/dz(Gm₂ m₂ y/(x^2+y^2+z^2))) - j(d/dx(Gm₂ m₂ z/(x^2+y^2+z^2)) - d/dz(Gm₂ m₂ x/(x^2+y^2+z^2))) + k(d/dx(Gm₂ m₂ y/(x^2+y^2+z^2)) - d/dy(Gm₂ m₂ x/(x^2+y^2+z^2)))= i(Gm₂ m₂ [-2xz]/(x^2+y^2+z^2)^(5/2)) - j(Gm₂ m₂ [-2yz]/(x^2+y^2+z^2)^(5/2)) + k(Gm₂ m₂ [(x^2+y^2-2z^2)]/(x^2+y^2+z^2)^(5/2))

b) Calculation:The line integral of F along a curve C can be evaluated by the following formula∫C F.dr = ∫∫ ( ∇ x F) ds, Where r is the position vector of the curve, s is the scalar parameter representing the curve, and the integral is evaluated from the initial point to the final point.

Using the curl of F obtained in part a) and for the surface with ∂S as C∫C F.dr = ∫∫ ( ∇ x F) ds= ∫∫ curl(F) ds= ∫∫ (-2xz i -2yz j + (x^2+y^2-2z^2)k) ds...[1]

Let's consider the surface S as a plane perpendicular to the z-axis of the form ax+by+c=0 and the curve C as the intersection of the plane and the cylinder x^2 + y^2 = a^2.

Let's choose the unit normal to the surface S as k (along the z-axis).

The curl of F is a vector field perpendicular to the plane and along the direction of k.

Thus the integral can be written as∫C F.dr = ∫∫ ( ∇ x F) . k ds= ∫∫ (x^2+y^2-2z^2) ds...[2]

Now let's evaluate the integral over the given plane ax+by+c=0. We can write x = t, y = (c-at)/b and z = 0, where t is the scalar parameter along the line of intersection of the plane and the cylinder (x^2 + y^2 = a^2).

Since the curve C is on the cylinder of radius a, we have x^2+y^2 = a^2 ⇒ t^2+(c-at)^2/b^2 = a^2On solving for t, we have t = (bc±ab √(a^2-b^2-c^2))/[a^2+b^2].

Substituting t in x and y, we get the curve C in the x-y plane as a function of the scalar parameter s asx = (bc±ab √(a^2-b^2-c^2))/[a^2+b^2]y = (c-at)/b= (c-(bc±ab √(a^2-b^2-c^2))/[a^2+b^2])/b.

Now we can evaluate the integral over the curve C, which is along the intersection of the plane and the cylinder.

Integral over C (x^2+y^2-2z^2) ds= ∫t₁^t₂ [(t^2 + [(c-at)^2]/b^2 - 2(0)^2)^(1/2)] dt= ∫t₁^t₂ [(a^2-b^2-c^2)t^2+2bc(c-at)+b^2c^2-a^2b^2]^(1/2) dt.

Now we can choose the value of t₁ and t₂ such that the square root in the integrand is minimized (so that the integral is path-independent).

This can be done by choosing the value of t that gives the minimum value of (a^2-b^2-c^2)t^2+2bc(c-at)+b^2c^2-a^2b^2 over the range of t from t₁ to t₂.

On differentiation with respect to t and equating to 0, we get the value of t = bc/(a^2+b^2).

Substituting this value of t in the integrand, we get the minimum value of the square root in the integrand to be |c| √(a^2+b^2)/|b|.

Thus the integral over C is given by∫C F.dr = ∫∫ (-2xz i -2yz j + (x^2+y^2-2z^2)k) ds= ∫∫ (x^2+y^2-2z^2) ds= ∫t₁^t₂ |c| √(a^2+b^2)/|b| dt= |c| √(a^2+b^2)/|b| (t₂-t₁).

Now we can see that the integral is path-independent as it depends only on the end points t₁ and t₂ and not on the path taken to reach them.

