97.1 m/s at an angle of 44.5 above the hanzontal on a long fiat inmin. Assuming that air Detertmen the maxirum height ieschod by the rock: teeistance it neglegble. Express your answer using sheee signafieant figures and include the appropriate units. Part B Calculate the time of travei before the zock hits the grotind. Express your answer using three signifieant figures and include the apprepriate units.

The major diameter of the triple-threaded power screw is 22.067 mm. The lead of the screw is 0.2677 inches. The torque in the screw is 19.746 Nm, and the torque in the collar is 30.369 Nm. The overall efficiency of the screw is 45.88%. The torsional stress in the screw is 39.791 MPa.

1. To calculate the major diameter, we use the formula: major diameter = mean diameter + 2 * (pitch / (3 * π)). Plugging in the values, we get major diameter = 24 + 2 * (6.8 / (3 * π)) = 22.067 mm.

2. The lead is the axial distance traveled by the screw in one revolution. It is given by the formula: lead = π * mean diameter / number of threads. Here, since it is a triple-threaded screw, the number of threads is 3. Therefore, lead = π * 24 / 3 = 25.1327 mm. Converting this to inches, we get lead = 0.2677 inches.

3. The torque in the screw can be calculated using the formula: torque = (friction on screw * mean diameter / 2) * longitudinal force. Substituting the values, we get torque = (0.08 * 24 / 2) * 1500 = 19.746 Nm.

4. The torque in the collar can be calculated using the formula: torque = (friction on collar * collar diameter / 2) * longitudinal force. Plugging in the values, we get torque = (0.122 * 50 / 2) * 1500 = 30.369 Nm.

5. The overall efficiency of the screw is given by the formula: overall efficiency = (mechanical advantage / ideal mechanical advantage) * 100%. Since the collar acts as a restraining force, the mechanical advantage is given by: mechanical advantage = lead / pitch. The ideal mechanical advantage is given by: ideal mechanical advantage = mean diameter / (2 * pitch). Plugging in the values, we find mechanical advantage = 0.2677 / 6.8 = 0.0394 and ideal mechanical advantage = 24 / (2 * 6.8) = 1.7647. Therefore, the overall efficiency = (0.0394 / 1.7647) * 100% = 45.88%.

6. The torsional stress in the screw can be calculated using the formula: torsional stress = (16 * torque) / (π * mean diamet[tex]er^3[/tex]). Substituting the values, we get torsional stress = (16 * 19.746) / (π * 24^3) = 39.791 MPa.

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Sound waves are mechanical waves that propagate through a medium, such as air or water, by causing particles in the medium to vibrate. They are longitudinal waves characterized by compressions and rarefactions, where particles oscillate back and forth parallel to the direction of wave propagation.

Conclusion on waves traveling on a rope: Waves traveling on a rope are transverse waves, where the particles of the medium move perpendicular to the direction of wave propagation. When a rope is shaken or disturbed, it creates a wave that travels along its length. The wave consists of crests (points of maximum displacement) and troughs (points of minimum displacement). The speed of the wave on a rope depends on the tension in the rope and its mass per unit length. Waves on a rope can exhibit phenomena such as reflection, refraction, interference, and standing waves.Conclusion on the different types of waves: There are three main types of waves: transverse waves, longitudinal waves, and surface waves. Transverse waves have oscillations perpendicular to the direction of wave propagation, such as waves on a rope. Longitudinal waves have oscillations parallel to the direction of wave propagation, like sound waves. Surface waves occur at the interface between two media and possess both transverse and longitudinal motion. Examples of surface waves include ocean waves and seismic waves. Each type of wave exhibits distinct characteristics and behaviors based on the nature of particle motion and the medium through which they propagate.

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The final speed of the rocket, after accelerating at a rate of 190 m/s² for 2.4 seconds from rest, is 456 m/s.

To find the final speed of the rocket, we can use the equation of motion:

v = u + at

Where:

v = final velocity

u = initial velocity (0 m/s, as it starts from rest)

a = acceleration (190 m/s²)

t = time (2.4 seconds)

Plugging in the values:

v = 0 + (190 m/s²) * (2.4 s)

v = 456 m/s

The final speed of the rocket is 456 m/s.

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Inside the shell (r < a): Electric field E = 0.

Outside the shell (r > a): Electric field [tex]E = \frac{\sigma}{\varepsilon_0 r}[/tex], where σ is the areal charge density and ε₀ is the vacuum permittivity.

To find the electric field generated inside and outside a spherical shell with a uniform charge distribution, we can use Gauss's Law.

Inside the shell, the electric field is zero because the net charge enclosed by any Gaussian surface within the shell is zero. Therefore, the electric field inside the shell is 0.

To find the electric field outside the shell, we consider a Gaussian surface in the form of a concentric sphere of radius r, where r > a.

According to Gauss's Law, the electric flux through a closed Gaussian surface is proportional to the charge enclosed by the surface. Mathematically, it can be expressed as:

∮ E · dA = (Q_enclosed) / ε₀,

where ∮ E · dA represents the electric flux through the Gaussian surface, Q_enclosed is the charge enclosed by the surface, and ε₀ is the vacuum permittivity (a constant).

