In ideal flow, a liquid of density 850 kg/m3 moves from a horizontal tube of radius 1.00 cm into a second horizontal tube of radius 0.500 cm at the same elevation as the first tube. The pressure differs by DP between the liquid in one tube and the liquid in the second tube. (a) Find the volume flow rate as a function of DP. Evaluate the volume flow rate for (b) DP 5 6.00 kPa and (c) DP 5 12.0 kPa. 49. The Venturi tube discussed

Answer:

a)V = 25.1 m/s

b)V = 4.226 m/s

Explanation:

Given that

x(t)=A t + B t​²

A = -4.3 m/s

B = 4.9 m/s​²

x(t)=  - 4.3 t +4.9 t​²

The velocity of the particle is given as

[tex]V=\dfrac{dx}{dt}[/tex]

V=-4.3 + 4.9 x 2 t

V= - 4.3 + 9.8  t m/s

Velocity at point t= 3 s

V= - 4. 3 + 9.8 x 3 m/s

V= - 4.3 + 29 .4 m/s

V = 25.1 m/s

At origin :

x= 0 m

0 =  - 4.3 t +4.9 t​²

0 = - 4.3 + 4.9 t

[tex]t=\dfrac{4.3}{4.9}\ s[/tex]

t=0.87 s

The velocity at t= 0.87 s

V= - 4.3 + 9.8  t m/s

V= - 4. 3 + 9.8 x 0.87 m/s

V= - 4.3 + 8.526 m/s

V = 4.226 m/s

a)V = 25.1 m/s

b)V = 4.226 m/s

The velocity of the particle at t = 3.0s is 25.1 m/s.

The velocity of the particle when it is at the origin is 4.324 m/s.

The given parameters:

The velocity of the particle at t = 3.0s is calculated as follows;

[tex]v = \frac{dx}{dt} = A + 2Bt\\\\v = -4.3 + 2(4.9\times 3)\\\\v = 25.1 \ m/s[/tex]

The velocity of the particle when it is at the origin (x = 0)

[tex]0 = -4.3t + 4.9t^2\\\\0 = t(-4.3 + 4.9t)\\\\t = 0 \ \ \ or \ \ -4.3 + 4.9t = 0\\\\4.9t = 4.3\\\\t = \frac{4.3}{4.9} \\\\t = 0.88 \ s\\\\v = A + 2Bt\\\\ v = -4.3 + (2\times 4.9 \times 0.88)\\\\v = 4.324 \ m/s[/tex]

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Answer:

Heat required to raise the temperature will be 563.625 J

Explanation:

We have given number of moles n = 2.70 mole

Temperature is raises from 22°C to 32°C

So increase in temperature [tex]\Delta T=32-22=10^{\circ}C[/tex]

It is given that air is diatomic so [tex]c_v=\frac{5}{2}R=2.5\times 8.314=20.875[/tex]

We know that heat is given by [tex]Q=nc_v\Delta T[/tex]

So heat will be equal to [tex]Q=2.70\times 20.875\times 10=563.625J[/tex]

So heat required to raise the temperature will be 563.625 J

Explanation:

Given that,

Charge acting on the object, [tex]q=-4\ nC=-4\times 10^{-9}\ C[/tex]

Force acting on the object, [tex]F=19\ nC=19\times 10^{-9}\ C[/tex] (in downward direction)

(a) The electric force acting in the electric field is given by :

[tex]F=qE[/tex]

E is the electric field

[tex]E=\dfrac{F}{q}[/tex]

[tex]E=\dfrac{19\times 10^{-9}\ N}{4\times 10^{-9}\ C}[/tex]

E = 4.75 N/C

The direction of electric field is as same as electric force. But it is negative charge. So, the direction of electric field is in upward direction.

(b) The charge on the proton is, [tex]q=1.6\times 10^{-19}\ C[/tex]

The force acting on the proton is :

[tex]F=qE[/tex]

[tex]F=1.6\times 10^{-19}\times 4.75[/tex]

[tex]F=7.6\times 10^{-19}\ N[/tex]

If the charge on the proton is positive, the force on the proton is in upward direction.

Hence, this is the required solution.

Use the relation between electric force and electric field and the concept of gravitational force to calculcate the electric field.  The electric force is given by

[tex]F_e = qE[/tex]

And the gravitational force is

[tex]F_g = mg[/tex]

For the ball to float above the ground, the magnitude of electric force on the ball must be equal to the magnitude of the gravitational force. That is must be a equilibrium condition, so,

[tex]F_e = F_g[/tex]

[tex]qE = mg[/tex]

[tex]E = \frac{mg}{q}[/tex]

Replacing the values we have that,

[tex]E = \frac{(0.010)(9.8)}{-24*10^{-6}}[/tex]

[tex]E = -4.083*10^{3} N/C[/tex]

Therefore the electric field magnitude that would cause the ball to float above the ground is [tex]-4.083*10^{3} N/C[/tex]

Answer:

a) 8.8 sec b) -86 m/sec

Explanation:

Assuming  King Kong started from rest its fall, once in the air, neglecting air resistance, is only affected by gravity, which accelerates it downward.

As this acceleration is constant, we can use the following equation in order to get how long it was falling:

Δh = 1/2*g*t² ⇒ -380 m = 1/2 (-9.8 m/s²)*t²

Solving for t:

t = √((2*380)/9.8) s² = 8.8 sec.

b) In order to know the value of the velocity in the instant just before it hits the ground, we can apply acceleration definition, as follows:

a = (vf-v₀) /t

In our case, a = -g (assuming positive direction is upward) and v₀=0, so, we can solve for vf as follows:

vf = -g*t = -9.8 m/s²*8.8 sec = -86 m/s.