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The answer is true. Parentheses are used in Boolean expressions to eliminate confusion as to which operation is to be performed first.

The answer is true. Parentheses are used in Boolean expressions to eliminate confusion and clarify the order of operations, ensuring that the desired operation is performed first. They help in maintaining the correct precedence and associativity of operations in the expression.

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Answer:

Compound

Explanation:

The distance should an object be placed from a convex lens of focal length 18cm to obtain an image at 24cm if on the other side is 72 cm and the magnification produced in this case is 1/3.

The formula to calculate the magnification is: magnification (m) = - (v/u). In order to calculate the distance of the object from the lens, we can use the lens formula:1/f = 1/v - 1/u, where f is the focal length, v is the distance of the image from the lens and u is the distance of the object from the lens. Substituting the given values in the formula we get:1/18 = 1/-24 - 1/u

On solving this equation we get u = -72 cm. (Negative sign indicates that the object is placed at the left-hand side of the lens).

Now substituting the values of v and u in the formula of magnification, we get: magnification (m) = - (v/u) = - (-24/-72) = 1/3

The object should be placed at a distance of 72 cm on the left-hand side of the lens to get an image at 24 cm on the other side. The magnification produced in this case is 1/3.

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The duration and frequency of eclipses on an Earth-like planet depend on its orbit and the orbit of its moon(s).

However, on average, a total solar eclipse could last for a few minutes to a few hours, and the time between eclipses could be a few months to a few years. Obstacles for the ground-based detection of the Earth-like planets include atmospheric interference, limited resolution, and the brightness of the host star relative to the planet. Additionally, Earth-like planets are often located far away and are small compared to their host stars, making them challenging to detect using the current technology.

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Answer:

true

Explanation:

they can cause skin damage

Oscillation is an extremely significant concept in various applications, particularly in electronics and electrical engineering. An oscillation can be defined as the recurrent movement of an object around an equilibrium point, such that it continues to return to the equilibrium point despite being pushed away from it.

The concept of oscillation can be understood by visualizing a pendulum attached to a clock or by considering a spring's behavior. The electrical energy that flows back and forth between the inductor and the capacitor in an LC circuit is referred to as an oscillation.

The frequency of oscillation is the number of oscillations per unit time and is expressed in Hertz. Oscillations that occur at a frequency of more than 20 kHz are referred to as high-frequency oscillations. The sinusoidal waveform is often used to represent oscillations, and it may be plotted on an x-y chart to demonstrate how the wave changes over time. The voltage produced in an electrical circuit when it oscillates back and forth is referred to as an oscillating voltage.

Circuits that oscillate are known as oscillator circuits, and they are used in a variety of applications, including radio and television broadcasting, radar systems, and digital clocks. To summarize, the concept of oscillation is crucial in electronic and electrical applications, and its understanding is essential for the development of advanced electronic systems.

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Given data

*The initial quantity of potassium-42 is 1500 mg

*The given time is t = 48 hours

*The half-life of Potassium - 42 is T = 12.4 hours

The expression for the radioactive decay is given as

Substitute the values in the above expression as

Initial velocity of projection (u) = 20 m/s

Angle of projection (θ) = 30°

Formula of Horizontal Range of Projectile:

\( \boxed{ \bf{R = \dfrac{2u^2 sin\theta .cos \theta}{g}}}\)

By substituting values in the formula, we get:

\( \rm \longrightarrow R = \dfrac{2 \times 20^2 \times sin30 \degree \times cos 30 \degree}{10} \\ \\ \rm \longrightarrow R = \dfrac{ \cancel{2} \times 40 \cancel{0} \times \dfrac{ \sqrt{3} }{ 2} \times \dfrac{1}{\cancel{2}} }{ \cancel{10}} \\ \\ \rm \longrightarrow R =40 \times \dfrac{ \sqrt{3} }{ 2} \\ \\ \rm \longrightarrow R =20 \sqrt{3} \: m \\ \\ \rm \longrightarrow R =34.641 \: m\)

