A worker pushes a 1.50 x 10^3 N crate with a horizontal force of 345 N a distance of 24.0 m. Assume the coefficient of kinetic friction between the crate and the floor is 0.220.

a) How much work is done by the worker on the crate?
b) How much work is fone by the floor on the crate?
c) What is the net work done on the crate?

Final answer:

To find the acceleration, the kinematic equation is used. The kinetic friction coefficient is derived from force equations, and the friction force can be calculated with this coefficient and the normal force. Finally, the block's final speed after sliding down the incline is determined using another kinematic equation.

Explanation:

Finding the Magnitude of Acceleration, Coefficient of Kinetic Friction, Friction Force, and Final Speed

To solve for the acceleration of a 2.10-kg block sliding down a 30.0° incline, we can use the kinematic equation s = ut + (1/2)at², where s is the distance slid which is 1.90 m, u is the initial velocity which is 0 m/s, t is the time which is 1.40 s, and a is the acceleration. Solving for acceleration, a = 2s/t², we find:

a = (2 * 1.90 m) / (1.40 s)²

a = 3.80 m / 1.96 s²

a = 1.9388 m/s²

The coefficient of kinetic friction μ can be determined by using the equation f_k = μ * N, where f_k is the kinetic friction force and N is the normal force. The gravitational component parallel to the incline is mg sin(θ) and perpendicular to the incline is mg cos(θ). Knowing the total force equation, mg sin(θ) - μ*mg cos(θ) = ma, we can isolate μ and solve for it.

The friction force then can be calculated using f_k = μ * mg cos(θ), and the direction of this force is up the incline since it opposes the motion of the block sliding down.

Finally, to find the final speed v of the block after sliding 1.90 m, we use the kinematic equation v² = u² + 2as. Since the initial speed u is 0, this simplifies to v = √(2as), and plugging in the known values for a and s gives us the final speed.

The distance from the top of the wood block to the water can be calculated using the buoyancy force. For fresh water, the distance is 0.1 cm and for sea water, it is 0.09709 cm.

To determine the distance from the top of the wood block to the water, we need to consider the buoyancy force acting on the block. When an object is submerged in a fluid, it experiences an upward force called buoyant force equal to the weight of the fluid displaced by the object.

The density of the wood block is given as 700 kg/m^3. Since the block floats, the density of the wood is less than the density of water (1000 kg/m^3). Therefore, the volume of the wood block is the same as the volume of water it displaces.

To find the distance from the top of the block to the water, we first need to calculate the volume of the block. The volume formula for a cube is V = s^3, where s is the length of a side. In this case, the side length of the block is 10 cm, so the volume is 10 cm x 10 cm x 10 cm = 1000 cm^3.

Next, we can calculate the weight of the water displaced by the block. The weight of an object is given by the formula W = m x g, where m is the mass and g is the acceleration due to gravity. The mass of the displaced water can be found using the density formula m = d x V, where d is the density and V is the volume. In this case, the density of water is 1000 kg/m^3 and the volume of the block is 1000 cm^3. Converting the volume to m^3, we get 0.001 m^3.

Finally, we can calculate the distance from the top of the block to the water using the formula h = (W/Water density) x (100 cm/1 m). Plugging in the values, we get h = (0.001 m^3 x 1000 kg/m^3) / (1000 kg/m^3) x (100 cm/1 m) = 0.001 m x 100 cm = 0.1 cm.

Therefore, the distance from the top of the block to the water is 0.1 cm when the water is fresh. For sea water, the density is higher at around 1030 kg/m^3. Using the same calculations, we get h = (0.001 m^3 x 1000 kg/m^3) / (1030 kg/m^3) x (100 cm/1 m) = 0.001 m x 97.09 cm = 0.09709 cm.

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Stars do not crash into each other because the gravitational force acts as a centripetal force, maintaining circular orbits and keeping them apart. The masses of the stars can be calculated using Kepler's third law for circular orbits.