The answer explains the estimated time for King Kong to fall from the Empire State Building and his velocity just before landing using physics formulas.

a) Estimate: Using the formula h = 0.5 * g * t^2 where h is the height, g is gravity (9.8 m/s^2), we find t = sqrt(2h/g) = sqrt(2*380/9.8) = 8.7 s. Therefore, King Kong took approximately 8.7 seconds to fall.

b) Velocity: To find the velocity just before landing, we use v = g*t where v is the final velocity. Substituting values, v = 9.8 * 8.7 = 85.3 m/s. Therefore, his velocity just before 'landing' was around 85 m/s.

Answer:

Albedo Alto: Snow

Albedo bass: ocean

Explanation:

Albedo is the fraction of solar radiation reflected by a surface. The term has its origin from the Latin word albus, which means "white." It is quantified as the proportion or percentage of solar radiation of all wavelengths reflected by a body or surface to the amount incident on it.

The color of the soil certainly affects the reflectivity, the lighter colors that have greater albedo than the dark colors and, therefore, present greater albedo. Soil texture is also a factor that affects albedo. Some studies have shown that snow-covered soils, being a light color, reflect most of the light and therefore do not absorb it and present the greatest amount of terrestrial surfaces. While with the ocean and seas most of the radiation is absorbed it has a dark color increasing with the depth of the place and therefore has the lowest surface albedo. In the end of the year the snow has 86% of reflected light while the oceans 8% of reflected light.

Answer:

The range is 1881.8 m.

Explanation:

Given that,

Time [tex]t=15\times10^{-6}\sec[/tex]

Frequency range [tex]\Delta f= f_{f}-f_{i}[/tex]

[tex]\Delta f= 429-383[/tex]

[tex]\Delta f=46\ MHz[/tex]

The value of [tex]\dfrac{df}{dt}[/tex]

[tex]\dfrac{df}{dt}=\dfrac{438-383}{15\times10^{-6}}[/tex]

We need to calculate the range

Using formula of range

[tex]R=\dfrac{c\Delta f}{2\times\dfrac{df}{dt}}[/tex]

Put the value into the formula

[tex]R=\dfrac{3\times10^{8}\times46}{2\times\dfrac{438-383}{15\times10^{-6}}}[/tex]

[tex]R=1881.8\ m[/tex]

Hence, The range is 1881.8 m.

Answer:

The expression of gravitational field due to mass [tex]m_![/tex] at a distance [tex]l_a[/tex]

Explanation:

We have given mass is [tex]m_1[/tex]

Distance of the point where we have to find the gravitational field is [tex]l_a[/tex]

Gravitational constant G

We have to find the gravitational filed

Gravitational field is given by [tex]g=\frac{Gm_1}{l_a^2}[/tex]

This will be the expression of gravitational field due to mass [tex]m_![/tex] at a distance [tex]l_a[/tex]

The expression for the gravitational field g₁ at position la in terms of m₁, la, and the gravitational constant G is:

[tex]g_1 = \frac{G\times m_1}{la^{2} }[/tex]

The gravitational field is the force field that exists in the space around every mass or group of masses.

The gravitational field of m₁ is denoted by g₁, and can be represented through the following expression.

[tex]g_1 = \frac{F_1}{m}[/tex]    [1]

where,

We can calculate the force (F₁) between m₁ and m that are at a distance "la" using Newton's law of universal gravitation.

[tex]F_1 = G \frac{m \times m_1 }{la^{2} }[/tex]   [2]

where,

If we replace [2] in [1], we get

[tex]g_1 = \frac{G \frac{m \times m_1 }{la^{2} }}{m} = \frac{G\times m_1}{la^{2} }[/tex]

The expression for the gravitational field g₁ at position la in terms of m₁, la, and the gravitational constant G is:

[tex]g_1 = \frac{G\times m_1}{la^{2} }[/tex]

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Answer:

D

Explanation:

The smallest possible allowable length that can be used:

Calculate area of contact between plates A

A = 2 * 6 * a

A = 12a in^2

Where a is an arbitrary constant length along the direction L.

Shear stress = Shear force / Area

Area = 15000 lb / 120 psi = 12*a

Evaluate a

a = 10.411 in

Total Length = 2*a + gap = 2*10.411 + 0.5 = 21.33 in

Answer: L = 21.33

To find the smallest allowable length for the splice plates, the required shear area was calculated using the provided load and the shear strength of the glue, then divided by the effective board width. The resulting length per plate was approximately 11.35 inches, making the smallest length that can safely be used 11.6 inches.

To determine the smallest allowable length L that can be used for the splice plates, we must first calculate the shear area required to resist the applied load P of 15,000 lb with a glue shear strength of 120 psi.

The total shear force that the splice plates must resist is equal to the applied load P. The shear strength per unit area provided by the glue is 120 psi. Therefore, the required shear area A can be found using the formula:

A = P / Shear Strength

A = 15,000 lb / 120 psi = 125 square inches

Each board has a width of 6 inches, but we account for a 0.5-inch gap between the boards, so the effective width w for the splice is (6 - 0.5) inches = 5.5 inches. Then, we can calculate the length L needed for each splice plate as follows:

L = A / w

L = 125 square inches / 5.5 inches ≈ 22.7 inches

But since the glue is applied to two plates, we divide this length by 2 to get the length required for one splice plate:

L = 22.7 inches / 2 ≈ 11.35 inches

To ensure safety, the smallest allowable length should be the next highest option available, which is 11.6 inches (Option E).

Answer:

Reduce the mass of the earth to one-fourth its normal value.

Reduce the mass of the sun to one-fourth its normal value.