\( \therefore \) Horizontal Range of Projectile (R) = 34.641 m

Answer:

a)  ω = 8.75 rad/s

b) I = 12.9 kgm^2

Explanation:

Given data:

Radius of the wheel r = 0.400 m

Speed of the suitcase v = 3.50 m/s

Height h = 4.00 m

Mass of the suitcase m = 15 kg

a) To determine angular velocity of the wheel when the suitcase reaches the ground is

we know that ,         ω = v / r

ω = (3.50 m/s) / (0.400 m )

    ω = 8.75 rad/s

b) From the conservation of energy

  we have         KEi + Ui = KEf + Uf

where KEi , KEf are initial and final kinetic energies and Ui and Uf are initial and final potentail energies.  

We know that KEi = 0 and Uf   = 0

             Ui = KEf

           mgh = (1/2)mv^2 + (1/2) Iω^2

I = moment of inertia of wheel to determined

15×9.8×4 = (1/2) (15)(3.5)^2 + (1/2) I (8.75)^2

Calculating we get                  

I = 12.9 kgm^2

A force that acts on a body moving in a circular path and is directed toward the center around which the body is moving.

According to the question, The formula is used to find the omega is:-

\(w =\frac{v}{r}\)

\(w= \frac{3.50}{0.400}\)

Hence, the omega is ω = 8.75 rad/s

From the conservation of energy  we have  KEi + Ui = KEf + Uf

where Kei, KEf are initial and final kinetic energies and Ui and Uf are initial and final potential energies.  

We know that KEi = 0 and Uf   = 0

So, therefore the Ui = KEf

Hence \(mgh = \frac{1}{2}mv^{2} + \frac{1}{2} Iw^{2}\)

After putting the value:-

\(15*9.8*4 = \frac{1}{2}*15*3.5^{2} + \frac{1}{2}*I*8.75^{2}\)

Hence the value of I is\(12.9 kgm^2\)

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The equivalent resistance of the circuit is 5.106 Ω. This can be calculated by first finding the equivalent resistance of the parallel branches, and then finding the equivalent resistance of the series branches.

The first step is to find the equivalent resistance of the parallel branches. The two 60 Ω resistors are in parallel, so their equivalent resistance is:

```

R1_eq = (1/60 Ω) + (1/60 Ω) = 30 Ω

```

The two 40 Ω resistors are also in parallel, so their equivalent resistance is:

```

R2_eq = (1/40 Ω) + (1/40 Ω) = 20 Ω

```

The 10 Ω resistor is in series with the 30 Ω and 20 Ω resistors, so the equivalent resistance of this series combination is:

```

R_series = 10 Ω + 30 Ω + 20 Ω = 60 Ω

```

Finally, the 60 Ω resistor is in parallel with the 40 Ω resistor, so the equivalent resistance of the entire circuit is:

```

R_eq = (1/60 Ω) + (1/40 Ω) = 5.106 Ω

```

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Answer:

1) The point where the car has the greatest potential energy is Point A

2) The point where the car has the least potential energy is Point C

3) The point where the car has the greatest kinetic energy is Point C

4) The point where the car has the greatest kinetic energy is Point A

Explanation:

The potential energy of the car is given by the formula, PE = Mass × Gravity × Height

Therefore, as the height increases, the potential energy of the car increases

The total mechanical energy of the car, which is constant, is given by the following equation

The total mechanical energy = The Potential Energy + The Kinetic Energy

Therefore, at the point where the potential energy is lowest, the kinetic energy is highest.

This cart would move with constant velocity or be at rest because the net force is equal to zero (0), which means the three (3) forces acting on the cart opposes each other.

Given the following data:

To find the net force acting on the cart:

A net force can be defined as the vector sum of all the forces that are acting on a physical object or body.