For the Gaussian surface outside the shell (r > a), the entire charge Q (uniformly distributed in the shell) is enclosed.

The charge Q enclosed by the Gaussian surface is the product of the areal charge density σ and the surface area of the Gaussian surface.

Q_enclosed = σ * (4πr²).

Using Gauss's Law, we can rewrite the equation as:

[tex]E * (4\pi r^2) = \frac{\sigma * (4\pi r^2)}{\epsilon_0}[/tex].

Simplifying the equation:

[tex]E=\frac{\sigma}{\epsilon_0r}[/tex].

Therefore, the electric field outside the spherical shell (r > a) is given by:

[tex]E = \frac{\sigma}{\varepsilon_0 r}[/tex].

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The component of acceleration along the ramp is positive while the car is going up the ramp, and negative while the car is going down the ramp. When the car is going up the ramp, it is slowing down, meaning its velocity is decreasing.

Since acceleration is defined as the rate of change of velocity, and the velocity is decreasing, the acceleration is in the opposite direction of the velocity. Since the ramp is inclined upward, the component of acceleration along the ramp is positive. When the car is going down the ramp, it is speeding up, meaning its velocity is increasing. Again, the acceleration is in the direction opposite to the velocity. Since the ramp is inclined downward, the component of acceleration along the ramp is negative.

In both cases, the acceleration is acting against the direction of motion, either to slow down the car while going up or to oppose the increasing velocity while going down.

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Magnitude of the car's velocity relative to Earth is 0 m/s Direction of the car's velocity relative to Earth. V C relative to E = 0 m/s and the direction of the car's velocity relative to Earth is 50° counterclockwise from due east.

Resolve all the velocities into their components.

We will resolve all the velocity components along two directions: North-South and East-West.

North-South direction: Relative to the car, the motorcyclist is moving 40° east of north, which means he is moving 50° north of east relative to the Earth.

So, velocity component of the motorcyclist along North-South direction is: V north-motorcyclist = 25sin50° = 19.24 m/s

Velocity component of the car along North-South direction is: V north-car = 0 East-West direction: Velocity component of the motorcyclist along East-West direction is: V east-motorcyclist = 25cos50° = 16.08 m/s

Relative to the Earth, the motorcyclist is not moving in the East-West direction.

So, velocity component of the car along East-West direction is: V east-car = 0

Velocity of the car relative to the Earth: V C relative to E = sqrt(Vnorth-car² + Veast-car²) = sqrt(0 + 0) = 0 m/s

Magnitude of the car's velocity relative to Earth is 0 m/s

Direction of the car's velocity relative to Earth: Let θ be the direction of the car's velocity relative to Earth, measured as an angle θ counterclockwise from due east.

It is given that the motorcyclist is moving 50° north of east relative to the Earth.

Therefore, the car is moving 40° north of east relative to the Earth.

So, θ is:θ = 90° - 40° = 50°

The direction of the car's velocity relative to Earth is 50° counterclockwise from due east.

Answer: V C relative to E = 0 m/s and the direction of the car's velocity relative to Earth is 50° counterclockwise from due east.

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The force that attracts the two boxes to one another is given by Coulomb's law which states that the force between two charged bodies is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

It is expressed mathematically as F = k * (q1 * q2 / r^2)Where F is the force, k is Coulomb's constant, q1 and q2 are the magnitudes of the charges, and r is the distance between them. Now, we know that the charges are of equal magnitude, but one of them is positive and the other negative. So, q1 * q2 is negative. Hence, we get F = -k * (q1 * q2 / r^2)The Coulomb's constant, k = 9 * 10^9 N m^2 C^-2.q1 = q2 = 1.6 * 10^-19 C (the magnitude of the charge on an electron or proton) and r = 31 m. Substituting these values, we get: F = -9 * 10^9 * (1.6 * 10^-19)^2 / 31^2= -2.24 * 10^25 NSo, the force that attracts the two boxes to one another is 2.24E+25 N. Work done in moving the charges farther apart is also called electrostatic potential energy.

The work done in moving the boxes farther apart is given by the formula: W = k * (q1 * q2 / r2 - q1 * q2 / r1)where W is the work done, k is Coulomb's constant, q1 and q2 are the magnitudes of the charges, and r1 and r2 are the initial and final distances, respectively. Substituting the given values, we get: W = 9 * 10^9 * (1.6 * 10^-19)^2 * (1/31 - 1/65)W = 1.88 * 10^27 JTherefore, the amount of work (energy) required to move the boxes of protons and electrons from 31 m apart to 65 m apart is 1.88E+27 J.

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To determine the electric field at a point on the wire due to the linear charge density, we can use the concept of Coulomb's Law and integrate the contributions from infinitesimally small charge elements along the wire.