Reduce the mass of the earth to one-half its normal value and the mass of the sun to one-half its normal value.

Explanation:

Every particle in the universe attracts any other particle with a force that is directly proportional to the product of its masses and inversely proportional to the square of the distance between them. So, in this case we have:

[tex]F=\frac{Gm_Em_S}{d^2}[/tex]

If [tex]m'_E=\frac{m_E}{4}[/tex]:

[tex]F'=\frac{Gm'_Em_S}{d^2}\\F'=\frac{G(\frac{m_E}{4})m_S}{d^2}\\F'=\frac{1}{4}\frac{Gm_Em_S}{d^2}\\F'=\frac{1}{4}F[/tex]

If [tex]m'_S=\frac{m_S}{4}[/tex]

[tex]F'=\frac{Gm_Em'_S}{d^2}\\F'=\frac{Gm_E(\frac{m_S}{4})}{d^2}\\F'=\frac{1}{4}\frac{Gm_Em_S}{d^2}\\F'=\frac{1}{4}F[/tex]

If [tex]m'_E=\frac{m_E}{2}[/tex] and [tex]m'_S=\frac{m_S}{2}[/tex]:

[tex]F'=\frac{Gm'_Em'_S}{d^2}\\F'=\frac{G(\frac{m_E}{2})(\frac{m_S}{2})}{d^2}\\F'=\frac{1}{4}\frac{Gm_Em_S}{d^2}\\F'=\frac{1}{4}F[/tex]

Final answer:

To reduce the gravitational force between the Earth and the Sun to one-fourth, you can either reduce the mass of either body to one-fourth, or increase the distance between them by a factor of four. Reducing the mass of both bodies to one-half would also achieve the same result, as the product of their masses would be one-fourth of the original value.

Explanation:

The force between the Earth and the Sun can be described by Newton's law of universal gravitation, which states that the force (F) is proportional to the product of the two masses (m1 and m2) divided by the square of the distance (r) between their centers of mass. The law is formulated as F = G × m1 × m2 / r2, where G is the gravitational constant.

To reduce the magnitude of the gravitational force between the Earth and the Sun to one-fourth, one could either reduce the product of the masses by one-fourth or increase the separation distance by four times, because the force is inversely proportional to the square of the distance. Thus, the correct answers would be:

Reduce the mass of the Earth to one-fourth its normal value.

Reduce the mass of the Sun to one-fourth its normal value.

Increase the separation between the Earth and the Sun to four times its normal value.

However, the third option offered in the question, reducing the mass of both the Earth and the Sun to one-half their normal values, would also result in reducing the force to one-fourth, because (1/2) × (1/2) = 1/4.

Answer:

A) t₁ = 0.56 s t₂ =3.26 s

B) vh₁=vh₂ = 25.1 m/s

C) v₁ = 13.2 m/s  v₂ = -13.2 m/s

D) v = 31. 3 m/s

E) 36.7º below horizontal.  

Explanation:

A) As the only acceleration of the baseball is due to gravity, as it is constant, we can apply the kinematic equations in order to get times.

First, we can get the horizontal and vertical components of the velocity, as the movements  along these directions are independent each other.

v₀ₓ = v* cos 36.7º = 31.3 m/s * cos 36.7º = 25.1 m/s

v₀y = v* sin 36.7º =  31.3 m/s * sin 36.7º = 18.7 m/s

As in the horizontal direction, movement is at constant speed, the time, at any point of the trajectory, is defined by the vertical direction.

We can apply to this direction the kinematic equation that relates the displacement, the initial velocity and time, as follows:

Δy = v₀y*t -1/2*g*t²

We can replace Δy, v₀y and g for the values given, solving a quadratic equation for t, as follows:

4.9*t²-18.7t + 9 = 0

The two solutions for  t, are just the two times at which the baseball is at a height of 9.00 m above the point at which it left the bat:

t = 1.91 sec +/- 1.35 sec.

⇒ t₁ = 0.56 sec   t₂= 3.26 sec.

B) As we have already told, in the horizontal direction (as gravity is always downward) the movement is along a straight path, at a constant speed, equal to the x component of the initial velocity.

⇒ vₓ = v₀ₓ = 25.1 m/s

C) In order to get the value of  the vertical components at the two times that we have just found, we can apply the definition of acceleration (g in this case), solving for vfy, as follows:

vf1 = v₀y - g*t₁ = 18.7 m/s - (9.8m/s²*0.56 sec) = 13.2 m/s

vf₂ = v₀y -g*t₂ = 18.7 m/s - (9.8 m/s²*3.26 sec) = -13.2 m/s

D) In order to get the magnitude of  the baseball's velocity when it returns to the level at which it left the bat, we need to know the value of the vertical component at this time.

We could do in different ways, but the easiest way is using the following kinematic equation:

vfy² - v₀y² = 2*g*Δh

If we take the upward path, we know that at the highest point, the baseball will come momentarily to an stop, so at this point, vfy = 0

We can solve for Δh, as follows:

Δh = v₀y² / (2*g) = (18.7m/s)² / 2*9.8 m/s² = 17.8 m

Now, we can use the same equation, for the downward part, knowing that after reaching to the highest point, the baseball will start to fall, starting from rest:

vfy² = 2*g*(-Δh) ⇒ vfy = -√2*g*Δh = -√348.9 = -18. 7 m/s

The horizontal component is the same horizontal component of the initial velocity:

vx = 25.1 m/s

We can get the magnitude of the baseball's velocity when it returns to the level at which it left the bat, just applying Pythagorean Theorem, as follows:

v = √(vx)² +(vfy)² = 31.3 m/s

E) The direction below horizontal of the velocity vector, is given by the tangent of the angle with the horizontal, that can be obtained as follows:

tg Ф = vfy/ vx = -18.7 / 25. 1 =- 0.745

⇒ Ф = tg⁻¹ (-0.745) = -36.7º

The minus sign tell us that the velocity vector is at a 36.7º angle below the horizontal.