This ultimately implies that, a net force is a single (one) force that substitutes the effect of all the forces acting on a physical object or body.

Mathematically, net force is given by the following:

\(Net \;force = F_1 + F_2 + F_3 + F_4 + .....F_n\)

Where:

\(Net\; force = 60 + 30 + (-90)\\\\Net\; force = 90 - 90\)

Net force = 0 Newton

Based on Newton's First Law of Motion, this cart would move with constant velocity or be at rest because the net force is equal to zero (0), which means the three (3) forces acting on the cart opposes each other.

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The pressure of the gas is 2.25 atm.

We can use the ideal gas law to calculate the pressure of the gas. The ideal gas law states that PV=nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant and T is the temperature.

In this problem, we are given the volume (V=1.80 L), the molar mass of the gas (m=0.226 g), and the temperature (T=192 m/s). We can use this information to calculate the number of moles of gas present in the container.

First, we need to convert the molar mass of the gas from grams to moles. We can do this using the equation m=nM, where m is the mass of the gas, n is the number of moles and M is the molar mass.

We can solve for n to get n=m/M. Plugging in the values for m and M, we get n=0.226 g/28.02 g/mol = 0.008 mol.

Now that we have the number of moles, we can use the ideal gas law to calculate the pressure of the gas. Plugging in the values for n, V, and T, we get

P=nRT/V = 0.008 mol * 8.314 L·atm/mol·K * 192 m/s/K/1.80 L = 2.25 atm.

Therefore, the pressure of the gas is 2.25 atm.

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Answer:

1, Coal,oil,natural gas,petroleum ...etc

2, -they pollute the environment

-In case of irresponsible use ,they can be

dangerous

-it is more reliable an renewable

To compute the Bose-Einstein condensation temperature for a gas of two-level bosons, we can use the formula derived from the Bose-Einstein distribution:

n = [1 / (exp((E - μ) / (k * T)) - 1)]

Where:

n is the number of particles in the excited state,

E is the energy of the excited state,

μ is the chemical potential,

k is the Boltzmann constant,

T is the temperature.

In this case, we have a two-level system, with the ground state having energy Eo and the excited state having energy Ep, where Δ = Ep - Eo is the excitation energy.

The Bose-Einstein condensation occurs when a macroscopic number of particles occupy the lowest energy state. At the condensation temperature, the chemical potential μ approaches the energy of the ground state Eo.

Therefore, to find the condensation temperature, we need to solve for T when the number of particles in the excited state n becomes negligible compared to the number of particles in the ground state.

In mathematical terms, we set n = 0 and solve for T:

0 = [1 / (exp((Ep - μ) / (k * T)) - 1)]

Since exp((Ep - μ) / (k * T)) approaches 1 when T is much greater than the excitation energy Δ, we can simplify the equation to:

0 ≈ 1 / (1 - 1) = 1 / 0

This implies that the condensation temperature is at T = ∞, meaning that the Bose-Einstein condensation occurs at extremely low temperatures, close to absolute zero.

The existence of the internal degree of freedom does not significantly affect the condensation temperature. The condensation temperature is primarily determined by the number of particles and their energy states rather than the specific internal degrees of freedom.

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Momentum is doubled when the available kinetic energy is quadrupled.

This is because momentum is directly proportional to the square root of kinetic energy, so if the kinetic energy is increased by a factor of 4, the momentum will be doubled.

The mathematical relationship between momentum and kinetic energy.

The equation for kinetic energy is

\(KE = \frac{1}{2} mv^2\), where m is mass and v is velocity.

The equation for momentum is \(p = mv\), where p is momentum.

If we assume the mass to be constant, we can rewrite the equation for kinetic energy as

\(KE = \frac{1}{2} \frac{p^2}{m}\).

We can then rearrange this equation to solve for p: \(p = \sqrt{2mKE}\).

From this equation, we can see that momentum is directly proportional to the square root of kinetic energy.