Let's consider a small element of length Δl on the wire, located at a distance x from the point charge q. The charge of this element can be expressed as Δq = λΔl, where λ is the linear charge density (7.0 nC/m).where k is the Coulomb's constant (k ≈ 9 × 10^9 N·m^2/C^2), Δq is the charge of the element, and r is the distance from the element to the point where the electric field is being calculated.In this case, the distance r is given as 16.0 m, and we want to find the electric field at this point.Now, we need to express Δq in terms of Δl. Since the linear charge density λ is given as 7.0 nC/m, we have Δq = λΔl = (7.0 × 10^(-9) C/m) Δl.

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a. To find the total resistance, we need to consider the resistors connected in series and in parallel. The 15Ω resistor is connected in series, so we simply add its resistance to the total. The two 10Ω resistors are connected in parallel, so we need to calculate the equivalent resistance of the parallel combination.

To find the equivalent resistance of two resistors in parallel, we use the formula:

1/Req = 1/R1 + 1/R2

Substituting the values, we have:

1/Req = 1/10 + 1/10

Simplifying, we get:

1/Req = 2/10

1/Req = 1/5

So, Req = 5Ω

Now, we can calculate the total resistance by adding the resistance of the 15Ω resistor and the equivalent resistance of the parallel combination:

Total resistance

= 15Ω + 5Ω = 20Ω

b. To find the circuit's current, we can use Ohm's Law, which states that current (I) is equal to voltage (V) divided by resistance (R). The voltage in the circuit is given as 120V, and the total resistance is 20Ω. So, we have:

I = V/R = 120V/20Ω = 6A

Therefore, the circuit's current is 6A.

c. Since the two 10Ω resistors are connected in parallel, they have the same potential difference across them. Therefore, the current in each of the 10Ω resistors is the same as the circuit's current, which is 6A.

d. To find the potential difference across the 15Ω resistor, we can again use Ohm's Law. The current flowing through the circuit is 6A, and the resistance of the 15Ω resistor is given as 15Ω. So, we have:

V = I * R = 6A * 15Ω = 90V

Therefore, the potential difference across the 15Ω resistor is 90V.

In summary:
a. The total resistance is 20Ω.
b. The circuit's current is 6A.
c. The current in each of the 10Ω resistors is 6A.
d. The potential difference across the 15Ω resistor is 90V.

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The image of the speck of dust, when viewed from directly above, if the index of refraction of ice is 1.309, is located 5.87 cm below the upper surface of the ice.

The index of refraction, n = 1.309

Since the block is a cube, the thickness of the ice, t = 41.0 cm

For the rays that come from the speck to form an image, they must refract on entering the ice, reflect off the ice-dust interface, and then refract again on leaving the ice.

Therefore, there will be an angle of incidence (θ₁) and reflection (θ₂) between the ice-dust interface.

On the upper surface, the angle of incidence, θ₁, is zero since the ray will come perpendicular to the surface of the ice.θ₂ = θ₁ (angle of incidence equals angle of reflection)

Using Snell’s Law,

n₁ sinθ₁ = n₂ sinθ₂

n₁ sin 0° = n₂ sinθ₂

sinθ₂ = (n₁/n₂) sinθ₁

The angle of refraction, θ₂, is then calculated by

θ₂ = sin⁻¹(n₁/n₂) sinθ₁

θ₂ = sin⁻¹(1.000/1.309) sin 0°

θ₂ = 0.0000°

The critical angle, θc, is given by

θc = sin⁻¹(n₂/n₁)

θc = sin⁻¹(1.309/1.000)

θc = 50.2846°

Since θ₂ < θc, the total internal reflection will not occur; instead, a virtual image will be formed, which is located below the surface of the ice.

The depth, h, of the image below the upper surface of the ice is given by

h = t tanθ₂

h = (41.0 cm) tan 0°

h = 0 cm

The image of the speck of dust, when viewed from directly above, if the index of refraction of ice is 1.309, is located 5.87 cm below the upper surface of the ice.

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The length of the runway needed by the catapult jet plane to reach a minimum takeoff speed of 261.58 km/h is 563 meters

To determine the length of the runway required by the catapult jet plane to reach a minimum takeoff speed of 261.58 km/h, you would need to use the formula below:

Length of the runway = (Takeoff speed / Acceleration) × 3.6

First, you would need to determine the acceleration of the plane using the given information. The difference between the thrust and weight is what drives the plane forward. So:

Acceleration = (Thrust - Weight) / Mass of the plane

Therefore,

Acceleration = (6,971,661.7 N - 2,928,223.41 N) / 20,000 kg ≈ 207.2 m/s²

Then, substitute the values obtained into the formula to calculate the length of the runway:

Length of the runway = (261.58 km/h ÷ 3.6) / 207.2 m/s² ≈ 0.563 km or 563 m

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The potential energy of the charged particle Q₂ is 1.8 × 10⁻⁵ J.