The scenario provided describes a system in which two equal-mass stars maintain a constant distance apart and revolve around their common center of mass due to the gravitational force between them. The system is in balance when the gravitational force provides the necessary centripetal force to keep the stars in their circular orbits without them crashing into each other.

Part (a)

The two stars don't crash into one another because the gravitational attraction is providing the centripetal force that keeps them in stable, circular orbits around their common center of mass. This balance between gravitational forces and the stars' inertia maintains the separation between them.

Part (b)

To determine the mass of each star, we can use Kepler's third law and Newton's version of it for two bodies orbiting each other, which states:

[tex]T^2 = (4*pi^2*r^3)/(G*(m1+m2))[/tex]

Since m1 = m2 (equal masses), we can solve for a single mass. By substituting the given values (radius and period), we can calculate the mass of each star.

(a) The two stars do not crash into each other because they are in a stable orbit, rotating around their common center of mass (barycenter). (b) The mass of each star must be approximately [tex]9.61 \times 10^{11} \, \text{kg}[/tex].

To address the questions about the two stars, we need to use concepts from orbital mechanics and Newtonian gravity. Here's the step-by-step approach:

(a) Why don’t the two stars crash into one another due to the gravitational force between them?

The two stars do not crash into each other because they are in a stable orbit, rotating around their common center of mass (barycenter). The gravitational force between the stars provides the necessary centripetal force to keep them in this orbit.

(b) What must be the mass of each star?

To find the mass of each star, we can use the following relationship between gravitational force and centripetal force. First, let's summarize the given data:

- Distance between the stars, [tex]d = 8.0 \times 10^{10} \, \text{m}[/tex]

- Period of revolution,  T = 12.6 yr

Convert the period  T  into seconds:

[tex]T = 12.6 \, \text{yr} \times \left(365.25 \, \text{days/yr}\right) \times \left(24 \, \text{hours/day}\right) \times \left(3600 \, \text{seconds/hour}\right) \\\\T \approx 12.6 \times 365.25 \times 24 \times 3600 \, \text{s} \approx 3.97 \times 10^8 \, \text{s}[/tex]

Each star rotates around the common center of mass with a radius  r  that is half the distance between the stars, so:

[tex]r = \frac{d}{2} = \frac{8.0 \times 10^{10} \, \text{m}}{2} = 4.0 \times 10^{10} \, \text{m}[/tex]

The gravitational force  F  providing the centripetal force [tex]F_c[/tex] can be equated as follows:

F =  [tex]F_c[/tex]

The gravitational force between the two stars is given by Newton's law of gravitation:

[tex]F = \frac{G m_1 m_2}{d^2}[/tex]

Since the stars have equal mass ( m₁ = m₂ = m ):

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

The centripetal force required to keep each star in orbit is:

[tex]F_c = m \frac{v^2}{r}[/tex]

Where  v  is the orbital velocity of each star. The orbital velocity  v  can be related to the period  T  and the radius r by:

[tex]v = \frac{2 \pi r}{T}[/tex]

Now substituting  v  into the centripetal force equation:

[tex]F_c = m \left( \frac{2 \pi r}{T} \right)^2 \frac{1}{r} \\\\F_c = m \frac{4 \pi^2 r^2}{T^2 r} \\\\F_c = m \frac{4 \pi^2 r}{T^2}[/tex]

Since  F =  [tex]F_c[/tex] :

[tex]\frac{G m^2}{d^2} = m \frac{4 \pi^2 r}{T^2}[/tex]

We can cancel one  m  from each side:

[tex]\frac{G m}{d^2} = \frac{4 \pi^2 r}{T^2}[/tex]

Solving for  m :

[tex]m = \frac{4 \pi^2 r d^2}{G T^2}[/tex]

We know:

[tex]r = 4.0 \times 10^{10} \, \text{m} \\\\d = 8.0 \times 10^{10} \, \text{m} \\\\G = 6.674 \times 10^{-11} \, \text{N} \cdot \text{m}^2 \cdot \text{kg}^{-2} \\\\T = 3.97 \times 10^8 \, \text{s}[/tex]

Substitute these values in:

[tex]m = \frac{4 \pi^2 (4.0 \times 10^{10} \, \text{m}) (8.0 \times 10^{10} \, \text{m})^2}{(6.674 \times 10^{-11} \, \text{N} \cdot \text{m}^2 \cdot \text{kg}^{-2}) (3.97 \times 10^8 \, \text{s})^2}[/tex]

First calculate [tex]r \cdot d^2[/tex]:

[tex]r \cdot d^2 = 4.0 \times 10^{10} \, \text{m} \times (8.0 \times 10^{10} \, \text{m})^2 = 4.0 \times 10^{10} \, \text{m} \times 64.0 \times 10^{20} \, \text{m}^2 = 256.0 \times 10^{30} \, \text{m}^3[/tex]

Then calculate  T² :

[tex]T^2 = (3.97 \times 10^8 \, \text{s})^2 = 15.76 \times 10^{16} \, \text{s}^2[/tex]

Now, plug in the values:

[tex]m = \frac{4 \pi^2 \times 256.0 \times 10^{30}}{6.674 \times 10^{-11} \times 15.76 \times 10^{16}} \\\\m = \frac{4 \pi^2 \times 256.0 \times 10^{30}}{1.05136 \times 10^6 \times 10^{16}} \\\\m = \frac{4 \pi^2 \times 256.0 \times 10^{30}}{1.05136 \times 10^{22}} \\\\m = \frac{4 \pi^2 \times 256.0 \times 10^{8}}{1.05136} \\\\m = \frac{4 \times 9.8696 \times 256.0 \times 10^{8}}{1.05136} \\\\m = \frac{10102.4 \times 10^{8}}{1.05136} \\\\m \approx 9.61 \times 10^{11} \, \text{kg}[/tex]

Thus, the mass of each star must be approximately [tex]9.61 \times 10^{11} \, \text{kg}[/tex].

Answer:

Part a)

[tex]\theta = 45 degree[/tex]

Part b)

[tex]R = 39.2[/tex]

Part c)

[tex]v_x = 13.86 m/s[/tex]

[tex]v_y = 13.86 m/s[/tex]

Explanation:

Since it took 4.0 s to come back at the same position so we can say

[tex]\Delta y = v_y t + \frac{1}{2}at^2[/tex]

[tex]0 = v_y(4.0) - \frac{1}{2}(9.8)(4.0)^2[/tex]

[tex]v_y = 19.6 m/s[/tex]

Part a)

Now we know that horizontal range of projectile is given as

[tex]R = \frac{v^2 sin2\theta}{g}[/tex]

Now this range would be maximum if the angle of the projectile is giving maximum value of sine

so we have

[tex]sin(2\theta) = 1[/tex]

[tex]\theta = 45 degree[/tex]

Part b)

For maximum range we have

[tex]R = \frac{v^2}{g}[/tex]

[tex]R = \frac{19.6^2}{9.8}[/tex]

[tex]R = 39.2[/tex]

Part c)

Since we projected at an angle of 45 degree

so the components are given as

[tex]v_x = 19.6 sin45 = 13.86 m/s[/tex]

[tex]v_y = 19.6 cos45 = 13.86 m/s[/tex]

The width of the slit can be determined using the formula θ = λ / (2d), where θ is the angle, λ is the wavelength, and d is the width of the slit. By rearranging the formula and substituting the given values, we find that the width of the slit is approximately 244 nm.

