f the large ring, small ring, and each of the spokes weigh 100 lb, 15 lb, and 20 lb, respectively. what is the mass moment of inertia of the wheel about an axis perpendicular to the page and passing through point a?

Paraffin has a specific heat capacity of 1000 J/kg degrees Celsius.

The massic heat capacity, also known as the specific heat capacity or symbol c in thermodynamics, is the quotient of the heat capacity of a sample of a substance and its mass.

Informally, it is the quantity of heat needed to raise a substance's temperature by one degree for every unit of mass that is added to it.

SI unit of specific heat capacity is J-kg⁻¹-K⁻¹.

The  specific heat capacity of paraffin is about 1000 J/kg degrees Celsius. Hence, paraffin is the liquid.

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To provide an estimate of a car's MPG (miles per gallon) at average values of the inputs, we need specific information regarding the car's fuel efficiency, driving conditions, and engine specifications.

The MPG value can vary significantly based on factors such as the car's make, model, engine type, transmission, weight, aerodynamics, driving style, and road conditions.

However, as a rough approximation, the average MPG for a typical gasoline-powered car is around 25-30 MPG in mixed driving conditions. For a hybrid vehicle, the average MPG can range from 40-50 MPG. Electric vehicles (EVs) do not use MPG as a metric since they are powered by electricity and typically measured in terms of miles per kilowatt-hour (miles/kWh).

It's important to note that the actual MPG a car achieves can vary from these average values based on various factors. For a more accurate estimate, specific details about the car's make, model, and any additional parameters would be necessary.

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The sum of all resistances in the series circuit is known as Equivalent resistance.

where R is the equivalent resistance.

Hence, option 1 is the correct answer.

Answer:

The nuclear division processes of mitosis and meiosis occur during cell division. During mitosis, body cells are divided, while during meiosis, s3x cells are divided. At mitosis, a cell divides once; at meiosis, it divides twice.

Explanation:

Answer:

carpet

Explanation:

Diffuse reflection is the reflection of light from a surface such that an incident ray is reflected at many angles rather than at just one angle as in the case of specular reflection.

The structure of carpet's surface is as shown. Thus it shows large amount of diffuse reflection.

Answer:

50%

Explanation:

Given

Initial power = 200W

Final power = 300W

Increment = 300 - 200 = 100W

percentage increase = increment/initial power * 100

percentage increase = 100/200 * 100%

percentage increase = 0.5 * 100

percentage increase = 50%

Hence the percentage increase in speed is 50%

The value of force is 1.7 × 10⁻³ N, with the direction opposite to that of the bat's motion.

When an object collides with another object, they exchange energy. For example, a baseball and bat collision or a car collision. When two objects collide, the force of the collision has to be equal on both sides of the collision according to Newton's Third Law. So, to find the value of force, we will apply the equation:

F = ΔP / ΔT

where F is the force, ΔP is the change in momentum, and ΔT is the time of collision. The equation represents the impulse momentum theorem.

Now, let's apply the given values to the above equation.

Final momentum (p2) = mass × final velocity (v2)

p2 = 0.15 kg × (-50 m/s)

p2 = -7.5 kg.m/s

Initial momentum (p1) = mass × initial velocity (v1)

p1 = 0.15 kg × (40 m/s)

p1 = 6 kg.m/s

Change in momentum (ΔP) = p2 - p1

ΔP = -7.5 kg.m/s - 6 kg.m/s

ΔP = -13.5 kg.m/s

Time of collision (ΔT) = 8.0 × 10³ s

Now, putting the values of ΔP and ΔT in the equation of impulse momentum theorem, we get:

F = ΔP / ΔT

F = -13.5 kg.m/s ÷ 8.0 × 10³ s

F = -1.7 × 10⁻³ N

Thus, the value of force is 1.7 × 10⁻³ N, with the direction opposite to that of the bat's motion.

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If the Sun were replaced by a star that is 3 times brighter, the habitable zone in our solar system would undergo significant changes. The habitable zone, also known as the Goldilocks zone, refers to the region around a star where conditions are suitable for the presence of liquid water and potentially life as we know it.

A star that is 3 times brighter than the Sun would emit more radiation, including visible light and other forms of electromagnetic radiation. This increase in brightness would have several effects on the habitable zone:

Expansion of the habitable zone: The increased brightness of the star would lead to the habitable zone expanding farther away from the star. This means that planets previously located outside the habitable zone might now fall within it, potentially increasing the number of planets with the potential for liquid water and habitability.

Increased radiation and heat: With a brighter star, the habitable zone would experience higher levels of radiation and heat. This could lead to higher surface temperatures on planets within the habitable zone, potentially pushing the upper limits of what is considered habitable for life as we know it.

Changes in planetary atmospheres: The increased radiation and heat from a brighter star could impact the composition and stability of planetary atmospheres. There could be an increase in atmospheric evaporation, loss of volatile compounds, and changes in the balance of greenhouse gases, potentially affecting the overall habitability of planets.