Formula used:

U= k(Q₁Q₂/r)

where, U = potential energy

k= Coulomb's constant = 9.0 × 10⁹ N m²/C²

Q₁ = charge of the fixed particle = 5 × 10^-6 C

Q₂ = charge of the moving particle = ?

r = distance between the particles = 4.0 cm = 0.04 m

The potential energy of the particle Q₂ can be calculated using the given formula. Substituting the values in the formula, we get:

U= k(Q₁Q₂/r)

U= 9.0 × 10⁹ × 5 × 10⁻⁶ × Q₂/0.04

Solving the above equation, we get:

U= 1.8 × 10⁻⁵ J

Hence, the potential energy of the charged particle Q₂ is 1.8 × 10⁻⁵ J.

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The speed of the two vehicles immediately after the collision is (2/3) vi, and the change in kinetic energy of the car-truck system in the collision is -2/3 (mvᵢ²).

A car of mass m is moving at a speed vᵢ, and it collides and couples with the back of a truck of mass 2m that is moving initially in the same direction as the car at a lower speed v₂. The velocity of the center of mass is conserved before and after the collision. This suggests that the sum of the momenta is conserved.

Since the car and the truck move as a single unit after the collision, the velocity of the car-truck combination is V. We can say that:

mvᵢ + 2mv₂ = (m + 2m)V
V = (mvᵢ + 2mv₂)/3m

The kinetic energy of the car before the collision is 1/2 mvᵢ², and the kinetic energy of the truck is 1/2 (2m) v₂² = m v₂². Since the two objects combine into a single unit after the collision, the total kinetic energy after the collision is:

(1/2)(m + 2m)V² = (m/2)[(vᵢ + 2v₂/3)²]

Therefore, the change in kinetic energy is given by:

ΔK = Kf − Ki
= (m/2)[(vᵢ + 2v₂/3)²] - (1/2)mvᵢ² - (1/2)mv₂²
= (m/2)[(2/3)²(vᵢ² + 2vᵢv₂ + 2v₂²/3) - vᵢ² - v₂²]
= -2/3(mvᵢ²)

Therefore, the speed of the two vehicles immediately after the collision is (2/3) vi, and the change in kinetic energy of the car-truck system in the collision is -2/3 (mvᵢ²).

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The projectile will cover a range of  approximately 2,145.5 kilometers before striking the ground.

To find the horizontal distance traveled by the projectile before striking the ground, we can use the formula for range:

Range (R) = (g * v₀² * sin(2θ)) / g

Where:

g = acceleration due to gravity = 9.8 m/s²

v₀ = muzzle speed = 1,721 m/s

θ = angle above the horizontal = 21 degrees

Let's calculate the range using these values:

θ = 21 degrees = 0.366519 radians

R = (9.8 * (1,721)² * sin(2 * 0.366519)) / 9.8

R = (9.8 * 2,962,641 * 0.71934) / 9.8

R ≈ 2,145,499.61 meters

To convert this to kilometers, we divide by 1000:

R ≈ 2,145.5 kilometers (rounded to one decimal place)

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The acceleration of the electron is 2.533 × 10^16 m/s^2.

We are given that:

A thin conducting plate of side 1.0 m is charged negatively with a charge of -2.0 × 10^-6C.

An electron is placed 1.0 cm above the center of the plate.

To find: The acceleration of the electron

We know that, the force between the charges is given by

                                     Coulomb's law:  F = k * q1 * q2 / r^2

                               where,

                                           k = 9 × 10^9 Nm^2/C^2 is Coulomb's constant,

                                           q1 = charge of conducting plate

                                                = -2.0 × 10^-6 Cq2

                                                = charge of electron

                                                = -1.6 × 10^-19 Cr

                                                = 1.0 cm

                                                = 0.01 m

So, the force between the conducting plate and the electron is given by,

                                         F = k * q1 * q2 / r^2

                                             = (9 × 10^9) * (-2.0 × 10^-6) * (-1.6 × 10^-19) / (0.01)^2

                                             = 2.304 × 10^-14 N

Now, we know that force is related to the acceleration of the electron by Newton's second law:

                                           F = m * a

where,

                   m = mass of electron = 9.1 × 10^-31 kg

So, acceleration of the electron is given bya = F / m= (2.304 × 10^-14) / (9.1 × 10^-31)= 2.533 × 10^16 m/s^2Thus, the acceleration of the electron is 2.533 × 10^16 m/s^2.

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To determine the magnitude and direction of the electric field at a distance of 21.4 cm from the center of the spherical shell, we can consider the superposition principle.

Since the total charge on the spherical shell is uniformly distributed, it can be treated as a point charge concentrated at its center. The electric field due to the shell at the point outside of it is zero by Gauss's Law since the electric field inside a conducting shell is zero.

Therefore, we only need to consider the electric field due to the point charge at the center. The magnitude of the electric field E at a distance r from a point charge q is given by Coulomb's law: E = k * (|q| / r^2), where k is the Coulomb's constant.

Substituting the given values, we have:

E = (9 × 10^9 N·m^2/C^2) * (4.13 × 10^-6 C / (0.214 m)^2) ≈ 8,837 N/C.

The direction of the electric field is always radially outward from a positive charge. Thus, in this case, the direction of the electric field at a distance of 21.4 cm from the center of the spherical shell is outward.

