Two infinite nonconducting sheets of charge are parallel to each other, with sheet A in the x = -2.15 plane and sheet B in the x = +2.15 m plane. Find the electric field in the region x < -2.15 m, in the region x > +2.15 m, and between the sheets for the following situations.

(a) when each sheet has a uniform surface charge density equal to +3.25 µC/m2 region (m) electric field (N/C)
x < -2.15 ________________

x > +2.15 ________________

-2.15 < x < +2.15 _________________

(b) when sheet A has a uniform surface charge density equal to +3.25 µC/m2 and sheet B has a uniform surface charge density equal to -3.25 µC/m2

region (m) electric field (N/C)
x < -2.15 __________________

x > +2.15 __________________

-2.15 < x < +2.15 _____________________

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:

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:

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

Answer:

(a) T1 = 938.3lb

(b) R = 665.5lb

The detail solution to this problem can be found in the attachment below.

This problem was solved by resolving the forces along the vertical and horizontal and equating to Rx and Ry respectively. Rx = 0 because the resultant is directed along the vertical.

Explanation:

See attachment below for the full solution.

Thank you for reading.

Final answer:

To find the tension T1 and the vertical resultant R, use trigonometry, considering that T1 must have an equal and opposite horizontal force component to T2 and a vertical component that combines with T2's vertical component.

Explanation:

Understanding the Tension in Two Portions of a Cable

The student's question pertains to the tensions in a telephone cable clamped at point A to a pole. We are given that the tension in the right-hand portion of the cable (T2) is 1000 lb and are asked to determine two things:

The required tension T1 in the left-hand portion if the resultant R of the forces exerted by the cable at A is to be vertical.

The corresponding magnitude of R.

To solve for the required tension T1 in the left-hand portion so that the resultant force R is vertical, we need to use trigonometry. Given that the resultant R must be vertical and that we already have a horizontal force component from T2, T1 must provide an equal and opposite horizontal force component to cancel it out, as well as contribute to the resultant vertical force. The magnitude of the horizontal component of T1 must equal that of T2, and the vertical component of T1 will combine with T2 to form the vertical resultant R.

To determine the magnitude of the resultant force R, which is vertical, we will sum up the vertical components of T1 and T2. This magnitude can be found using the Pythagorean theorem if necessary, or simply as the sum of the vertical components if T1 is already calculated and has the proper direction to ensure a vertical R.

For this specific scenario, without additional information such as angles or the tension division between T1 and T2, we cannot provide numerical answers but can describe the method to find the solution.

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:

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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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:

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.

The amount of charge on the plates increases during this process.

Explanation:

The relationship between charge and potential difference through a capacitor is

[tex]C=\frac{Q}{\Delta V}[/tex]

where

C is the capacitance

Q is the charge stored

[tex]\Delta V[/tex] is the potential difference

The capacitance of a parallel-plate capacitor is given by

[tex]C=\frac{\epsilon_0 A}{d}[/tex]

where

[tex]\epsilon_0[/tex] is the vacuum permittivity

A is the area of the plates

d is the distance between the plates

Combining the two equations, we get:

[tex]Q=\frac{\epsilon_0 A \Delta V}{d}[/tex]

In this problem:

- The potential difference between the plates, [tex]\Delta V[/tex], is kept constant

- The area of the plates, A, remains  constant

- The distance between the plates, d, is decreased

Since Q is inversely proportional to d, this means that as the plates are pushed together, the amount of charge on the plates increases during the process.

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Final Answer:

By sustaining a constant voltage across two conducting plates, the parallel-plate capacitor creates an electric field, gathers positive and negative charges, and preserves the battery's electric potential difference.

Explanation:

The amount of charge on the plates remains constant during the process of pushing the plates of a parallel-plate capacitor together. When a constant voltage (potential difference) is applied across the plates by a battery, it establishes an electric field between the plates. When you push the plates together without touching, you are essentially changing the distance between them while keeping the voltage constant.

The key point to understand is that the amount of charge on the plates is determined by the voltage (potential difference) and the capacitance of the capacitor, the charge on the plates is denoted by the equation Q = C * V.

In summary, when the plates of a parallel-plate capacitor are maintained with a constant voltage by a battery and pushed together without touching, the amount of charge on the plates remains constant throughout the process.

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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.

Final answer:

The magnitude of the electric force of attraction between an iron nucleus and its innermost electron can be calculated using Coulomb's law.

Explanation:

The magnitude of the electric force of attraction between an iron nucleus and its innermost electron can be calculated using Coulomb's law. Coulomb's law states that the magnitude of the electric force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. In this case, the charge of the iron nucleus (+26e) is given as 26 times the charge of an electron (e).

The formula to calculate the electric force is: F = k * |q1 * q2| / r^2, where F is the electric force, k is the electrostatic constant, q1 and q2 are the charges of the particles, and r is the distance between them.

Using this formula, we can plug in the values: F = (8.99 × 10^9 Nm^2/C^2) * |(26e) * (-e)| / (1.5 × 10^(-12)m)^2. Simplifying this equation will give you the magnitude of the electric force.