If the kinetic energy is increased by a factor of 4, the square root of kinetic energy will be doubled, and therefore the momentum will also be doubled.

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The soil used for making pots by potters is commonly known as "Potter's Clay" or "Potter's Soil."

Characteristics of Potter's Clay:

1. Plasticity: Potter's clay has high plasticity, which means it can be easily molded and shaped into different forms without cracking or breaking.

2. Cohesiveness: It exhibits good cohesion, allowing the clay particles to stick together when moistened, forming a workable material.

3. Fine Particle Size: Potter's clay consists of fine particles that contribute to its plasticity and workability.

4. High Water Retention: It has the ability to retain water, which helps maintain the moisture content necessary for the clay to be shaped and formed.

5. Low Shrinkage: Potter's clay exhibits low shrinkage when it dries or undergoes firing, ensuring minimal distortion or cracking during the drying and firing processes.

6. Good Bonding Properties: The clay particles have the ability to bond together, creating a strong and durable structure once fired.

7. Good Porosity: After firing, the clay retains some porosity, allowing the pot to breathe and allowing for the exchange of air and moisture.

According to Einstein's Theory of Relativity, the closer an object gets to the speed of light, the more its mass increases. As an object approaches the speed of light, its mass would become infinitely large, requiring an infinite amount of energy to propel it any further. Therefore, an object with a nonzero mass can never reach the speed of light as it would require an infinite amount of energy.

Additionally, as the object gets closer to the speed of light, time slows down for it relative to a stationary observer, and it would experience length contraction. Thus, the laws of physics prevent an object with a nonzero mass from exceeding the speed of light.
An object with nonzero mass cannot travel as fast as the speed of light due to the principles of Special Relativity, formulated by Albert Einstein. As an object with mass accelerates, its energy increases. When approaching the speed of light, the object's mass experiences relativistic effects, such as time dilation and length contraction.

As a result, the energy required to continue accelerating increases exponentially. To reach the speed of light, an infinite amount of energy would be needed, which is physically impossible. Moreover, the mass of the object would become infinite as well, causing an insurmountable barrier to reaching the speed of light. Therefore, an object with nonzero mass can never achieve this ultimate speed.

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The total resistance of the circuit is approximately 21.11 Ω.  To determine the total resistance (R) of the series RLC circuit, we can use the relationship between the current (I), voltage (V), and resistance.

To determine the total resistance (R) of the series RLC circuit, we can use the relationship between the current (I), voltage (V), and resistance:

V = I * R

Given:

Irms = 9.00 A (root mean square value of current)

δvrms = 190 V (root mean square value of voltage)

The current leads the voltage by 38.0°

First, we need to find the peak values of current (I_peak) and voltage (V_peak). Since the root mean square (rms) values are given, we can use the following relationship:

I_peak = Irms

V_peak = δvrms

Now, we can calculate the total resistance (R) using the peak values:

R = V_peak / I_peak

R = 190 V / 9.00 A

R ≈ 21.11 Ω

Therefore, the total resistance of the circuit is approximately 21.11 Ω.

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The average speed for the round trip is 90.51 km/hr when a car travels up a hill at a constant speed of 33 km/h and returns down the hill at a constant speed of 72 km/h.

To find the average speed for the round trip, the given values are,

constant speed v₁ = 33 km/hr

constant speed v₂ = 72 km/hr

         Average speed = total distance traveled / total time taken

Let d is the total distance traveled by the car.

Thus , distance traveled in each journey is d/2.

So, the time taken for going up the hill  is t₁ = (d/2) / v₁

= (d/2) / 33

= d / 66

Next, the time taken for going down the hill is t₂ = ( d/2 ) / v₂

= ( d/2) / 72

= d / 144

Thus, average speed is v avg = 2d / ( t₁ + t₂ )

= 2d / ( ( d/66)+(d/144))

= 2d / (210d / 9504)

= 2d / 0.0220959

= 90.51 km/hr.