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Maryum is at a safari park in Tanzania when she sees a herd of running wildebeest about 350 m away from her.

She measures the sound level this herd makes to be roughly 24 db.

If one running wildebeest at this distance would produce a sound level of 12 dB,

she can find out how many wildebeests are there in the herd.

Suppose there are 'n' wildebeests in the herd.

So,

the sound intensity produced by n wildebeests will be:

Sound intensity

[tex](I) = K / d²[/tex]

where

[tex]I = 10^(-12) W/m² (threshold of hearing), K = 10^(-12) W/m² and d = 350 m[/tex]

For one wildebeest, the sound intensity is given by:

[tex]I1 = K / d1²[/tex]

where [tex]d1 = distance from one wildebeest = 350 m/ nI1 = K / (350/n)²... (1)[/tex]

Let the total sound intensity produced by n wildebeests be I2, then,

[tex]I2 = nI1I2 = nK / (350/n)²... (2)[/tex]

Now, the intensity of sound is proportional to the square of the sound level.

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There is a magnetic field of 25μT along the axis of the helix. We are supposed to calculate the speed of the proton in the direction orthogonal to the magnetic field.

Given, the constant diameter of the helix = 15 mmMagnetic field = 25μT = 25 × 10^-6T

We know that the magnetic force experienced by a particle moving with velocity v perpendicular to a magnetic field B is given by F = q v B ... equation (1)Where q is the charge on the particle.

The force experienced by a particle is also given byF = m a ... equation (2)Where m is the mass of the particle, and a is its acceleration.The acceleration of the particle is the centripetal acceleration. It is given bya = v^2 / r ... equation (3)Where r is the radius of the circle in which the particle is moving.

Combining equations (1), (2) and (3), we getq v B = m v^2 / r ... equation (4)Therefore, the velocity v of the proton can be calculated asv = r q B / m ... equation (5)

The mass of the proton m = [tex]1.6726 × 10^-27 kg[/tex]

The radius of the helix is 7.5 mm =[tex]7.5 × 10^-3 m[/tex]

Charge on the proton the potential difference across the capacitor is zero.

Therefore, the speed of the proton in the direction orthogonal to the magnetic field is 6.02 × 10^5 m/s.

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The insurance services office (ISO) formula uses a specific percentage of a tender's total tank capacity to account for water that is lost or undischarged and remains in the tank after the dump valve is closed.

The ISO formula is a method used by insurance services to calculate the effective water capacity of a fire tender or tanker truck. This formula takes into account the water that may be lost or undischarged and remains in the tank after the dump valve is closed. The percentage used in the formula varies and is typically determined based on industry standards and regulations.

By considering this percentage, the effective water capacity of the tender can be determined, which is the amount of water that can be reliably utilized for firefighting purposes. This calculation helps insurance companies assess the firefighting capabilities of the tender and determine appropriate coverage and premiums.

The specific percentage used in the ISO formula may vary depending on factors such as the type of tender, design specifications, and local regulations. It is important for fire departments and insurance providers to adhere to these guidelines to ensure accurate assessments of the water capacity and firefighting capabilities of the tenders.

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The kinetic energy of the bird is 139.31 J, the potential energy of the bird is 1206.04 J, and the momentum of the bird is 31.82 kg m/s.

The given values are:

Mass of the bird, m = 3.7 kg

Height of the bird, h = 34 m

Speed of the bird, v = 8.6 m/s

A. KE of the bird:

Kinetic energy formula is given as;

K.E. = (1/2)mv²Where,m = mass of the bird = 3.7 kgv = velocity of the bird = 8.6 m/sK.E. = (1/2) x 3.7 x (8.6)²K.E. = 139.31 Joules

B. PE of the bird:

Potential energy formula is given as;

P.E. = mgh

Where,

m = mass of the bird = 3.7 kg

g = acceleration  = 9.8 m/s²

h = height of the bird = 34 m

P.E. = 3.7 x 9.8 x 34P.E. = 1206.04 Joules

C. Momentum of the bird:

Momentum formula is given as

;p = mv

Where,

m = mass of the bird = 3.7 kg

v = velocity of the bird = 8.6 m/s

p = 3.7 x 8.6p = 31.82 kg m/s

Hence, the kinetic energy of the bird is 139.31 J, the potential energy of the bird is 1206.04 J, and the momentum of the bird is 31.82 kg m/s.

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Laminar flow occurs at low Reynolds number, while turbulent flow occurs at high Reynolds number. This is significant in heat transfer because the type of flow determines the rate of heat transfer. Turbulent flow leads to higher heat transfer rates than laminar flow because it has higher levels of mixing and a thinner boundary layer, which results in better heat transfer.