It's important to note that the specific effects on the habitable zone would depend on various factors, including the characteristics of the star and the composition and properties of the planets within the system. Additionally, the concept of habitability is complex and depends on many factors beyond just the presence of liquid water, such as atmospheric conditions and the presence of suitable chemical building blocks for life.

In summary, if the Sun were replaced by a star 3 times brighter, the habitable zone in our solar system would likely expand and experience increased radiation and heat. This would have implications for the potential habitability of planets within the zone and could lead to changes in planetary atmospheres.

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The frequency of the wave at which it is most strongly reflected from the crystal is 200 GHz. Hence, the correct option is (b) 200 GHz.

Given, Side of a FCC cubic unit cell of a monatomic crystal = a = 5.6 A = 5.6 × 10^-8cm Density of the crystal = ρ = 5 g/cc = 5 × 10^3 kg/m³Force constant between two atoms = k = 1.5 × 10^7 N/m Young's modulus in the [100] direction = Y = 5 × 10¹¹ N/m²The wave is traveling along the [100] direction.

Here, λ = 2a = 2 × 5.6 × 10^-8 = 1.12 × 10^-7m, since the wave is traveling along the [100] direction.

The mass of each atom in the unit cell of the crystal is given by, m = (ρ × a³)/N where N is the Avogadro's number. Then, m = (5 × 10³ × (5.6 × 10^-8)³)/6.022 × 10²³

= 1.77 × 10^-25kg.

The velocity of sound in the crystal in the [100] direction is given by, v = √(Y/ρ) = √(5 × 10¹¹/5 × 10³)

= 2.24 × 10^3m/s.

The frequency of the wave at which it is most strongly reflected from the crystal is given by, f = (v/λ) = 2.24 × 10^3/(1.12 × 10^-7) = 2 × 10¹⁰ Hz or 200 GHz.

Approximately, the frequency of the wave at which it is most strongly reflected from the crystal is 200 GHz. Hence, the correct option is (b) 200 GHz.

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Answer:

Down Below

Explanation:

As nouns the difference between speed and strength

is that speed is the state of moving quickly or the capacity for rapid motion; rapidity while strength is the quality or degree of being strong.

Answer:

Explanation:is that speed is the state of moving quickly or the capacity for rapid motion; rapidity while strength is the quality or degree of being strong.

Answer:They have 10

Explanation:

Answer:

In order to convert density to grams, you have to put the mass on one side of the equation, and the density and the volume on the other. Therefore, d * v = m. Multiply the density by the volume. Using the example in step 1, you would multiply 2 g/mL by 4mL.

Explanation:

ok

Answer:

7

Explanation:

The magnitude of the acceleration of the body is approximately 1.802 m/s^2 when forces of magnitude 5 Newton and 4 Newton act on a body of mass 4.0 kg at a right angle to each other.

To determine the magnitude of the acceleration of a body when two forces act on it at a right angle, we can use the Pythagorean theorem and Newton's second law of motion.Newton's second law states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration (F = ma). In this case, the net force is the vector sum of the two forces acting on the body.Using the Pythagorean theorem, we can find the magnitude of the net force. Let's call the force with magnitude 5 Newton as F1 and the force with magnitude 4 Newton as F2. The net force (Fnet) can be calculated as the square root of the sum of the squares of the individual forces: Fnet = sqrt(F1^2 + F2^2).Now we can substitute the net force into Newton's second law to find the acceleration (a). Rearranging the formula, we have a = Fnet / m.Plugging in the values, Fnet = sqrt(5^2 + 4^2) = sqrt(25 + 16) = sqrt(41) Newtons. The mass (m) is given as 4.0 kg.

Finally, the magnitude of the acceleration of the body is calculated as a = Fnet / m = sqrt(41) / 4.0 ≈ 1.802 m/s^2.

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the magnitude of the average force needed to bring the rhino to a stop in 0.50s is 6800

The mass of the rhino is 1700 kg and it charges at a speed of 50.0 km/h. The magnitude of the average force needed to bring the rhino to a stop in 0.50s is 6800 N.

The kinetic energy of the rhino before stopping is given by:

K.E = ½ mv²where,m = 1700 kg

v = 50 km/h = (50 x 1000) / 3600 = 13.89 m/s

K.E = ½ x 1700 x (13.89)²= 1,282,587.35 J

Using the formula; force = change in momentum / time

We have,

Change in momentum = m(v - u)

where,

m = 1700 kg

v = final velocity = 0m/s

u = initial velocity = 13.89 m/s

Change in momentum = 1700 x (0 - 13.89)Change in momentum = -23613 N s

Force = -23613 / 0.50sForce = -47226 N

The force needed to stop the rhino is 47226 N (negative sign denotes that the force is opposite to the direction of motion)

Hence, the magnitude of the average force needed to bring the rhino to a stop in 0.50s is 6800 N.

