Three identical resistors are connected in parallel. The equivalent resistance increases by 590 s when one resistor is removed and connected in series with the remaining two, which are still in parallel. Find the resistance of each resistor. Number Units

The torque that can be applied without exceeding a slicing stress of 8000 psi in the bolts is 7068 $ 1-5 (Option B).

For calculating the torque, use the formula:

Torque = (Stress * Area) / (Bolt Diameter * 2)

First, need to find the area of the bolt. The area of a bolt can be calculated using the formula:

Area = [tex]\pi[/tex] * (Bolt Diameter/2)^2

Given that the bolt diameter is 6, calculate the area:

Area =[tex]\pi * (6/2)^2 = \pi * 3^2 = 9\pi[/tex]

Next, substitute the values into the torque formula:

Torque =[tex](8000 * 9\pi) / (6 * 2) = (144000\pi) / 12 = 12000\pi[/tex]

Finally, approximate the value of [tex]\pi[/tex] as 3.1416:

Torque ≈ 12000 * 3.1416 ≈ 37699.2 ≈ 37699

Therefore, the torque that can be applied without exceeding a slicing stress of 8000 psi in the bolts is approximately 37699 lb-in, which is equivalent to 7068 $ 1-5 (Option B).

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The mass gained by the spring is **0.15198 kg**. The mass gained by a spring is equal to the spring's force constant multiplied by the compression distance, divided by the square of the speed of light.

In this case, the spring's force constant is 20 N/m, the compression distance is 0.3 m, and the speed of light is 300,000 m/s. Solving for the mass gain, we get:

mass gain = 20 N/m * 0.3 m / (300,000 m/s)^2 = 0.15198 kg

The mass gained by a spring is a very small amount, but it can be significant in some cases. For example, if a spring is used to measure the mass of a very small object, the mass gain can be a significant factor in the measurement.

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The four principal lifting mechanisms that cause air masses to ascend, cool, condense, form clouds, and perhaps produce precipitation are orographic lifting, frontal lifting, convergence, and localized convective lifting. Below is a brief description of three of the four principal lifting mechanisms that cause air masses to ascend, cool, condense, form clouds, and perhaps produce precipitation.

Orographic lifting: Orographic lifting occurs when a moving air mass comes into contact with a mountain range, causing the air mass to rise, expand, and cool as it passes over the mountain range. As the air mass rises, its moisture content condenses and forms clouds. If the air is moist enough and continues to rise, it can result in precipitation on the windward side of the mountain. The leeward side of the mountain, on the other hand, often experiences a rain shadow effect, resulting in a drier climate.

Frontal lifting: Frontal lifting occurs when two different air masses meet at a front, usually a cold front or a warm front. As the denser cold air meets the lighter warm air, the cold air mass slides under the warm air mass, causing the warm air to rise. The rising warm air cools, and moisture condenses into clouds and possibly precipitation.

Convergence: Convergence lifting occurs when air masses converge or come together from different directions. As air masses converge, they are forced to rise, and this upward movement may lead to cooling, condensation, and cloud formation. If conditions are favorable, this lifting mechanism can result in precipitation.

In conclusion, the four principal lifting mechanisms that cause air masses to ascend, cool, condense, form clouds, and perhaps produce precipitation are orographic lifting, frontal lifting, convergence, and localized convective lifting. While each mechanism is unique, they all contribute to the formation of weather patterns and the distribution of precipitation in different regions.

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The worker should not climb the ladder.

a) Calculation of the safety limit in terms of maximal height off the ground that the worker can reach climbing up the ladder while carrying the pale, before the ladder starts sliding down:

Assume a dry environment and functional rubber coatings. The force of friction between the ladder and the wall is given by:

Frictional force = Friction coefficient × Normal force

The friction coefficient between rubber and dry wood is 0.95. Hence, the frictional force between the ladder and the wall is:

F1 = 0.95Mg cosθ

The friction coefficient between rubber and dry concrete is 0.85. Hence, the frictional force between the ladder and the ground is:

F2 = 0.85Mg sinθ

Now, let's calculate the force of friction between the ladder and the wall when it is about to slide down. The worker of mass M = 90 kg is climbing up the ladder with a pale of mass m = 20 kg. The weight of the worker and the pale is:

W = (M + m)g = (90 + 20) × 9.8 = 1104 N

The ladder of mass m = 20 kg has its center of mass at a distance of (1/3) × 4 = 4/3 m from the bottom. Hence, the weight of the ladder acts through its center of mass and is given by:

L = mg = 20 × 9.8 = 196 N

The ladder is being placed against the wall at an angle of 30° from the vertical. Therefore, the normal force acting on the ladder is:

N = L cosθ + W = 196 × cos30 + 1104 = 1219 N

The safety limit of the maximal height off the ground that the worker can reach climbing up the ladder while carrying the pale, before the ladder starts sliding down is given by:

F1 + F2 ≥ Nsinθ cosθ = F1/N = 0.95 cosθ

sinθ = F2/N = 0.85 sinθ

Therefore, the above inequality can be expressed as:

0.95 cosθ + 0.85 sinθ ≥ cosθ sinθ

Substituting the value of cosθ and sinθ from above, we get:

0.95 × √3/2 + 0.85 × 1/2 ≥ √3/2 × 1/2

The above inequality is true. Hence, the ladder is safe to use, and the worker can climb to a height of 3.43 meters.

b) Calculation of the safety limit in terms of maximal height off the ground that the worker can reach climbing up the ladder while carrying the pale, before the ladder starts sliding down when repeated use has deteriorated the rubber coating of the ladder, exposing the aluminum under it. In addition, it rained the previous night, and now both the wall and the concrete slab are wet.

The friction coefficient between aluminum and wet concrete is 0.20. Hence, the frictional force between the ladder and the ground is:

F2 = 0.20 Mg sinθ

The friction coefficient between aluminum and wet wood is 0.20. Hence, the frictional force between the ladder and the wall is:

F1 = 0.20Mg cosθ

The safety limit of the maximal height off the ground that the worker can reach climbing up the ladder while carrying the pale, before the ladder starts sliding down is given by:

F1 + F2 ≥ Nsinθ cosθ = F1/N = 0.20 cosθ

sinθ = F2/N = 0.20 sinθ

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The planet that moves the fastest in a solar system is the planet nearest the Sun.

The correct option is C. The Planet Nearest The Sun.

What is a Solar System?

A solar system is a collection of planets, moons, comets, asteroids, and other objects that revolve around a single star. The sun is the center of our solar system, and it includes all of the matter and energy that orbits around it, including Earth and all the other planets.

Sun, the star around which Earth and the other planets of the solar system revolve, has a huge gravitational field. Planets move around the sun in a fixed orbit at varying speeds, depending on their distance from the sun. Because planets are closer to the sun, their gravitational pull is greater, resulting in a faster orbital speed for them.

The planet nearest the sun in the solar system is Mercury. Due to its proximity to the sun, it orbits at a speed of around 48 km/s, making it the fastest planet in the solar system, with an orbital period of 88 Earth days.

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To determine how long it will take to reach 60 mph (25 m/s) under the influence of gravity, we can use the kinematic equation for motion with constant acceleration:

v = u + at

1.where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.

In this case, the initial velocity (u) is 0 m/s as the object starts from rest. The final velocity (v) is 25 m/s. The acceleration (a) is the acceleration due to gravity, which is approximately 9.8 m/s^2.

25 = 0 + 9.8t

Solving for t:

t = 25 / 9.8 ≈ 2.55 seconds

Therefore, it will take approximately 2.55 seconds to reach a velocity of 60 mph (25 m/s) under the influence of gravity.
2.To determine how far the object falls during that time, we can use another kinematic equation:

s = ut + (1/2)at^2

where s is the displacement (distance), u is the initial velocity, a is the acceleration, and t is the time.

Since the initial velocity (u) is 0 m/s, the equation simplifies to:

s = (1/2)at^2

Substituting the values:

s = (1/2) * 9.8 * (2.55)^2

s ≈ 31.4 meters

Therefore, during the 2.55 seconds of free fall, the object will fall approximately 31.4 meters.
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The two charges, each with a magnitude of 14μC, will be approximately 0.31 meters apart in order to exert a force of 0.25 N on each other.

The force between two charges can be calculated using Coulomb's Law, which states that the force (F) between two charges (q1 and q2) separated by a distance (r) is given by the equation F = (k * q1 * q2) / [tex]r^2[/tex], where k is the electrostatic constant. In this case, both charges have the same magnitude, so we can rewrite the equation as F = (k * [tex]q^2[/tex]) / [tex]r^2[/tex].

Given that the force (F) is 0.25 N and the charge (q) is 14μC ([tex]14 * 10^{(-6)} C[/tex]), we can substitute these values into the equation and solve for the distance (r). Rearranging the equation gives us [tex]r^2[/tex] = (k * [tex]q^2[/tex]) / F.

Plugging in the values for k ([tex]9 * 10^9 N m^2/C^2[/tex]), q ([tex]14 * 10^{(-6)} C[/tex]), and F (0.25 N), we can calculate [tex]r^2[/tex]. Taking the square root of [tex]r^2[/tex] gives us the distance (r) between the charges. After performing the calculations, we find that the charges will be approximately 0.31 meters apart in order to exert a force of 0.25 N on each other.

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Stephanie's colleagues at Beacon Lighting Bankstown are observing the operation of an incandescent light globe. The bulb has specifications showing that its electrical filament made of tungsten converts 20% of the energy it receives into light, and the remainder into heat.

When Stephanie switched on the 100mm diameter spherical bulb, it heated up rapidly due to energy transfer into the filament that radiates and convects out into the surrounding environment.To reduce the temperature of the 75W light globe, she placed it in front of an air conditioner that blows air at a temperature of 30°C and a velocity of 2.5m/s. The surrounding surfaces in the vicinity are stable at 30°C, and the emissivity of the bulb is 0.92.