The width of the slit can be determined using the formula for the angle of the first minimum in a single-slit diffraction pattern, which is given by θ = λ / (2d), where θ is the angle, λ is the wavelength, and d is the width of the slit. In this case, the angle is 45.0 degrees and the wavelength is 470 nm. Converting the angle to radians and rearranging the formula, we can solve for the width of the slit:

d = λ / (2 * tan(θ))

Substituting the given values, we have:

d = (470 nm) / (2 * tan(45.0 degrees))

Calculating this, we find that the width of the slit is approximately 244 nm.

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Answer:

diameter is decreased by factor 0.91

Explanation:

As we know by the equation of trajectory

[tex]y = xtan\theta - \frac{gx^2}{2v^2cos^2\theta}[/tex]

here as per first given situation we know that

x = 36.9 m

y = 8.85 m

[tex]\theta = 45^0[/tex]

now from above equation we have

[tex]8.85 = 36.9 tan45 - \frac{(9.8)(36.9)^2}{2(v^2)cos^245}[/tex]

[tex]8.85 = 36.9 - \frac{13343.8}{v^2}[/tex]

[tex]\frac{13343.8}{v^2} = 28.05[/tex]

[tex]v = 21.8 m/s[/tex]

now similarly after nozzle is adjusted we have

y = 17.9 m

x = 36.9 m

[tex]\theta = 45^0[/tex]

now again from equation we have

[tex]17.9 = 36.9 tan45 - \frac{(9.8)(36.9)^2}{2(v'^2)cos^245}[/tex]

[tex]17.9 = 36.9 - \frac{13343.8}{v'^2}[/tex]

[tex]\frac{13343.8}{v'^2} = 19[/tex]

[tex]v' = 26.5 m/s[/tex]

Now by equation of continuity we can find the change in diameter

as we know that

[tex]A_1v_1 = A_2v_2[/tex]

now we have

[tex]\pi d_1^2 v_1 = \pi d_2^2 v_2[/tex]

[tex]d_1^2 (21.8) = d_2^2(26.5)[/tex]

[tex]\frac{d_1}{d_2} = \sqrt{\frac{26.5}{21.8}}[/tex]

[tex]\frac{d_1}{d_2} = 1.10[/tex]

so we have

[tex]\frac{d_2}{d_1} = 0.91[/tex]

so diameter is decreased by factor of 0.91

The nozzle diameter changed by a factor of 0.907

[tex]\texttt{ }[/tex]

Acceleration is rate of change of velocity.

[tex]\large {\boxed {a = \frac{v - u}{t} } }[/tex]

[tex]\large {\boxed {d = \frac{v + u}{2}~t } }[/tex]

a = acceleration (m / s²)v = final velocity (m / s)

u = initial velocity (m / s)

t = time taken (s)

d = distance (m)

Let us now tackle the problem!

[tex]\texttt{ }[/tex]

Given:

horizontal distance = x = 36.9 m

angle of projection = θ = 45°

initial height = y₁ = 8.85 m

final height = y₂ = 17.9 m

Asked:

ratio of nozzle diameter = d₂ : d₁ = ?

Solution:

The motion of the water is a parabolic motion.

Firstly, we will calculate the time taken for the water to reach the hotspot:

[tex]x = (u \cos \theta) t[/tex]

[tex]t = x \div ( u \cos \theta )[/tex]

[tex]t = x \div ( u \cos 45^o )[/tex]

[tex]\boxed {t = \frac{\sqrt{2}x}{u}}[/tex]

[tex]\texttt{ }[/tex]

Next , we could calculate the initial speed (u) of the water as it leaves the nozzle:

[tex]y = (u \sin \theta) t - \frac{1}{2}gt^2[/tex]

[tex]y = (u \sin 45^o)( \frac{\sqrt{2}x}{u} ) - \frac{1}{2}g ( \frac{\sqrt{2}x}{u} )^2[/tex]

[tex]y = x - \frac{gx^2}{u^2}[/tex]

[tex]\frac{gx^2}{u^2} = x - y[/tex]