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:

Force acting on the basketball will be 351.85 N

Explanation:

We have given velocity of baseball is changes from 50 m/sec to 45 m /sec in 0.00135 sec

So initial; velocity of the ball u = 50 m /sec

And final velocity of the ball v = 45 m /sec

Mass of the basketball m = 95 gram = 0.095 kg

Time taken t = 0.00135 sec

From third equation of motion v = u+at

So [tex]45=50+a\times 0.00135[/tex]

a = 3703.70 [tex]m/sec^2[/tex]

We have to find the force needed to change the speed of the basketball

According to newtons law force is given by F = ma

So F = 0.095×3703.70 = 351.85 N

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:

B = 1.67 μ T

Explanation:

given,

current, I = 1 x 10⁴ A

r = 120 m

treating lightning bolt as long straight conductor

  [tex]B=\dfrac{\mu_0I}{2\pi r}[/tex]

  [tex]B=\dfrac{4\pi \times 10^{-7}\times 1 \times 10^4}{2\pi\times 120}[/tex]

resulting magnitude would be equal to

    B = 16.67 x 10⁻⁶ T

    B = 1.67 μ T

The resulting magnetic field is equal to B = 1.67 μ T

Answer:

1.67 x 10^-5 Tesla

Explanation:

Current, i = 1 x 10^4 A

distance, d = 120 m

The formula for the magnetic field due to long straight current carrying conductor is given by

[tex]B =\frac{\mu _{0}}{4\pi }\times \frac{2i}{r}[/tex]

[tex]B =10^{-7}\times \frac{2\times 10^{4}}{120}[/tex]

B = 1.67 x 10^-5 Tesla

To develop this problem we will apply the concepts related to the kinematic equations of motion, specifically that of acceleration. Acceleration can be defined as the change of speed in an instant of time, mathematically this is

[tex]\vec{a} = \frac{\vec{v_2}-\vec{v_1}}{\Delta t}[/tex]

If a mobile is decreasing its speed (it is slowing down), then its acceleration is in the opposite direction to the movement. This would imply that the acceleration vector is opposite to the velocity vector.

Therefore the correct answer is B.

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:

B. Yes, at the end of the trajectory.

Explanation:

By the conservation of energy, in order the object to have the same speed, it must have the same potential and kinetic energy, therefore, it must be at the same height as its initial height.

Answer:

b. Yes, at the end of the arch (end of trajectory)

Explanation:

according to conservation of energy and considering no air resistance:

initially the object has some kinetic energy due to speed which gets converted to gravitational potential energy till its maximum height.

since there is no energy loss, the object again gets to same amount of kinetic energy and hence the same speed.

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:

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:

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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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:

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 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:

Light takes 14.12 minutes to travel from the Sun to tatoone.

Explanation:

The equation for the average velocity can be used to estimate the time that light will take to travel from the Sun to tatoone. The average velocity is defined as:

[tex]v = \frac{d}{t}[/tex]   (1)

Where v is the velocity, d is the covered distance and t is the time.  

Therefore, t can be isolated from equation 1:    

[tex]t = \frac{d}{v}[/tex]            

It is necessary to express the speed of light in terms of minutes:

[tex]3x10^{8} \frac{m}{s} . \frac{60 s}{1 min}[/tex] ⇒ [tex]1.8x10^{10} m/min[/tex]

An astronomical unit is defined as the distance between the Earth and the Sun ([tex]1.496 x 10^{11} m[/tex]).        

[tex]d = 1.7 AU \cdot \frac{1.496 x 10^{11} m}{1AU}[/tex] ⇒ [tex]2.5432x10^{11}[/tex]m

Finally, equation 2 can be used:

[tex]t = \frac{d}{v}[/tex]      

[tex]t = \frac{2.5432x10^{11}m}{1.8x10^{10} m/min}[/tex] 

t = 14.12 min                

Hence, light takes 14.12 minutes to travel from the Sun to tatoone.

Answer:

20 A

Explanation:

Given

Spring pulled from position

x = 5A

we need to calculate total distance of one full cycle  of spring motion

if you see image below, you understand easily

When cycle complete

Its total distance become 20 A

Answer:

0.067 seconds

Explanation:

The time needed for a person seated in the middle row to receive a sound from the stage, can be calculated assuming that the sound is moving in a straight line at constant speed, taking into account that the distance traveled will be half of the distance between the stage and the back wall:

t₁ = x₁ / v = 11.5 m / 343 m/s = 0.034 sec

The same sound, after reflecting from the back wall, will travel a distance equal to one and a half the distance between the stage and the back wall:

t₂ = 34.5 m /343 m/s = 0.101 sec

The time elapsed between both sounds will be equal to the difference between t₂ and t₁, as follows:

t₂ - t₁ = 0.101 sec - 0.034 sec = 0.067 sec.

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.

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:

Acceleration of the cart in the beginning is positive

Speed is positive when speeding up and slowing down.

Speed is zero when the cart stops.

Speed is negative when the cart speeds up in opposite direction.

Acceleration is negative when speeding up in opposite direction from rest.

Explanation:

When the cart is speeding up then its speed is increasing with time hence its acceleration is having a positive  value.