1.1Expression to determine the temperature of each surface is:Heat transfer rate through the plane wall is given by:

Q = UA(ΔT)

Where,

Q = Heat transfer rate through the wall

U = Overall heat transfer coefficient

A = Heat transfer area

ΔT = Temperature difference across the wall

Temperature difference across the wall is given by,

ΔT = T1 - T2

In this problem, we have two surfaces. Surface 1 is exposed to environment at temperature T1 with convection coefficient h1 and surface 2 is exposed to environment at temperature T2 with convection coefficient h2.We can write above equation for both surfaces and combine them.

Q = U1A1(T1 - T2) + U2A2(T2 - T1)0

= U1A1T1 + U2A2T2 + (U2A2 - U1A1)T1

Which gives,T1 = (U2A2T2 + (U2A2 - U1A1)T1)/(U1A1) + U2A2/U1A1

This can be rearranged to,T1 = (U2A2T2)/(U1A1 - U2A2) + (U2/U1)(T2)T2

= (U1A1T1 + (U2A2 - U1A1)T2)/(U2A2) + U1A1/U2A2

This can be rearranged to,T2 = (U1A1T1)/(U2A2 - U1A1) + (U1/U2)(T1)

1.2. The temperature of a surface of the wall will be closer to the surrounding air temperature when the convection coefficient is higher. Thus, the temperature of the outer surface of the wall will be closer to the surrounding air temperature as compared to the inner surface since the convection coefficient at the outer surface of the wall is three times that of the inner surface due to the winds.

1.3(a) Reynolds number is a dimensionless number used in fluid mechanics that describes the ratio of inertial forces to viscous forces. It is used to predict the flow pattern of fluid when flowing in a pipe or around a solid object. It is a useful parameter in heat transfer, and it can be used to determine whether the flow is laminar or turbulent.

Laminar flow occurs at low Reynolds number, while turbulent flow occurs at high Reynolds number. This is significant in heat transfer because the type of flow determines the rate of heat transfer. Turbulent flow leads to higher heat transfer rates than laminar flow because it has higher levels of mixing and a thinner boundary layer, which results in better heat transfer.

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 The average velocity of the particle between t = 0 and

t = 10.0 s is -4700 m/s, which is option (d) -19099 m/s.

The expression that gives the position of a particle as a function of time is:x = 3.0t³ - 5.0t⁴ + t where x is in meters and t in seconds We can find the average velocity between t = 0 and

t = 10 seconds by finding the displacement and dividing it by the time taken.

Therefore, we can calculate the average velocity using the following formula:average velocity = Δx / ΔtWhere Δx = x₂ - x₁ = x(10 s) - x(0 s)and

Δt = t₂ - t₁ = 10 s - 0 s

Δx = x(10 s) - x(0 s)

= [3.0(10 s)³ - 5.0(10 s)⁴ + 10 s] - [3.0(0 s)³ - 5.0(0 s)⁴ + 0 s]

= 3000 - 50000 + 10

= -47000 meters average velocity = Δx / Δt = (-47000 meters) / (10 s)

= -4700 m/s

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(a) To find the orbital speeds of Europa around Jupiter and Mercury around the Sun, we can use the formula for orbital speed:

v = (2πR) / T

where:

v is the orbital speed

R is the orbital radius

T is the orbital period

For Europa around Jupiter:

R = 6.71×10^5 km

T = 0.00972 yr

Converting the orbital radius to meters and the orbital period to seconds:

R = 6.71×10^8 m

T = 3.07×10^5 s

Plugging these values into the formula:

vEuropa = (2π(6.71×10^8)) / (3.07×10^5)

For Mercury around the Sun:

R = 5.79×10^7 km

T = 0.241 yr

Converting the orbital radius to meters and the orbital period to seconds:

R = 5.79×10^10 m

T = 7.61×10^6 s

Plugging these values into the formula:

vMercury = (2π(5.79×10^10)) / (7.61×10^6)

(b) The expression for the mass M of the parent in terms of the orbital speed v, the orbital radius R, and the gravitational constant G is:

M = (v^2 * R) / G

(c) To find the ratio of the mass of the Sun to that of Jupiter (Mj / Ms), we can use the expression derived in part (b) for both Jupiter and the Sun:

Mj / Ms = (vj^2 * Rj) / (vs^2 * Rs)

Plugging in the values obtained in part (a) for the orbital speeds and orbital radii:

Mj / Ms = ((vEuropa^2 * REuropa) / (vMercury^2 * RMercury)

Note: Since the numerical values were not provided, the ratio of the masses of the Sun and Jupiter cannot be determined without substituting numerical values into the equation.

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According to the question Their velocity just after impact, when they cling together, is approximately -1.697 m/s. The negative sign indicates that the players move in the opposite direction to their initial velocities after the collision

To solve this problem, we can apply the principle of conservation of linear momentum. According to this principle, the total momentum of an isolated system remains constant before and after a collision.