A major leaguer hits a baseball and the question asks for the time, velocity components, velocity magnitude, and direction at different points of the ball's trajectory when it reaches a certain height.

A) Let's consider the vertical motion of the baseball. The equation to determine the time it takes for an object to reach a certain height is:

h = v0t + 0.5gt2

Where:

By rearranging the equation, we can solve for t:

t2 + (2v0/g)t - (2h/g) = 0

Using the quadratic formula, we can plug in the values and solve for t. The two possible values of t will correspond to the two times when the ball is at a height of 9.00 m above the point at which it left the bat.

B) The horizontal component of velocity remains constant throughout the motion. Therefore, the horizontal component of velocity at each of the two times found in part (a) will be the same as the initial horizontal velocity, which is 31.3 m/s * cos(36.7º).

C) The vertical component of velocity changes due to the acceleration of gravity. At each of the two times found in part (a), the vertical component of velocity can be found using the equation:

vv1,2 = v0 + gt

Where:

D) When the baseball returns to the level at which it left the bat, its vertical velocity will be -v0, which means it will be moving downward with the same magnitude as the initial velocity but in the opposite direction. The horizontal component of velocity remains unchanged. Therefore, the magnitude of the baseball's velocity when it returns to the level at which it left the bat will be the square root of the sum of the squares of the horizontal and vertical components, which can be calculated using the formula:

v = √(vh2 + vv2)

Where:

E) The direction of the baseball's velocity when it returns to the level at which it left the bat can be determined using trigonometry. The angle below the horizontal can be found as:

θ = tan-1(vv/vh)

Where:

Answer:

U= 23.544 KJ

Explanation:

Given that

mass ,m = 120 kg

Height above the datum ,h = 20 m

Take g = 9.81 m/s²

The potential energy U is given as

U= m g h

m=mass

h=Height above the datum

g=Gravitational constant

Now by putting the value in the above values we get

U= 120 x 9.81 x 20 J

U=23544 J

Potential energy in KJ

U= 23.544 KJ

Therefore the answer will be 23.544 KJ.

Answer:

v = 0.969 m/s

Explanation:

See attachment for FBD

p = 0.25*(4/5) = 0.2 m

Sum of normal forces

Ns (3/5) - 0.2*Ns*(4/5) = 2 * (v^2 / 0.2)

Sum of vertical forces

Ns (4/5) - 0.2*Ns*(3/5) = 2*9.81

Solve both equations simultaneously to get Ns and v

Ns = 21.3 N

v = 0.969 m/s

Final answer:

The oscillation frequency of an ideal harmonic oscillator can be determined using the formula f = (1/2π) * √(k/m). By solving the equation based on the given data, we find that the oscillation frequency is approximately 2.27 Hz.

Explanation:

The oscillation frequency of a harmonic oscillator can be determined using the equation:

f = (1/2π) * √(k/m)

Where:

f is the oscillation frequency

k is the spring constant

m is the mass of the oscillator

In this case, the total mechanical energy of the oscillator is given as 9.8 J. Since the oscillation amplitude is 20.0 cm (or 0.20 m), the total mechanical energy can be equated to 11/13 * m * (Aω)² = 9.8 J, where ω is the angular frequency.

Solving for ω, we find ω = √(13 * 9.8 / (11 * 0.2²)). Using the relationship between angular frequency ω and oscillation frequency f (f = ω / (2π)), we can calculate the oscillation frequency:

f = √(13 * 9.8 / (11 * 0.2²)) / (2π)

Calculating this value gives us f ≈ 2.27 Hz. Therefore, the correct option is c. 2.3 Hz.

Answer:

The horizontal force that will be needed to move the pickup along the same road at the same speed is 230 N

Explanation:

given information:

horizontal force, F = 200 N

speed, v = 2.4 m/s

total weight increase 42%

coefficient of friction decrease by 19%

now, lets take a look of horizontal system

F - F(friction) = 0

F = F(friction)

F = μ N

F = μ m g

μ m g = 200 N

now find the force to move the pickup along the same road at the same speed(F(move))

total weight increase 42%, m g = 1+0.42 = 1.42

he coefficient of rolling friction decreases by 19%, μ = 1 - 0.19 = 0.81

F (move) = (0.81μ) (1.42 m g)

               = (0.81) (1.42) (μ m g)

                = (0.81) (1.42) (200)

                = 230 N

Answer:

The distance of the bee from the hive is 740 m.

Explanation:

Given that,

Bee starts fly 500 m due east, 430 m west and 670 m east.

The direction of the bee

500 m in positive direction

430 m in negative direction

670 m in positive direction

We need to calculate the net distance

Using formula of distance

[tex]D=500-430+670[/tex]

[tex]D=740\ m[/tex]

Hence, The distance of the bee from the hive is 740 m.

Answer:

216000 W or 216 kW

Explanation:

Power: This can be defined as the rate at which energy is consumed or used, The S.I unit of power is Watt (W)

Generally,

Power = Energy/time

P = E/t........................ Equation 1.

But

E = 1/2mv²..................... Equation 2

Where m = mass, v = velocity.

Substitute equation 2 into  equation 1

P = 1/2mv²/t...................... Equation 3

Let flow rate (Q) = m/t

Q = m/t................ Equation 4

Substitute equation 4 into equation 3

P = Qv²/2........................ Equation 5

Where Q = flow rate, v = velocity, P = power.