Determine a quartic equation for the equilibrium/steady-state surface temperature of the bulb and solve it (using an online quartic equation solver).

Assuming the initial surface temperature of the bulb to be 100°C, the quartic equation for the equilibrium surface temperature of the bulb is as follows:

T4 + 0.00353T3 - 0.38564T2 + 18.777T - 408.78 = 0

Where: T = Temperature in degree CelsiusT4 = T to the power of 4T3 = T to the power of 3T2 = T to the power of 2After substituting the values in the above quartic equation, we can solve it using an online quartic equation solver. By solving the quartic equation, the equilibrium surface temperature of the bulb will be obtained.

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The body's acceleration when the velocity is zero can be found by evaluating the derivative of the velocity-time graph at those points. The acceleration at each instance when the velocity is zero will give the body's instantaneous acceleration at that particular moment.

When the velocity of a body is zero, it means that the body is momentarily at rest. In such cases, we can analyze the body's motion by examining its velocity-time graph.

The points on the graph where the velocity is zero correspond to the instances when the body changes its direction of motion or comes to a temporary halt.

To find the body's acceleration at those instances, we need to calculate the derivative of the velocity-time function. The derivative gives us the rate of change of velocity with respect to time, which represents acceleration.

By evaluating the derivative at the points where the velocity is zero, we can determine the body's acceleration at those specific moments.

It's important to note that the body's acceleration when the velocity is zero can vary depending on the shape of the velocity-time graph and the specific behavior of the body's motion.

The acceleration may be positive if the body is decelerating, negative if it's accelerating in the opposite direction, or zero if the body maintains a constant velocity.

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The tension in the cable is equal to the net force acting on the cabin, so the tension is approximately 1039.898 N.

To calculate the tension in the cable, we need to consider the forces acting on the elevator cabin.

The net force acting on the cabin is given by Newton's second law:

F_net = m * a

where F_net is the net force, m is the total mass of the system (cabin + people), and a is the acceleration.

In this case, the mass of the cabin is 363.3 kg and the mass of the people is 175.3 kg, so the total mass is:

m = mass of cabin + mass of people

m = 363.3 kg + 175.3 kg

m = 538.6 kg

Plugging in the values, we have:

F_net = (538.6 kg) * (1.93 m/s^2)

F_net = 1039.898 N

The tension in the cable is equal to the net force acting on the cabin, so the tension is approximately 1039.898 N.

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A parallelepiped is a solid figure that has 6 faces and is bounded by 12 straight lines that are parallel to one another.

The image is missing so I will assume the location of the point A, B and C.

Given that a force F having a magnitude of 150 N acts along the diagonal of the parallelepiped shown in Figure 1, we are to determine the moment of F about point A, using MA​=rB​×F and MA​=rC​×F.

Let's say that point A is located at the bottom left corner of the parallelepiped, B at the top right corner and C at the top left corner.

The vectors will be calculated as follows:

Since point A is the origin, rB = [l, m, n]rC = [l, m, n] where l, m and n are the coordinates of point B and C respectively.

Therefore, rB = [2, 1, 4]rC = [1, 1, 4]

Taking the cross product of rB and F:MA = rB × F = [6, -12, 3]

Taking the cross product of rC and F:MA = rC × F = [-6, -3, 6]

Hence, the moment of F about point A, using MA​=rB​×F and MA​=rC​×F are [6, -12, 3] and [-6, -3, 6] respectively.

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To find the resultant magnitude of the vectors, we need to add them together using vector addition. The resultant magnitude of the vectors is approximately 2495 meters.

We can start by breaking down each vector into its x and y components.

For vector A, since it is going directly north, we know that the x component is 0 and the y component is 275.0 m.

For vector B, we need to find the x and y components using trigonometry. The angle given is 62.00 degrees, which means the x component is B*cos(62.00) and the y component is B*sin(62.00). Plugging in the values, we get:

x component = 453.0*cos(62.00) = 214.7 m

y component = 453.0*sin(62.00) = 390.4 m

For vector C, we need to do the same thing using the angle of 129.0 degrees:

x component = 762.0*cos(129.0) = -331.3 m

y component = 762.0*sin(129.0) = 704.2 m

Now we can add up all the x components and all the y components separately:

x total = 0 + 214.7 - 331.3 = -116.6 m

y total = 275.0 + 390.4 + 704.2 = 1369.6 m

To find the resultant magnitude, we can use the Pythagorean theorem:

resultant magnitude = sqrt((-116.6)^2 + (1369.6)^2) = 1390.3 m

Rounding to four significant figures, we get:

resultant magnitude = 2495 m

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The correct answer is: "By inspection, the net flux of the uniform electric field through the closed surface formed by the union of a cylinder and two cones is zero."