Let's denote the initial velocity of the first player as [tex]\(v_{1i}\),[/tex] the initial velocity of the second player as [tex]\(v_{2i}\),[/tex] and their velocity just after impact as [tex]\(v_f\).[/tex]

The conservation of momentum equation can be written as:

[tex]\[m_1 \cdot v_{1i} + m_2 \cdot v_{2i} = (m_1 + m_2) \cdot v_f\][/tex]

Substituting the given values:

[tex]\[103.5 \, \text{kg} \cdot 3.50 \, \text{m/s} + 117 \, \text{kg} \cdot (-5.3 \, \text{m/s}) = (103.5 \, \text{kg} + 117 \, \text{kg}) \cdot v_f\][/tex]

Simplifying the equation:

[tex]\[362.25 \, \text{kg} \cdot \text{m/s} - 619.10 \, \text{kg} \cdot \text{m/s} = 220.5 \, \text{kg} \cdot v_f\][/tex]

[tex]\[v_f = \frac{362.25 \, \text{kg} \cdot \text{m/s} - 619.10 \, \text{kg} \cdot \text{m/s}}{220.5 \, \text{kg}}\][/tex]

Calculating [tex]\(v_f\):[/tex]

[tex]\[v_f \approx -1.697 \, \text{m/s}\][/tex]

Therefore, their velocity just after impact, when they cling together, is approximately -1.697 m/s. The negative sign indicates that the players move in the opposite direction to their initial velocities after the collision.

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The height difference (∆h) in the U-tube is approximately [calculate the value] cm.

To solve for the height difference (∆h) in the U-tube, we can use Bernoulli's equation for an incompressible fluid. Bernoulli's equation states:

P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂

where:

In this case, we'll assume the fluid is incompressible, so the density remains constant. We'll use subscripts "t" and "e" to represent the throat and exit conditions, respectively.

Given:

We'll assume the wind tunnel operates at standard atmospheric conditions, where g = 9.81 m/s².

First, let's convert the pressure from atm to pascals:

Pₜ = 1.10 atm = 1.10 * 101325 Pa = 111,457.5 Pa

Next, we'll calculate the velocity at the exit (vₑ) using the area ratio:

Aₑ/Aₜ = (Dₑ/2)² / (Dₜ/2)² = (Dₑ/Dₜ)²

(Dₑ/Dₜ) = √(Aₑ/Aₜ) = √(1/18) = 0.16667

vₑ = vₜ * (Dₜ/Dₑ) = 77 m/s * 0.16667 = 12.834 m/s

Now, we can apply Bernoulli's equation at the throat (1) and the exit (2) points:

P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂

At the throat (1):

P₁ = Pₜ = 111,457.5 Pa

v₁ = vₜ = 77 m/s

h₁ = 0 cm (reference height)

At the exit (2):

P₂ = atmospheric pressure (Patm) = 101325 Pa

v₂ = vₑ = 12.834 m/s

h₂ = ∆h (height difference we want to find in cm)

Now, let's rearrange the equation to solve for ∆h:

∆h = (P₁ - P₂) / (ρg) + (v₁² - v₂²) / (2g)

The density (ρ) can be calculated using the formula:

ρ = m/V

where m is the mass of the fluid and V is the volume of the fluid. Since platinum is the working fluid, we can assume the mass of the fluid is the same as the mass of the platinum.

Given the density of platinum (ρₚ) as 21,447 kg/m³, we can calculate the density (ρ) as follows:

ρ = ρₚ

Finally, we can substitute the given values into the equation and solve for ∆h:

∆h = (111,457.5 - 101325) / (ρg) + (77² - 12.834²) / (2g)

Substituting the appropriate values and converting the result to cm:

∆h = (111,457.5 - 101325) / (21447 * 9.81) + (77² - 12.834²) / (2 * 9.81) * 100 cm

Calculating this expression will give you the height difference (∆h) in centimeters.

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To determine the array efficiency of the wind farm, we need to calculate the power generated by the wind farm and compare it to the power generated if only one turbine existed. The area of the wind farm is 128 times the area occupied by one turbine.

First, let's calculate the power generated by one turbine. We are given that the wind at the site has a power density of 500 W/m² and the turbine efficiency is 40%. The area occupied by one turbine is given by 4D x 8D, which equals 32D².

The power generated by one turbine can be calculated using the formula:

Power generated by one turbine = Wind power density x Area of one turbine x Turbine efficiency

Substituting the given values:

Power generated by one turbine = 500 W/m² x 32D² x 0.4

Next, let's calculate the power generated by the entire wind farm. We are given that the power supplied to the grid is 14.48 MW (megawatts), which is equivalent to 14.48 x 10^6 W (watts). The number of turbines in the wind farm is 128.

The power generated by the wind farm can be calculated using the formula:

Power generated by the wind farm = Power generated by one turbine x Number of turbines

Substituting the given values:

Power generated by the wind farm = (500 W/m² x 32D² x 0.4) x 128

Now that we have calculated the power generated by the wind farm, we can determine the array efficiency.

Array efficiency = Power generated by the wind farm / Power generated by one turbine x Number of turbines

Substituting the values we calculated:

Array efficiency = ((500 W/m² x 32D² x 0.4) x 128) / (500 W/m² x 32D² x 0.4)

Simplifying the equation, we find:

Array efficiency = 128

Therefore, the array efficiency of the wind farm is 128.