[tex]u^2 = \frac{gx^2}{x - y }[/tex]

[tex]u = \sqrt{ \frac{gx^2}{x - y } }[/tex]

[tex]\boxed {u = x \sqrt{ \frac{g}{x - y} }}[/tex]

[tex]\texttt{ }[/tex]

Finally , we could find the ratio of the diameter by using Continuity Equation as follows:

[tex]u_1 A_1 = u_2 A_2[/tex]

[tex]u_1 \frac{1}{4} \pi (d_1)^2 = u_2 \frac{1}{4} \pi (d_2)^2[/tex]

[tex](d_2)^2 : (d_1)^2 = u_1 : u_2[/tex]

[tex](d_2)^2 : (d_1)^2 = x \sqrt{ \frac{g}{x - y_1} } : x \sqrt{ \frac{g}{x - y_2} }[/tex]

[tex](d_2)^2 : (d_1)^2 = \sqrt { x - y_2 } : \sqrt { x - y_1}[/tex]

[tex]\frac {d_2}{d_1} = \sqrt[4] { \frac {x - y_2} {x - y_1} }[/tex]

[tex]\frac {d_2}{d_1} = \sqrt[4] { \frac {36.9 - 17.9} {36.9 - 8.85} }[/tex]

[tex]\frac {d_2}{d_1} \approx 0.907[/tex]

[tex]d_2 \approx 0.907 \times d_1[/tex]

[tex]\texttt{ }[/tex]

[tex]\texttt{ }[/tex]

Grade: High School

Subject: Physics

Chapter: Kinematics

50 years converted into seconds would be equivalent to 1.58×10⁹ seconds.

A unit of measurement is a specified magnitude of a quantity that is established and used as a standard for measuring other quantities of the same kind. It is determined by convention or regulation. Any additional quantity of that type can be stated as a multiple of the measurement unit.

As given in the problem we have to convert 50 years into seconds,

1 year = 365 days

50 years = 50×365 days

50 years = 50×365×24 hours

50 years =50×365 ×24×60 minutes

50 years =50×365 ×60×60 seconds

               =1576800000 seconds

               =1.58×10⁹ seconds

Thus,50 years converted into seconds would be 1.58×10⁹ seconds

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A 15.0 lbs block rests on the floor. Then, the force exerted on the block is in the upward direction to counterbalance the Earth's force on the block. If a rope is tied to it, then the force from the floor and the rope together balance the weight.

A force is an external agent which is capable of changing an object's state of rest or motion. Force has both the magnitude and the direction, hence it is a vector quantity. The direction towards which this force is applied is known as the direction of the force, and the application of force is the point at which the force is being applied.

The force exerted on the block by the floor is 15.0lb upwards, to counterbalance the Earth’s force on the block. If a rope is tied to the block, and is running vertically over a pulley, and other object of 10lb is attached to it, then a force of 5.00lb towards upward direction is exerted by the floor on the block. The Earth pulling it down with 15.0lb and the rope pulling it up with a force of 10.0lb. The forces from the floor and rope together balance the weight of objects.

If the 10 lb object is replaced with a 20lb object, then the force exerted by the floor on the 15lb block is 0, because the block now accelerates up away from the floor.

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1. The weight of the suitcase is approximately 1448.32 N.

2. The apparent weight of the 86-kg astronaut aboard the rocket is approximately 2504 N.

To find the weight and mass of the suitcase, we'll use Newton's second law of motion, which states that the net force acting on an object is equal to the product of its mass and acceleration.

Given:

Force (F) = 105 N

Acceleration (a) = 0.710 m/[tex]s^2[/tex]

We can use the equation F = ma to solve for mass (m):

105 N = m * 0.710 m/s^2

Rearranging the equation, we have:

m = F / a

m = 105 N / 0.710 m/[tex]s^2[/tex]

m ≈ 147.89 kg

So, the mass of the suitcase is approximately 147.89 kg.