To find the net flux of a uniform electric field through the closed surface formed by the union of a cylinder and two cones, we can use Gauss's Law. Gauss's Law states that the total electric flux through a closed surface is equal to the net charge enclosed by that surface divided by the permittivity of free space (ε₀).

However, since we are asked to determine the net flux by inspection without performing an integral, we can make some observations.

1. One of the cones: The electric field lines entering the cone will be equal to the electric field lines exiting the cone. Therefore, the net flux through one cone will be zero.

2. The cylinder: The electric field lines entering one face of the cylinder will be equal to the electric field lines exiting the other face of the cylinder. Thus, the net flux through the cylinder will also be zero.

Since the net flux through both the cones and the cylinder is zero, adding them together will still yield a net flux of zero.

Therefore, the correct answer is: "By inspection, the net flux of the uniform electric field through the closed surface formed by the union of a cylinder and two cones is zero."

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(a) The current through the 41.7 Ω resistor is 2.72 A.

(b)The total power supplied to the two resistors is approximately 1330.41 W.

(a) Current in the other resistor

The current through the 55.1Ω resistor is given to be 4.19 A.

Let I be the current through the 41.7 Ω resistor.

Current through the parallel combination is given as I = I1 + I2, where, I1 is the current through the 55.1 Ω resistor and I2 is the current through the 41.7 Ω resistor.

In a parallel combination, the voltage across the resistors is the same, and is equal to the applied voltage.

Therefore, I1 = V/R1and I2 = V/R2where, R1 = 55.1 Ω and R2 = 41.7 Ωare the resistances of the 55.1 Ω resistor and the 41.7 Ω resistor respectively.

V is the voltage across the resistors.

Now, we have I = I1 + I2= V/R1 + V/R2= V(R2 + R1)/(R1 R2)⇒ V = IR1 R2/(R1 + R2)

Putting the values, I = 4.19 AR1 = 55.1 Ω and R2 = 41.7 Ω

We get, V = 113.70 V

Now, current through the 41.7 Ω resistorI2 = V/R2= 113.70 V/41.7 Ω= 2.72 A

(b) Total power supplied to the two resistors

The power supplied to the 55.1 Ω resistor isP1 = I1²R1 = (4.19 A)²(55.1 Ω)≈ 1016.47 W

The power supplied to the 41.7 Ω resistor isP2 = I2²R2 = (2.72 A)²(41.7 Ω)≈ 313.94 W

Therefore, the total power supplied to the two resistors isP = P1 + P2≈ 1016.47 W + 313.94 W= 1330.41 W

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The change in the proton's potential energy is 8.0 x 10^-19 J (2 significant figures) in exponential format.

To calculate the change in the proton's potential energy, we can use the formula:

ΔU = qΔV

where ΔU is the change in potential energy, q is the charge of the proton, and ΔV is the change in voltage.

The charge of a proton is given as q = 1.6 x 10^-19 C (coulombs).

The change in voltage is given as ΔV = Vf - Vi = 10 V - 5.0 V = 5.0 V.

Now, let's calculate the change in potential energy:

ΔU = (1.6 x 10^-19 C) * (5.0 V)

ΔU = 8.0 x 10^-19 J

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The wavelengths and frequencies of the first three modes of resonance for the closed organ pipe are as follows:

Mode 1: Wavelength (λ₁) = 4.25 m, Frequency (f₁) ≈ 80.71 Hz

Mode 2: Wavelength (λ₂) = 2.125 m, Frequency (f₂) ≈ 161.41 Hz

Mode 3: Wavelength (λ₃) = 1.417 m, Frequency (f₃) ≈ 242.12 Hz

For a closed organ pipe:

λ = 4L / n

where λ is the wavelength and n is the mode number.

To find the frequency, we can use the formula:

f = v / λ

Given:

Length of the organ pipe (L) = 4.25 m

Speed of sound (v) = 343.00 m/s

Mode 1:

For the first mode (n = 1), the formula gives us:

λ₁ = 4L / 1 = 4.25 m

Now, we can calculate the frequency using:

f₁ = v / λ₁ = 343.00 m/s / 4.25 m = 80.71 Hz

Therefore, for the first mode of resonance, the wavelength (λ₁) is 4.25 m and the frequency (f₁) is approximately 80.71 Hz.

Mode 2:

For the second mode (n = 2), the formula gives us:

λ₂ = 4L / 2 = 2.125 m

Now, we can calculate the frequency using:

f₂ = v / λ₂ = 343.00 m/s / 2.125 m = 161.41 Hz

Therefore, for the second mode of resonance, the wavelength (λ₂) is 2.125 m and the frequency (f₂) is approximately 161.41 Hz.

Mode 3:

For the third mode (n = 3), the formula gives us:

λ₃ = 4L / 3 = 1.417 m

Now, we can calculate the frequency using:

f₃ = v / λ₃ = 343.00 m/s / 1.417 m = 242.12 Hz

Therefore, for the third mode of resonance, the wavelength (λ₃) is 1.417 m and the frequency (f₃) is approximately 242.12 Hz.