Given: Q = 120 kg/s, v = 60 m/s

Substitute into equation 5

P = 120(60)²/2

P = 60(60)²

P = 60×3600

P = 216000 W.

Thus the power generation potential of the water jet = 216000 W or 216 kW

The water jet could potentially generate 432 kW of power under ideal circumstances, calculated based on the given flow rate and velocity of the water jet.

To calculate the power generation potential of the water jet, you will use the formula for power: Power = Work/time. Since work can be translated into a measure of force times distance, we substitute Force * Distance into the equation for work. We can find the force exerted by the water jet with the equation for force: Force = mass * acceleration. The water's acceleration is its velocity out of the nozzle, or 60 m/s, and the mass flow rate is given as 120 kg/s. This gives us a force of 120 kg/s * 60 m/s = 7200 N.

However, distance is not given in the problem, so it's more helpful in this case to use an alternative equation for Power, given as the product of force and velocity: Power = Force * Velocity. Thus, our power becomes 7200 N * 60 m/s = 432,000 Watts, or 432 kW, assuming 100% efficiency.

Realistically, some power will be lost due to friction and inefficiencies in the system, but under ideal circumstances the water jet could potentially generate 432 kW of power.

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Final answer:

To calculate the pressure of the water in the hose at the spigot, you can use Bernoulli's equation, which relates pressure, density, velocity, and height of a fluid. By assuming the height at the spigot is the same as the nozzle and plugging in the given values, the pressure can be calculated.

Explanation:

The pressure of the water in the hose right after it comes out of the spigot can be calculated using Bernoulli's equation. Bernoulli's equation relates the pressure, density, velocity, and height of a fluid. In this case, the pressure can be found using the equation:

P1 + 1/2 ρv1² + ρgh1 = P2 + 1/2 ρv2²+ ρgh2

Where P1 is the pressure at the spigot, P2 is the pressure at the nozzle, ρ is the density of water, v1 is the initial velocity of water at the spigot, v2 is the velocity at the nozzle, h1 is the height of the spigot, and h2 is the height of the nozzle.

Since the height of the spigot is not given, we can assume it is at the same level as the nozzle, which means h1 = h2 = 7.25 m. The density of water, ρ, is 1000 kg/m³. The velocity at the nozzle, v2, is given as 11.2 m/s. Given these values, we can solve for the pressure at the spigot, P1.

Final answer:

The pressure of the water in the hose right after it comes out of the spigot can be calculated using Bernoulli's equation, which relates the pressure, velocity, and height of a fluid. Assuming the flow is laminar and neglecting viscosity, the pressure is found to be 229000 N/m².

Explanation:

The pressure of the water in the hose right after it comes out of the spigot can be calculated using Bernoulli's equation, which relates the pressure, velocity, and height of a fluid. Assuming the flow is laminar and neglecting viscosity, we can use the equation:

P + 1/2ρv² + ρgh = constant

Where P is the pressure, ρ is the density of water, v is the velocity of the water, g is the acceleration due to gravity, and h is the height difference between two points. Since the hose and nozzle are connected, we can assume that the pressures at these two points are the same. Also, the velocity of the water inside the hose can be considered negligible compared to the velocity at the nozzle. Therefore, we can simplify the equation to:

1/2ρv²+ 0 + ρgh = constant

The pressure at the spigot is atmospheric pressure, which is given as 1.013x10⁵ N/m². Rearranging the equation and solving for P, we have:

P = 1.013x10⁵ N/m² + 1/2ρv² + ρgh

Using the given values, we can substitute them into the equation and calculate the pressure

P = 1.013x10⁵ N/m² + 0.5x1000 kg/m³x(11.2 m/s)² + 1000 kg/m³x9.8 m/s²x7.25 m

P = 1.013x10⁵ N/m²+ 62656.96 N/m² + 68300 N/m²

P = 228957.96 N/m²

Rounding to three significant digits, the pressure of the water in the hose right after it comes out of the spigot is 229000 N/m².

Answer:

mass remains the same

Explanation:

Mass is the amount of matter in a substance which is independent from the external environment; hence, fields!

The weight of the object changes but mass remains same!

Final answer:

The question incorrectly asks for an object's mass on the moon based on gravitational acceleration. Mass is a constant and doesn't change with location. The gravitational acceleration on the moon (about 1.62 m/s²) affects an object's weight, not its mass.

Explanation:

The question seems to be asking for the mass of an object on the moon's surface based on the gravitational acceleration at the moon, which is a misunderstanding. The gravitational attraction on the surface of the moon is given as 5.37 ft/s2 (equivalent to about 1.63 m/s2, since the accurate value is 1.62 m/s2), but to find the mass of an object, we need to discuss how weight and mass are related in a gravitational field, not to calculate the mass based solely on the gravitational acceleration. To find an object's mass on the moon, one would use the formula Weight = Mass × Gravitational acceleration (W = mg). However, the mass of an object is a constant and does not change depending on its location, whether on Earth, the moon, or elsewhere. What changes is the weight of the object due to the difference in gravitational force exerted on it. For example, if an object weighs 9.8 N on Earth, it would weigh approximately 1.6 N on the moon due to the moon's lower gravity.

Answer:

1. A1, B2, C3

2. 47.1°

Explanation:

Sum of forces in the x direction:

∑Fₓ = ma

f − Fᵥᵥ = 0

f = Fᵥᵥ

Sum of forces in the y direction:

∑Fᵧ = ma

N − W = 0

N = W

Sum of moments about the base of the ladder:

∑τ = Iα

Fᵥᵥ h − W (b/2) = 0

Fᵥᵥ h = ½ W b

Fᵥᵥ (l sin θ) = ½ W (l cos θ)

l Fᵥᵥ sin θ = ½ l W cos θ

The correct set of equations is A1, B2, C3.