To find the weight of the suitcase, we can use the equation W = mg, where W is the weight and g is the acceleration due to gravity (approximately 9.8 m/[tex]s^2[/tex]).

W = 147.89 kg * 9.8 m/[tex]s^2[/tex]

W ≈ 1448.32 N

Therefore, the weight of the suitcase is approximately 1448.32 N.

Now let's calculate the apparent weight of the astronaut aboard the rocket.

Given:

Mass (m) = 86 kg

Acceleration (a) = 29.0 m/[tex]s^2[/tex]

Using the same formula W = mg, we can find the apparent weight (W) of the astronaut:

W = m * a

W = 86 kg * 29.0 m/[tex]s^2[/tex]

W ≈ 2504 N

Therefore, the apparent weight of the 86-kg astronaut aboard the rocket is approximately 2504 N.

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The common velocity of the two-car train after the collision, given that the 15000 kg train car was initially moving due east at 23 m/s, is +13.8 m/s

The question given above talks about conservation of linear momentum of colliding objects.

Thus, we can calculate the common velocity of the two-car train after the collision by using the formula of conservation of linear momentum as shown below:

Momentum before = Momentum after

[tex]m_1u_1 + m_2u_2 = v(m_1 + m_2)[/tex]

(15000 × 23) + (10000 × 0) = v(15000 + 10000)

345000 + 0 = 25000v

345000 = 25000v

Divide both sides by 25000

v = 345000 / 25000

= +13.8 m/s

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Planetary geologists suspect that the original surface of a certain celestial body has been covered by younger volcanic activity, based on the small number of impact craters. Evidence of past impact events is gradually erased by the constant renewal of Earth's crust through plate tectonics.

Based on the small number of impact craters, planetary geologists suspect that virtually all the original surface of a certain celestial body has been covered by younger volcanic activity.

This can be observed in the case of Earth, where the constant renewal of its crust through plate tectonics gradually erases evidence of past impact events.

Geologists have identified eroded remnants of impact craters on Earth, providing evidence for the influence of these impacts on the planet's evolution over time.

The student’s questions involve the principles of thermal radiation and heat conduction in Physics, relating to the determination of the equilibrium temperature in a radiative system and the electrical power needed to maintain a temperature. Calculations require additional data on materials and geometry to provide accurate answers.

The student's questions pertain to the topics of thermal radiation and heat conduction in Physics, which are within the realm of thermodynamics. Specifically, they are asking about (a) the equilibrium temperature of a conical surface and (b) the power required to maintain a certain temperature on a circular plate. These calculations involve understanding Stefan-Boltzmann law, conduction, and power calculations.

Unfortunately, without further information or input parameters for materials and geometries, providing a definitive answer for a system like the one described (circular plate and conical surface in an enclosure) isn't possible. To determine the electrical power required, one would need to know properties like thermal conductivity, surface area, and possibly the geometry of the objects in question to find out the heat transfer rates.

The index of refraction of the medium is 1.5.

The speed of light in a medium is given by the equation c = v/n, where c is the speed of light in a vacuum, v is the speed of light in the medium, and n is the index of refraction of the medium. Rearranging the equation, we can find the index of refraction as n = c/v. Substituting the given values, the index of refraction is n = (3.00 x 10^8 m/s) / (2.00 x 10^8 m/s) = 1.5.

To find the angle of refraction in water when light propagates in a glass wall, we can use Snell's law.

To find the angle of refraction in water, we can use Snell's law, which states that the ratio of the sines of the angles of incidence and refraction is equal to the ratio of the indices of refraction of the two media:

sin(θ1) / sin(θ2) = n2 / n1

Given that the incidence angle in glass is 19.0 degrees and the index of refraction of glass is 1.50, we can substitute these values into the equation:

sin(19.0) / sin(θ2) = 1.333 / 1.50

Solving for θ2, the angle of refraction in water, we find:

θ2 ≈ 13.14 degrees

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The angle of refraction in the water is approximately 24.21 degrees.