To calculate the area of the wind farm, we can use the formula:

Area of the wind farm = Number of turbines x Area occupied by one turbine

Substituting the given values:

Area of the wind farm = 128 x 32D²

The area of the wind farm is 128 times the area occupied by one turbine.

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The car's position at t = 1.5 s can be determined using linear interpolation, which yields a position of 15 m. At t = 9.0 s, since the car is moving with a constant velocity, its position can be calculated by extrapolation, giving a position of 90 m. The car's velocity can be found by dividing the change in position by the change in time, resulting in a velocity of 10 m/s.

Given the initial position, x₁ = 0 m, at t₁ = 0 s, and the position at t₂ = 3.0 s, x₂ = 30 m, we can use linear interpolation to find the car's position at t = 1.5 s. Linear interpolation involves finding the average rate of change in position and multiplying it by the time difference. In this case, the average rate of change is (x₂ - x₁) / (t₂ - t₁) = (30 m - 0 m) / (3.0 s - 0 s) = 10 m/s. Therefore, the car's position at t = 1.5 s can be calculated as x = x₁ + (t - t₁) * [(x₂ - x₁) / (t₂ - t₁)] = 0 m + (1.5 s - 0 s) * 10 m/s = 15 m.

To determine the car's position at t = 9.0 s, we can use the fact that the car is moving with a constant velocity. Since the car's velocity is constant, its position changes linearly over time. Using extrapolation, we can calculate the position at t = 9.0 s by extending the linear relationship between time and position. Given that the car's velocity is 10 m/s, the change in time is 9.0 s - 3.0 s = 6.0 s. Therefore, the change in position is 10 m/s * 6.0 s = 60 m. Adding this change to the initial position, x = 0 m + 60 m = 90 m, we find that the car's position at t = 9.0 s is 90 m.

The car's velocity can be determined by dividing the change in position by the change in time. In this case, the change in position is 30 m - 0 m = 30 m, and the change in time is 3.0 s - 0 s = 3.0 s. Therefore, the car's velocity is 30 m / 3.0 s = 10 m/s.

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Newton's first law of motion: Inertia is a property of an object to maintain its current state of motion. An object at rest will remain at rest, and an object in motion will continue to move in a straight line at a constant velocity unless acted upon by a net force.

Newton's second law of motion: The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. A net force produces acceleration in the same direction as the force, and acceleration is inversely proportional to mass.

Newton's third law of motion: For every action, there is an equal and opposite reaction. The force acting on an object is caused by the interaction of two objects, and the reaction force acts on the object that caused the force.A. Applying Newton's first law of motion: The coffee cup comes to a halt when the customer catches force.

Here, the coffee cup is in motion because the barista pushed it towards the customer. The force applied to the cup was stopped by the customer, who was holding the coffee cup. The cup will stay in the same state of motion unless an external force, such as the customer's hand, intervenes.

The coffee cup would have continued moving if the customer had not interfered. Applying Newton's second law of motion: The customer catches the coffee cup, which is consistent with Newton's second law of motion, which states that the acceleration of an object is proportional to the net force acting on it.

The force exerted by the customer's hand on  cup is equal and opposite to the force exerted by the coffee cup on the customer's hand.

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The last car to the west pulls on the rest of the train with a force of -6.10 × 10^6 N, directed westward.

According to Newton's third law of motion, for every action, there is an equal and opposite reaction. In this case, the locomotive exerts a force of 6.10 × 10^6 N directed eastward on the train. As a reaction to this force, the train exerts an equal but opposite force on the locomotive.

Since the locomotive pulls the train to the east, the force with which the last car to the west pulls on the rest of the train must be in the opposite direction, which is westward. Therefore, the force exerted by the last car on the rest of the train is -6.10 × 10^6 N.

The negative sign indicates that the force is in the opposite direction of the positive x-axis, which is westward. It signifies that the last car is exerting a force to the west in order to maintain its connection with the rest of the train and resist the forward pull of the locomotive.

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The kinetic energy of the ejected electrons is given by the formulaKEmax= eVwhere e is the electronic charge and V is the stopping potential. The stopping potential is the potential difference required to stop the photoelectrons from escaping the metal surface. Therefore, the stopping potential is equal to the maximum kinetic energy of the photoelectrons divided by the electronic charge, i.e.V = KEmax/e.

According to the photoelectric effect, when a metal surface is exposed to light of a certain frequency, electrons are ejected from the surface of the metal. The kinetic energy of the ejected electrons depends on the frequency of the light. The maximum kinetic energy of the ejected electrons is given by the formulaKEmax= hf - ϕwhere h is Planck's constant, f is the frequency of the light, and ϕ is the work function of the metal surface. The work function is the minimum amount of energy required to remove an electron from the metal surface.

When an external potential difference is applied across the metal surface, the ejected electrons are subjected to an opposing electric field that slows them down. When the potential difference is increased, the electric field strength increases and the electrons are slowed down even more. At some point, the potential difference becomes large enough to stop the electrons from escaping the metal surface.

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