At the smallest angle θ, f = Nμ.  Substituting into the first equation, we get:

Nμ = Fᵥᵥ

Substituting the second equation into this equation, we get:

Wμ = Fᵥᵥ

Substituting this into the third equation, we get:

l (Wμ) sin θ = ½ l W cos θ

μ sin θ = ½ cos θ

tan θ = 1 / (2μ)

θ = atan(1 / (2μ))

θ = atan(1 / (2 × 0.464))

θ ≈ 47.1°

Equation of equilibrium is the static or dynamic equilibrium of all internal and external forces present in the system.

Given information-

A ladder rests against a vertical wall.

The coefficient of static friction between the ladder and the ground is 0.464.

Equation of equilibrium is the static or dynamic equilibrium of all internal and external forces present in the system.

From the equation of equilibrium of two bodies, the net force in x axis can be given as,

[tex]\sum F_x=0[/tex]

As the normal force and friction force acting on the x-axis. Thus,

[tex]-f_w+f=0[/tex]

[tex]f=f_w[/tex]

Thus option A1 is correct.

As the gravitational force(due to weight of the body) is acting in the vertical direction. Thus from the equation of equilibrium of two bodies, the net force in y axis can be given as,

[tex]\sum F_y=W[/tex]

As the normal force acting on the x-axis. Thus,

[tex]N=W[/tex]

Thus option B2 is correct.

Apply torque equation at the base of the ladder,

[tex]\sum \tau=Ia[/tex]

[tex]F_wh=\dfrac{1}{2} Wb[/tex]

Here the value of h and b can be changed in the form of I using the trigonometry formula. Thus,

[tex]F_w(I\sin \theta)=\dfrac{1}{2} W(\cos \theta[/tex]

Thus option C3 is correct.

Hence the set A1, B2, C3 is correct.

The equation of option C3 is,

[tex]F_w(I\sin \theta)=\dfrac{1}{2} W(\cos \theta[/tex],

As normal force is can be given as,

[tex]F_w=W\mu[/tex]

Thus,

[tex]W\mu(I\sin \theta)=\dfrac{1}{2} W(\cos \theta[/tex]

[tex]\dfrac{\sin \theta}{\cos \theta} =\dfrac{1}{2\mu} \\\tan \theta =\dfrac{1}{2\times 0.464} \\\theta =\tan^-(1.0770)\\\theta=47.1^o[/tex]

Hence the smallest angle θ for which the ladder remains stationary is [tex]47.1^o[/tex]

Hence,

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Answer:

d) Both a and b are correct

Explanation:

Displacement: It is defined as the distance between initial and final position during motion.

Distance: It is defined as the total path length traveled by object

Or

It is the distance of one place from other place.

Student said that the displacement between my dorm and the lecture hall is 1km.

It is not displacement .It is distance or we say path length.

Therefore, he is using  incorrect physical quantity for the information provided.

He should have called distance 1 km or path length 1 km.

Option d is true.

The student is accurately using the term 'displacement' to describe the 1 kilometer between their dorm and the lecture hall. Technically, 'distance' or 'path length' could also have been used assuming a straight-line path and ignoring direction.

In the context of this specific question, the student is accurately using the term displacement as a physical quantity. Displacement refers to the shortest distance between two points in a particular direction. In this case, the 1 kilometer refers to the displacement between the dorm and the lecture hall, assuming a straight-line path. So, technically, both a) Distance and b) Path length could also have been used to describe the 1 kilometer, if we don't consider the direction.

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Answer:

The reach of the tongue is 23 cm.

Explanation:

Hi there!

The equation of traveled distance is the following:

x = x0 + v0 · t + 1/2 · a · t²

Where:

x = traveled distance at time t.

x0 = initial position.

v0 = initial velocity.

a = acceleration.

t = time.

Let´s calculate the distance traveled by the tongue of the chameleon during the first 20 ms (0.020 s):

The initial position and velocity are zero (x0 = 0 and v0 = 0)

x = 1/2 · a · t²

x = 1/2 · 280 m/s² · (0.020 s)²

x = 0.056 m

Now, let´s find the distance traveled while the tongue moves at constant speed. But first, let´s find the velocity (v) of the tongue after the accelertation interval using the following equation:

v = v0 + a · t     (v0 = 0)

v = 280 m/s² · 0.020 s

v = 5.6 m/s

Then, the distance traveled at constant speed can be calculated:

x = v · t

x = 5.6 m/s · 0.030 s

x = 0.17 m

The reach of the tongue is 0.17 m + 0.056 m = 0.23 m = 23 cm.

Answer:

(B) Vertically upward

Explanation:

r = Side of triangle = 14 cm

q = Charge = -2.5 nC

Electrical force is given by

[tex]F=\dfrac{kq_1q_2}{r^2}\\\Rightarrow F=\dfrac{8.99\times 10^9\times (2.5\times 10^{-9})^2}{0.14^2}\\\Rightarrow F=2.86671\times 10^{-6}\ N[/tex]

For the top charge

Net force on both charges is given by

[tex]F_n=2Fcos\theta\\\Rightarrow F_n=2\times 2.86671\times 10^{-6}\times cos30\\\Rightarrow F_n=4.96529\times 10^{-6}\ N[/tex]

The net force acting on the top charge is [tex]4.96529\times 10^{-6}\ N[/tex]

Here the forces are symmetrical hence, the net force is along +y axis i.e., upward

The direction of the electrical force exerted on the object at the top corner due to the two objects at the horizontal base of the triangle is vertically upward.