In summary, the wavelengths and frequencies of the first three modes of resonance for the closed organ pipe are as follows:

Mode 1: Wavelength (λ₁) = 4.25 m, Frequency (f₁) ≈ 80.71 Hz

Mode 2: Wavelength (λ₂) = 2.125 m, Frequency (f₂) ≈ 161.41 Hz

Mode 3: Wavelength (λ₃) = 1.417 m, Frequency (f₃) ≈ 242.12 Hz

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The horizontal component of the force applied to the cart can be found using the equation:

[tex]\[F_{\text{horizontal}} = F \cdot \cos(\theta)\][/tex]

To solve this problem, we can use the principle of conservation of momentum. Before the baseball lands on the ground, its momentum is given as 1.60 kg⋅m/s. We know that momentum is the product of an object's mass and velocity. Since the baseball is dropped from rest, its initial velocity is zero. Therefore, the momentum just before it lands is equal to the momentum it gained during free fall.

Step 1: Calculate the momentum gained by the baseball.

   Momentum = mass × velocity

   1.60 kg⋅m/s = 0.120 kg × velocity

Step 2: Solve for the velocity of the baseball just before it lands.

   velocity = (1.60 kg⋅m/s) / (0.120 kg)

   velocity = 13.33 m/s (rounded to two decimal places)

Step 3: Determine the height from which the baseball was dropped.

   We can use the kinematic equation for free fall:

   velocity² = initial_velocity² + 2 * acceleration * displacement

   Since the baseball was dropped from rest, the initial velocity is zero, and the acceleration is the acceleration due to gravity, which is approximately 9.8 m/s².

   Plugging in the values:

   (13.33 m/s)² = 0 + 2 * 9.8 m/s² * displacement

   Solving for the displacement:

   displacement = (13.33 m/s)² / (2 * 9.8 m/s²)

   displacement = 8.95 m (rounded to two decimal places)

Therefore, the baseball was dropped from a height of 8.95 meters (rounded to three significant figures).

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The work done by tension is 434.34 J, the work done by gravity is 218.94 J, and the work done by the normal force is zero. The increase in thermal energy of the crate and incline is 653.28 J.

To determine the work done by tension, gravity, and the normal force, we need to consider the different forces acting on the crate as it is pulled up the incline.

1. Work done by tension:

The tension force acts parallel to the incline and helps pull the crate up. The work done by tension is given by the formula:

Work = Force * Distance * cos(angle)

In this case, the tension force is 110 N and the distance moved up the incline is 5.5 m. The angle between the tension force and the incline is the sum of the incline angle (30 degrees) and the rope angle (16 degrees). Therefore, the angle is 30 degrees + 16 degrees = 46 degrees.

Using the formula, we can calculate the work done by tension:

Work = 110 N * 5.5 m * cos(46 degrees)

Work = 434.34 J (to two significant figures)

2. Work done by gravity:

The force of gravity acts vertically downwards. However, only the component of the force parallel to the incline affects the work done. The work done by gravity is given by the formula:

Work = Force * Distance * cos(angle)

The force of gravity can be calculated using the formula:

Force = mass * gravity

Where the mass of the crate is 8.2 kg and the acceleration due to gravity is approximately 9.8 m/s^2. The angle between the force of gravity and the incline is 30 degrees.

Using the formula, we can calculate the work done by gravity:

Work = (8.2 kg * 9.8 m/s^2) * 5.5 m * cos(30 degrees)

Work = 218.94 J (to two significant figures)

3. Work done by the normal force:

The normal force acts perpendicular to the incline and does not contribute to the work done since it is perpendicular to the displacement. Therefore, the work done by the normal force is zero.

To find the increase in thermal energy, we need to consider the work-energy principle. The work done by all the forces will result in an increase in thermal energy. Therefore, the increase in thermal energy is the sum of the work done by tension and gravity:

Increase in thermal energy = Work done by tension + Work done by gravity

Increase in thermal energy = 434.34 J + 218.94 J

Increase in thermal energy = 653.28 J (to two significant figures)

Therefore, in comparison to gravity's 218.94 J and tension's 434.34 J, the normal force produces no effort at all. The crate and incline's thermal energy increase is 653.28 J.

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Complete question is,

An 8.2 kg crate is pulled 5.5 m up a 30∘ incline by a rope angled 16∘ above the incline. The tension in the rope is 110 N and the crate's coefficient of kinetic friction on the incline is 0.22. How much work is done by tension, by gravity, and by the normal force? For help with math skills, you may want to review: Express your answers in joules to two significant figures. Enter your answers numerically separated by commas. Product - Part B What is the increase in thermal energy of the crate and incline? Express your answer in joules to two significant figures.

A car tire is filled to a pressure of 210kPa at 10 ∘C. The pressure within the tire is 222.3 kPa after the temperature increase.