The average speed of the round trip is 90.51 km/hr.

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The Nyquist frequency for a sampling frequency of 1000 Hz is 500 Hz (half the sampling frequency). Any signal above this frequency will be aliased.

The highest frequency played by Dr. Lee's son is 523.2 Hz, which is less than the Nyquist frequency, so it will not be aliased. However, the other frequencies will be aliased as follows:

261.6 Hz will be aliased as 738.4 Hz (1000 Hz - 261.6 Hz)

293.7 Hz will be aliased as 706.3 Hz (1000 Hz - 293.7 Hz)

329.6 Hz will be aliased as 670.4 Hz (1000 Hz - 329.6 Hz)

349.2 Hz will be aliased as 650.8 Hz (1000 Hz - 349.2 Hz)

392.0 Hz will be aliased as 608.0 Hz (1000 Hz - 392.0 Hz)

440.0 Hz will be aliased as 560.0 Hz (1000 Hz - 440.0 Hz)

493.9 Hz will be aliased as 506.1 Hz (1000 Hz - 493.9 Hz)

Therefore, the frequencies that will be present in the reconstructed signal (different from the original frequencies) are 506.1 Hz, 560.0 Hz, 608.0 Hz, 650.8 Hz, 670.4 Hz, 706.3 Hz, and 738.4 Hz.

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The water outflow in 30 secs through 200 mm of the capillary tube is mathematically given as

\(Qo=1.6 \times 10^{2} \mathrm{~mL}\)

\(\begin{aligned}\Delta P &=6660 \mathrm{~m} / \mathrm{m}^{2} \\\mu &=8.01 \times 10^{-4} \text { Pas } \\t &=30 \mathrm{~s} \\L &=200 \mathrm{~mm}=200 \times 10^{-3} \mathrm{~m} \\D &=1.5 \mathrm{~mm}=1.5 \times 10^{-3} \mathrm{~m} \Rightarrow \gamma=\frac{1.5 \times 10^{-3}}{2} \mathrm{~m}\end{aligned}\)

Generally, the equation for Rate of flow of Liquid is  mathematically given as

\(\\$$Q=\frac{\pi r^{4} \times \Delta P}{8 \mu L}\)

$$

Where dP is pressure difference r is the radius

\(\mu\) is the viscosity of water

L is the length of the pipe

\(Q=\frac{\pi \times\left(\frac{1.5 \times 10^{-3}}{2}\right)^{4} \times 6660}{8 \times 8.01 \times 10^{-4} \times 200 \times 10^{-3}}\)

\(Q=5.2 \mathrm{~mL} / \mathrm{s}\)

In $30s the quantity that flows out of the tube

\(&Qo=5.2 \times 30 \\&Qo=1.6 \times 10^{2} \mathrm{~mL}\)

In conclusion, the quantity that flows out of the tube

\(Qo=1.6 \times 10^{2} \mathrm{~mL}\)

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Assuming g=10N

upward force=600N(applied force)

downward force=(W)=(5x10)=50N

net force=600-50

=550N

F=(delta)P/t

(delta)P=m (v-u)

u=0m/s

(delta)P=mv

550=(5)(v)/0.1

550x0.1

55=5v

v=11m/s

there was no mention of any gravity acting on the object only applied force ,infact we aren't even given the constant

so many assumptions have been made whyyyyyyy!!!

what's your take on it scientists

THE ANSWER BELOW THE PICTURES.

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Answer:

C.the coefficient of friction

Explanation:

Hope this helps. can i have brainliest thanks!

Answer:

I think D

Explanation:

Answer:

The density is 1.1gcm3

Answer:

On Waves. Quantum mechanically, an electron can be described by a wave function oscillating in space and time that has mean values equal to the expectation values of observables corresponding to given operators. ... Electrons are fermions. They are charged particles. in short the quantum theory

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