The direction of the electrical force exerted on the object at the top corner due to the two objects at the horizontal base of the triangle can be determined using Coulomb's Law. Since the objects at the base of the triangle have a negative charge and the object at the top corner has a negative charge as well, the force between them will be repulsive. As a result, the direction of the electrical force on the top object will be vertically upward.

Answer:

h = 6.39 m

Explanation:

radius = 6/2 =3in= 3/12 in/ft = 0.25 ft

apply continuity equation to evaluate the velocity

flow rate = area * velocity

        2.2 = π(0.25^2) * V  

⇒        V = 11.2 ft/s

*assuming there is no applied pressure difference along the pipe

applying Energy conservation law

initial potential energy = final kinetic energy

                              pgh = (1/2)*pv^2

where p  = density

            v = velocity  

plugging in the values:

p cancels out on both the sides

⇒ 9.81 * h = (11.2^2)/2

              h = 6.39 m

Answer with explanation :

The negative sign means that the potential energy decreases by the movement of the electron.

negative charge at rest in an electric field moves toward the region of an electric field , so that its potential energy will diminish and change into the kinetic energy of motion. The total energy remains constant.

Positive charges will move downhill because of convention. It is to stay in accordance with other potential theories, particularly gravity, where the "charge" is mass, that moves downwards in the gravitational potential field expressed by ϕ(r)=−GM|r|ϕ(r)=−GM|r|. In an electronic system, howbeit, positive charges are fixed in their position within a component (e.g., a wire), therefore instead of the mobile,the negative charges, electrons, move uphill.

Final answer:

A negative charge at rest in an electric field moves toward a region of higher potential, thus decreasing its electric potential energy. A positive charge moves toward a region of lower potential, similarly decreasing its electric potential energy. The strength of the electric field relates to potential difference and is zero where electric potential is constant.

Explanation:

If a negative charge is initially at rest in an electric field, it will move toward a region of higher potential. This is because negative charges are attracted to areas of positive potential and repelled from areas of negative potential. Conversely, a positive charge will move toward a region of lower potential, much like how fluid moves from a region of high pressure to low pressure.

The potential energy of a charge in an electric field depends on both the electric potential and the sign and magnitude of the charge. For a negative charge, moving to a higher potential region decreases its electric potential energy, while a positive charge moving to a lower potential region also decreases its electric potential energy. The electric field is defined as the force per unit charge, and its strength is directly related to the potential difference. If the electric potential is constant in a region, the electric field strength is zero there.

When discussing the potential difference and electric field strength, a good example is a parallel plate capacitor which has a uniform electric field. If an electron, being negatively charged, is released at rest between the plates, it will accelerate toward the positively charged plate due to the attractive electrostatic force, thereby moving to a region of higher potential.

Answer:

Explanation:

ocities: the velocity with which the wave moves in the medium (e.g., air or a string) and the velocity of the medium (the air or the string itNBself).Consider a transverse wave traveling in a string. The mathematical form of the wave is: y(x,t) = A sin(kx - ωt)Part AFind the velocity of propagation v_p of this wave.Express the velocity of propagation in terms ofNGHJGHHG some or all of the variables A, k, and ω.Part BFind the y velocity v_y(x,t) of a point on the string as a function of x and t.Express the y velocity in terms of ω, A, k, x, and t.Part CWhich of the following statements about v_x(x,t), the x component of the velocity of the string, is true?A) v_x(x;t) = v_pB) v_x(x;t) = v_y(x;t)C) v_x(x;t) has the same mathematical form as v_y(x;t) but is 180° out of phase.D) v_x(x;t)=0Part DFind the slope of the string ∂_y(x,t) / ∂_x as a function of position x and time t.Express your answer in terms of A,k, ω, x, and t.NNNNN

a) The velocity of propogation of the wave  V=w/k

b) The y velocity v_y(x,t) of a point on the string as a function of x v=-wAcos(kx-wt)

A wave can be described as a disturbance that travels through a medium from one location to another location

y(x,t)=Asin(kx−ωt) defines the wave equation.

a)The velocity of propogation of the wave

We are asked to find wave speed (v)

Recall that v = fλ

From the wave equation above,

k = 2π/ λ where k is the wave number and λ is the wavelength, λ = 2π /k

ω = 2πf where f is the frequency and ω is the angular frequency.

f = ω/ 2π.

By substituting for λ and ω into the wave speed formulae, we have that

v =( ω/ 2π) × (2π /k)

v = ω/k

b)The y velocity v_y(x,t) of a point on the string as a function of x

y(x,t)=Asin(kx−ωt)

The first derivative of y with respect to x give the velocity (vy)

By using chain rule, we have that

v = dy/dt = A cos( kx −ωt) × (−ω)

v = - ωAcos( kx −ωt)

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Answer:

Explanation:

a ) Energy of spring = 1/2 k A² where A is amplitude of oscillation and k is force constant .

So initial energy = 1/2 x 2.5 x (6.4 x 10⁻²)²

= 51.2 x 10⁻⁴ J

So final  energy = 1/2 x 2.5 x (3.7 x 10⁻²)²

= 17.11 x 10⁻⁴ J

energy lost

= 34.1 J .

This energy is dissipated in the form of heat,  sound etc.

Answer:

a. [tex]\Delta U=3.375\ N.cm=3.375\times 10^{-2}\ J[/tex]

b. The energy lost during the damping is converted into kinetic energy of the molecules which is heat primarily.

Explanation:

Given:

a)

the energy lost in damping:

[tex]\Delta U=\frac{1}{2} \times k\times A^2-\frac{1}{2} \times k\times x^2[/tex]

[tex]\Delta U=\frac{1}{2} \times 2.5\times (6.4-3.7)[/tex]

[tex]\Delta U=3.375\ N.cm=3.375\times 10^{-2}\ J[/tex]

b)

The energy lost during the damping is converted into kinetic energy of the molecules which is heat primarily.