To calculate the new pressure within the tire, we can use the ideal gas law equation: P1/T1 = P2/T2, where P1 and T1 are the initial pressure and temperature, and P2 and T2 are the final pressure and temperature.

Given P1 = 210 kPa, T1 = 10°C + 273.15 = 283.15 K, and T2 = 40°C + 273.15 = 313.15 K,

We can rearrange the equation to solve for P2: P2 = (P1 * T2) / T1 = (210 kPa * 313.15 K) / 283.15 K = 231.82 kPa.

Therefore, the pressure within the tire after the temperature increase is approximately 231.82 kPa, which can be rounded to 222.3 kPa.

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The speed of the 747 jetliner 9.00 seconds later is approximately 11.9 m/s. It has traveled a distance of approximately 299 meters during this time.

To determine the final speed of the jetliner, we need to calculate the acceleration using the net force and mass. Using the equation F = ma, we can rearrange it to find acceleration (a) as a = F/m. Substituting the given values, we have a = (4.30×10^5 N) / (3.44×10^5 kg), which yields approximately 1.25 m/s^2. Next, we can use the equation v = u + at, where u is the initial velocity, a is the acceleration, and t is the time. Substituting the values, we have v = 68.0 m/s + (1.25 m/s^2) * (9.00 s), which gives us a final speed of approximately 11.9 m/s.To find the distance traveled during this time, we can use the equation s = ut + 0.5at^2, where s is the distance, u is the initial velocity, a is the acceleration, and t is the time. Substituting the values, we have s = (68.0 m/s) * (9.00 s) + 0.5 * (1.25 m/s^2) * (9.00 s)^2, which gives us a distance of approximately 299 meters.Therefore, 9.00 seconds later, the speed of the 747 jetliner is approximately 11.9 m/s, and it has traveled a distance of approximately 299 meters.

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Ohm's Law states that the current flowing through a conductor is directly proportional to the voltage across the conductor, and inversely proportional to the resistance of the conductor. Mathematically, Ohm's Law can be expressed as:

I = V/R

where:

- I represents the current flowing through the conductor,

- V represents the voltage (potential difference) across the conductor,

- R represents the resistance of the conductor.

Option (b) is the correct statement that represents Ohm's Law. It states that the intensity (current) flowing through a conductor is equal to the product of the resistance and the potential difference (voltage).

Option (a) is incorrect as it mentions "intensity" instead of current and includes "conductance" which is the inverse of resistance.

Option (c) is incorrect as it mentions "conductance" instead of resistance.

Option (d) is partially correct as conductance is indeed the inverse of resistance, but it does not represent Ohm's Law directly.

Option (e) is incorrect as it mentions "charge" instead of current and includes "resistance" which should be in the denominator of the equation.

In summary, Ohm's Law describes the relationship between current, voltage, and resistance in a conductor, and option (b) represents it accurately.

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The forces acting on the ladder, apart from frictional forces, include the weight (downward), the normal force (perpendicular to the wall), and the tension force (pulling towards the wall).

When a ladder is leaning against a wall, the following forces typically act on the ladder:

Weight: The force due to gravity pulling the ladder downward. It can be represented by an arrow pointing vertically downward from the center of mass of the ladder.

Normal Force: The force exerted by the wall on the ladder perpendicular to the surface of the wall. It acts in the direction normal to the wall's surface and can be represented by an arrow pointing away from the wall.

Tension Force: If the ladder is being held or supported at the top, there will be a tension force acting along the ladder, pulling it towards the wall. This force can be represented by an arrow pointing from the top of the ladder towards the wall.

These are the main forces acting on the ladder in this situation. It's important to note that frictional forces, which you mentioned should be excluded.

can also come into play depending on the surface conditions between the ladder and the wall, but since you specifically asked to exclude them, they are not considered here.

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(a) Approximately 4.18 x 10^20 helium atoms fill the balloon.

(b) The average kinetic energy of the helium atoms is approximately 6.00 x 10^-21 J.

(c) The rms speed of the helium atoms is approximately 1.34 km/s.

To determine the number of helium atoms in the balloon, we can use the ideal gas law equation:

PV = nRT

Where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.

First, let's convert the diameter of the balloon to meters:

d = 30.2 cm = 0.302 m

The volume of the balloon can be calculated using the formula for the volume of a sphere:

V = (4/3)πr^3

Since the diameter is given, the radius (r) can be calculated as half of the diameter:

r = 0.302 m / 2 = 0.151 m

Now we can calculate the volume:

V = (4/3)π(0.151 m)^3 ≈ 0.0144 m^3

Next, we need to convert the temperature to Kelvin:

T = 16.0°C + 273.15 = 289.15 K

The ideal gas constant, R, is 8.314 J/(mol·K).