When light passes from one medium to another, it changes direction. This change in direction is called refraction. Snell's Law describes the relationship between the angles of incidence and refraction. The equation is: n1*sin(θ1) = n2*sin(θ2), where n1 and n2 are the indices of refraction of the two media, and θ1 and θ2 are the angles of incidence and refraction, respectively.

In this case, light is propagating in glass with an index of refraction of 1.65 and is striking the interface between the glass and water at an angle of 19.0 degrees. The index of refraction of water is 1.333. To find the angle of refraction in the water, we can use Snell's Law:

1.65*sin(19.0) = 1.333*sin(θ2).

Solving for θ2, we find that the angle of refraction in the water is approximately 24.21 degrees.

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The magnitude of the angular deceleration of the cylinder is -10.0 rad/s^2, and the magnitude of the force of friction applied by the brake shoe is -8.50 N.

(a) To find the magnitude of the angular deceleration of the cylinder, we can use the formula:

angular deceleration = (final angular velocity - initial angular velocity) / time

Given that the initial angular speed is 88.0 rad/s, the final angular speed is half of that, and the time is 4.40 s, we can calculate:

angular deceleration = (44.0 rad/s - 88.0 rad/s) / 4.40 s = -10.0 rad/s^2

(b) To find the magnitude of the force of friction applied by the brake shoe, we can use the formula:

force of friction = moment of inertia * angular deceleration

Given that the moment of inertia is 0.850 kg·m², and the angular deceleration is -10.0 rad/s², we can calculate:

force of friction = 0.850 kg·m² * -10.0 rad/s² = -8.50 N

The rpm of the second pulley is 15.

To find the rpm of the second pulley, we can use the formula:

Angular velocity of first pulley * diameter of first pulley = angular velocity of second pulley * diameter of second pulley

Plugging in the given values:

10 rpm * 15.0 cm = rpm of second pulley * 10.0 cm

Dividing both sides by 10.0 cm:

Rpm of second pulley = (10 rpm * 15.0 cm) / 10.0 cm = 15 rpm

The critical temperature of a compound is the point at which its gas phase cannot be liquefied by pressure and for binary mixtures, it represents the temperature where two immiscible liquids become fully miscible.

The critical temperature of a compound is the temperature above which the gas phase of the substance cannot be made to liquefy, no matter how much pressure is applied. For a binary mixture, this can also refer to the upper critical temperature where two immiscible liquids become completely miscible in all proportions. An example provided is that of methyl acetate and carbon disufide, which has a critical temperature of approximately 230 K at one atmosphere. A similar behavior occurs with hexane/nitrobenzene mixtures, which have a critical temperature of 293 K.

These temperatures represent the critical point at which there is no distinction between the liquid and the gas phase of a substance. This is shown by the fact that at the critical temperature and the associated critical pressure, the gas does not condense. Understanding and utilizing the critical temperature of compounds is important in various applications, including the development and use of high-temperature superconductors.

Below is a picture of the bohr atom and the three particles that comprise it.

When an electron moves from one energy level to another, it either absorbs or emits a photon of electromagnetic radiation.

The Bohr model of the atom is a description of the structure of atoms, especially that of hydrogen, proposed by the physicist Niels Bohr in 1913. The model was a radical departure from earlier, classical descriptions of the atom, and it was the first that incorporated quantum theory.

Therefore,  Bohr model of the atom consists of a small, dense nucleus surrounded by orbiting electrons. The nucleus is positively charged, and the electrons are negatively charged. The electrons are attracted to the nucleus by the electromagnetic force, but they are also moving around the nucleus, so they are not pulled in.

Answer:

A + C + D

Explanation:

A) in stars

C) in lightning

D) in light-bulbs

(P.S. I just took a test on this and got it correct)