Answer: (a) power output = 3.85×10²⁶W

(b). There is no relative change in power as it is independent from frequency

(c). 590 W/m²

Explanation:

given Radius between earth and sun to be = 1.50 × 10¹¹m

Intensity of the radiation from the sun measured on earth to be = 1360 W/m²

Frequency = 60 MHz

(a). surface area A of the sun on earth is = 4πR²

substituting value of R;

A = 4π(.50 × 10¹¹)² = 2.863 10²³×m²

A = 2.863 10²³×m²

now to get the power output of the sun we have;

P sun = I sun-earth A sun-earth

where A = 2.863 10²³×m², and I is 1360 W/m²

P sun =  2.863 10²³ × 1360

P sun = 3.85×10²⁶W

(c). surface area A of the sun on mars is = 4πR²

now we substitute value of 2.28 ×10¹¹ for R sun-mars, we have

A sun-mars = 4π(2.28× 10¹¹)²

A sun-mars = 6.53 × 10²³m²

now to calculate the intensity of the sun;

I sun-mars = P sun / A sun-mars

where P sun = 3.85×10²⁶W and A sun-mars = 6.53 × 10²³m²

I sun-mars =  3.85×10²⁶W / 6.53 × 10²³m²

I sun-mars = 589.6 ≈ 590 W/m²

I sun-mars = 590 W/m²

Answer

given,

initial speed of the rocket A = 100 m/s

height of explode = 300 m

acceleration due to gravity = 9.8 m/s²

rocket b is launched after t₁ time

now, using equation of motion to calculate time

[tex]s = ut + \dfrac{1}{2}gt^2[/tex]

[tex]300 = 100t + \dfrac{1}{2}(-9.8)t^2[/tex]

 4.9 t² - 100 t + 300 = 0

using quadratic equation

[tex]t = \dfrac{-(-100)\pm \sqrt{100^2-4\times 4.9 \times 300}}{2\times 4.9}[/tex]

t₁ = 3.65 s   and  t₂ = 16.75 s

now, rocket A reaches 300 m on return at 16.75 s

rocket B reaches 300 m after 3.65 s

time difference of launch:

t = 16.75 - 3.65

t = 13.1 s

velocity of rocket A

v_a = u + g t

v_a =100 - 9.8 x 16.75

v_a = -64.15 m/s

velocity of rocket B

v_b = u + g t

v_b =100 - 9.8 x 3.65

v_b =+64.23 m/s

relative velocity of B relative to A at the time of the explosion

[tex]V_{BA} = v_b - V_a[/tex]

[tex]V_{BA} = 64.23 -(-64.15)[/tex]

[tex]V_{BA} = 128.38\ m/s[/tex]

relative velocity is equal to 128.38 m/s

Answer:

8.8 cm

31.422 cm/s

Explanation:

m = Mass of block = 0.6 kg

k = Spring constant = 15 N/m

x = Compression of spring

v = Velocity of block

A = Amplitude

As the energy of the system is conserved we have

[tex]\dfrac{1}{2}mv^2=\dfrac{1}{2}kA^2\\\Rightarrow A=\sqrt{\dfrac{mv^2}{k}}\\\Rightarrow A=\sqrt{\dfrac{0.6\times 0.44^2}{15}}\\\Rightarrow A=0.088\ m\\\Rightarrow A=8.8\ cm[/tex]

Amplitude of the oscillations is 8.8 cm

At x = 0.7 A

Again, as the energy of the system is conserved we have

[tex]\dfrac{1}{2}kA^2=\dfrac{1}{2}mv^2+\dfrac{1}{2}kx^2\\\Rightarrow v=\sqrt{\dfrac{k(A^2-x^2)}{m}}\\\Rightarrow v=\sqrt{\dfrac{15(0.088^2-(0.7\times 0.088)^2)}{0.6}}\\\Rightarrow v=0.31422\ m/s[/tex]

The block's speed is 31.422 cm/s

Final answer:

This physics problem involves calculating the amplitude of oscillations and the speed of a block at a specific displacement by applying conservation of mechanical energy in harmonic motion. The amplitude is found to be approximately 6.37 cm, and the block's speed at 0.70 A (amplitude) is approximately 28.4 cm/s.

Explanation:

The problem involves a 0.600 kg block attached to a spring with a spring constant of 15 N/m that is initially at rest and is then given a speed of 44 cm/s. To find the amplitude of the subsequent oscillations and the block's speed when x = 0.70 A, where A is the amplitude, we first use the principle of conservation of mechanical energy in harmonic motion. The initial kinetic energy given to the block will be equal to the potential energy of the spring at maximum displacement, which allows us to calculate the amplitude. Next, we determine the speed of the block at a displacement of 0.70 A using the relationship between kinetic and potential energy at any point during the oscillation.

The amplitude can be found using: KE = 1/2 k A^2, where KE is the kinetic energy of the block. Converting 44 cm/s to m/s gives 0.44 m/s. The kinetic energy of the block is KE = 1/2 mv^2 = 1/2 (0.600 kg)(0.44 m/s)^2, and solving for A gives the amplitude in meters, which can then be converted back to centimeters. To find the block's speed at x = 0.70 A, we use the conversion of potential energy at this displacement back to kinetic energy, considering the total mechanical energy of the system remains constant.

Using these principles, calculations show the amplitude of the oscillations to be approximately 6.37 cm, and the block's speed at x = 0.70 A is about 28.4 cm/s.