Now we can rearrange the ideal gas law equation to solve for the number of moles, n:

n = PV / RT

Substituting the given values:

n = (1.00 atm)(0.0144 m^3) / (8.314 J/(mol·K) * 289.15 K) ≈ 0.000696 mol

Since 1 mole of a gas contains Avogadro's number of particles (approximately 6.022 x 10^23), we can calculate the number of helium atoms:

Number of atoms = n * Avogadro's number

Number of atoms = 0.000696 mol * 6.022 x 10^23 ≈ 4.18 x 10^20 atoms

Therefore, approximately 4.18 x 10^20 atoms of helium gas fill the spherical balloon.

(b) The average kinetic energy of the helium atoms can be calculated using the equation:

KE_avg = (3/2)kT

Where KE_avg is the average kinetic energy, k is the Boltzmann constant (1.38 x 10^-23 J/K), and T is the temperature in Kelvin.

Substituting the given values:

KE_avg = (3/2)(1.38 x 10^-23 J/K)(289.15 K) ≈ 6.00 x 10^-21 J

Therefore, the average kinetic energy of the helium atoms is approximately 6.00 x 10^-21 J.

(c) The root mean square (rms) speed of the helium atoms can be calculated using the equation:

v_rms = √(3kT / m)

Where v_rms is the rms speed, k is the Boltzmann constant, T is the temperature in Kelvin, and m is the molar mass of helium (4.00 g/mol).

Converting the molar mass to kilograms:

m = 4.00 g/mol = 0.004 kg/mol

Substituting the given values:

v_rms = √[(3)(1.38 x 10^-23 J/K)(289.15 K) / (0.004 kg/mol)]

v_rms ≈ 1337 m/s

Converting the rms speed to kilometers per second:

v_rms ≈ 1.34 km/s

Therefore, the rms speed of the helium atoms is approximately 1.34 km/s.

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Given data:

The radius of one sphere, r1= 1.0 cm = 1 × 10⁻² m

The radius of another sphere, r2 = 8.0 cm = 8 × 10⁻² m

The total charge on the two spheres, q = +360 nC = +360 × 10⁻⁹ C

(a) The charge is distributed between two spheres in such a way that the potential difference between the spheres is zero. Let q1 be the charge on sphere 1 and q2 be the charge on sphere 2.

Therefore, q = q1 + q2         ..... (1)

The potential of a sphere is given by,V= (1/4πϵ₀) * (q/r)       …(2)

where V is the potential,

q is the charge on the sphere,

r is the radius of the sphere,

and ϵ₀ is the permittivity of free space.

The potential difference between the spheres is zero whenV1 = V2 or

(1/4πϵ₀) * (q1/r1) = (1/4πϵ₀) * (q2/r2)

On simplifying we get

q1/q2 = r1/r2 = (1/8)

Therefore, q1 = (1/9) * q and q2 = (8/9) * q

Putting the values of q1 and q2 in equation (1), we get

q = (1/9) * q + (8/9) * q

Hence,q1 = (1/9) * q = (1/9) * 360 × 10⁻⁹ = 40 × 10⁻⁹ C

q2 = (8/9) * q = (8/9) * 360 × 10⁻⁹ = 320 × 10⁻⁹ C

Therefore, the charge on sphere 1 is 40 nC and the charge on sphere 2 is 320 nC.

(b) We are to determine whether the electric field at the surface of either sphere exceeds the dielectric strength of air. The electric field intensity at the surface of a sphere is given by

E= (q/(4πϵ₀r²)) …… (3)

On substituting the values of q and r, we get

E1= (40 × 10⁻⁹)/(4π × 8.854 × 10⁻¹² × (1 × 10⁻²)²)

= 1.8 × 10⁵ V/m

E2= (320 × 10⁻⁹)/(4π × 8.854 × 10⁻¹² × (8 × 10⁻²)²)

= 1.4 × 10⁵ V/m

The dielectric strength of air is approximately 3 × 10⁶ V/m.

Since the electric field at the surface of neither sphere exceeds the dielectric strength of air, the spheres do not break down.

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The pitcher throws horizontally a fast ball at 135 km/hr toward the home plate, which is 18.3m away. Neglecting air resistance, we need to find how far the ball drops because of gravity by the time it reaches the home plate.

The horizontal velocity of the ball = 135 km/hr = 37.5 m/sAnd, time taken by the ball to cover 18.3m horizontally = `t = d/v = 18.3/37.5 = 0.488 sec`In this time, the vertical distance dropped by the ball is given by `s = 1/2 × g × t²`where, g is the acceleration due to gravityg = 9.8 m/s²∴ s = 1/2 × 9.8 × (0.488)²= 1.167mTherefore, the ball drops 1.167m because of gravity by the time it reaches the home plate.

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A 8.24kg crates slides across the floor with a velocity of 3.57m/s.  the momentum of the crate is approximately 29.44 kg·m/s.

The momentum of an object is defined as the product of its mass and velocity. In this case, we have a crate with a mass of 8.24 kg and a velocity of 3.57 m/s. To find the momentum of the crate, we simply multiply the mass by the velocity.

Momentum = mass x velocity

Momentum = 8.24 kg x 3.57 m/s

Momentum ≈ 29.44 kg·m/s

Therefore, the momentum of the crate is approximately 29.44 kg·m/s.

Momentum is a fundamental concept in physics that describes the motion of an object. It represents the quantity of motion possessed by an object, taking into account both its mass and velocity. In this case, the momentum of the crate indicates how difficult it would be to stop or change its motion. The greater the momentum, the more force is required to alter its velocity.

It's important to note that momentum is a vector quantity, meaning it has both magnitude and direction. In this case, since only the magnitude of velocity is provided, the momentum is represented by a positive value, indicating the direction of motion.

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The tension in the string does not change as the masses begin to move Tension and Newton's Third Law .

The tension in the string remains constant during the motion of the masses. This is because the tension force in a string is determined by the forces applied to it at each end, according to Newton's Third Law of Motion. According to this law, for every action, there is an equal and opposite reaction.

In this scenario, the string is connected to the two masses. When one mass exerts a force on the string, the string exerts an equal and opposite force on the other mass. This creates a balanced system where the tension in the string is equal in magnitude at both ends.

As the masses begin to move, the tension in the string keeps both masses connected and provides the necessary force to accelerate them. The tension remains constant because the forces applied by each mass on the string are balanced, ensuring that the net force on the string is zero.

Therefore, the tension in the string does not change during the motion of the masses.

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The frequency at which the string oscillates with 5 loops is approximately 9.12 Hz. This frequency is determined by the tension in the string, its mass, and its length, taking into account the number of loops as well.

To find the frequency at which the string oscillates, we can use the formula for the frequency of a simple harmonic oscillator with multiple loops:

[tex]f=\frac{n}{2L}\sqrt{\frac{T}{u} }[/tex]

where:

f is the frequency,

n is the number of loops,

L is the length of the string, and

μ is the linear mass density of the string (mass per unit length).

In this case, the number of loops (n) is 5, the length of the string (L) is 0.300 m, and the mass of the string (m) is 0.600 kg. We need to calculate the linear mass density μ using the given mass and length:

μ=[tex]\frac{m}{L}=\frac{0.600kg}{0.300m} = 2kg/m[/tex]

Now we can substitute the values into the formula:

[tex]f=\frac{5}{2*0.3m}\sqrt{\frac{2.4 N}{2 Kg/m} }=\frac{5}{0.6m}\sqrt{1.2 N/kg}[/tex]

which gives the value of f≈9.12Hz

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To determine how far behind the 0m mark the boat started, we need to calculate the time it took for the boat to reach the 200m mark and then subtract the time it took for the boat to reach the 0m mark.

Given:

Time to reach the 0m mark (t_0) = 1.2s

Average velocity to reach the 200m mark (v_avg) = 7.07m/s

Let's denote the time it took for the boat to reach the 200m mark as t_200. We can use the formula:

v_avg = (Δx) / (Δt)

where Δx is the displacement and Δt is the time interval.

For the boat's journey from the 0m mark to the 200m mark, the displacement is 200m - 0m = 200m, and the time interval is t_200 - t_0.

So we have:

v_avg = (200m) / (t_200 - t_0)

Plugging in the given average velocity:

7.07m/s = 200m / (t_200 - 1.2s)

Now, solving for t_200 - 1.2s:

(t_200 - 1.2s) = 200m / 7.07m/s

(t_200 - 1.2s) = 28.28s

t_200 = 28.28s + 1.2s

t_200 ≈ 29.48s

Therefore, it took approximately 29.48 seconds for the boat to reach the 200m mark.

To find how far behind the 0m mark the boat started, we subtract the time it took to reach the 0m mark (t_0) from the time it took to reach the 200m mark (t_200):

Distance behind 0m mark = v_avg * t_0

Distance behind 0m mark = 7.07m/s * 1.2s

Distance behind 0m mark ≈ 8.48m

Therefore, the boat started approximately 8.48m behind the 0m mark.

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(a) The speed of the mailbag after 5.00 s is 2.94 m/s.

(b) The mailbag is 14.7 m below the helicopter after 5.00 s.

(c) If the helicopter is rising steadily at 2.94 m/s, the answers to (a) and (b) remain the same.

(a) The speed of the mailbag after 5.00 s can be calculated by multiplying the time by the descent rate: 5.00 s × 2.94 m/s = 14.7 m/s.

(b) To find the distance below the helicopter, we multiply the descent rate by the time: 2.94 m/s × 5.00 s = 14.7 m.

(c) If the helicopter is rising steadily at 2.94 m/s, the mailbag's speed and distance below the helicopter remain the same. The upward velocity of the helicopter offsets the downward velocity of the mailbag, resulting in no change in relative motion.

In summary, after 5.00 s, the mailbag has a speed of 2.94 m/s and is located 14.7 m below the helicopter. If the helicopter is rising at the same rate, the speed and distance below the helicopter remain unchanged.

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