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Answer:

the potential energy is 19.6J

the kinetic energy is 0.0.004J

Explanation:

using the formulas

P.E = mgh while

K.E =mv²/

The wavelength for which the center of the main diffraction maximum arising from the first star falls onto the first diffraction minimum from the second star is approximately 589 nm.

The angular separation of the two stars can be related to the wavelength of light and the diameter of the telescope lens by the equation:

θ = 1.22 λ / D

where θ is the angular separation, λ is the wavelength, and D is the diameter of the lens.

Solving for the wavelength, we get:

λ = θ D / 1.22

Plugging in the values given, we get:

λ = (4.23 × \(10^{-7}\) rad) × (1.7 m) / 1.22

λ = 5.89 ×\(10^{-7}\) m.

Therefore, the wavelength for which the center of the main diffraction maximum arising from the first star falls onto the first diffraction minimum from the second star is approximately 589 nm (nanometers).

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Answer:

I think B but I could be wrong

Explanation:

Answer:

Parallel Circuit

Explanation:

Check the image below

in series circuits currents can only flow in one direction

In parallel circuits currents can flow  through more than one path

Because the circuit shown has two pathways it is considered a parallel circuit

Answer:

Parallel Circuit

Explanation:

Hope this helps

Answer:

V = (S2 - S1) / 3600     average velocity of satellite

V = 10 m / 3600 sec = 2.78E-3 m/s

(b) is the correct answer

The orbital speeds would fall off sharply with increasing distance from the galactic center.

The plot of radial distance versus the orbital speed of objects gives us the galaxy rotation curve. The theoretical and practical curves have significant differences. A possible explanation for this difference could be the existence of dark matter.

According to the theoretical calculations, the curve should increase sharply and then decrease as the radial distance increases. The theoretical graph does not take the dark matter into account. But the plot made by observations shows the plot increasing first and then becoming constant as the radial distance increases.

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0N is the magnitude of the normal force. 0.002 m/s is the acceleration of the puck when we apply on the puck a horizontal and constant force F-0. 30 N (Newton).

the S.I unit of force is Newton (N) which is defined as 1N=1Kgm/\(s^{2}\).

Data given:

Mass of puck=150g

puck at rest means a (acceleration =0)

Now to find the magnitude of the normal force.

F=ma

F=150*0

F=0N

Hence, 0N is the magnitude of the normal force.

Now, When we apply on the puck a horizontal and constant force F-0. 30 N.

then to find the acceleration of the puck.

a=F/m

a=0.30N/150

a=0.002 m/s

Hence, 0.002 m/s is the acceleration of the puck when we apply on the puck a horizontal and constant force F-0. 30 N.

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Similarities between the compass needle (magnetism) and a charge to probe the electric field (electricity) is that Electric fields are produced by two kinds of charges, positive and negative similarly Magnetic fields are associated with two magnetic poles

Magnetic Field is the region around a magnetic material or a moving electric charge within which the force of magnetism acts. A pictorial representation of the magnetic field which describes how a magnetic force is distributed within and around a magnetic material.

Electric field, an electric property associated with each point in space when charge is present in any form. The magnitude and direction of the electric field are expressed by the value of E, called electric field strength or electric field intensity or simply the electric field.

Electric fields are produced by two kinds of charges, positive and negative. Magnetic fields are associated with two magnetic poles, north and south, although they are also produced by charges (but moving charges).

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a) The isocost curve equation is G = (20000 - 100C)/500. b) The isocost curve equation is G = (30000 - 100c)/500. c) The isocost curve equation is G = (20000 - 100C)/375. d)  The isocost curve equation is G = (20000 - 120C)/500.

To draw the isocost curve showing the different combinations of gas and coal, we need to use the cost per unit values for coal and gas, as well as the given expenditure (TC) and the changes in expenditure or prices.

Let's denote the quantity of coal as C and the quantity of gas as G. The cost per unit of coal is 100, and the cost per unit of gas is 500.

(a) Initial expenditure (TC) of 20000:

To find the combinations of gas and coal that can be purchased with an initial expenditure of 20000, we can use the following isocost equation

TC = 100C + 500G

We can rearrange the equation to solve for G in terms of C

G = (TC - 100C) / 500

Now we can plot the isocost curve with TC = 20000 using the equation above.

(b) Expenditure (TC) increases by 50%

If the expenditure increases by 50%, the new expenditure (TC_new) becomes 1.5 × TC = 1.5 × 20000 = 30000.

We can use the same isocost equation as before, but with the new expenditure value:

TC_new = 100C + 500G

Rearranging the equation to solve for G

G = (TC_new - 100C) / 500

Now we can plot the isocost curve with TC_new = 30000.

(c) Gas price reduced by 25%:

If the gas price is reduced by 25%, the new cost per unit of gas (Gas_new) becomes 0.75 × 500 = 375.

We can use the original isocost equation, but with the new cost per unit value:

TC = 100C + 375G

Rearranging the equation to solve for G

G = (TC - 100C) / 375

Now we can plot the isocost curve with the reduced gas price.

(d) Coal price rises by 20%

If the coal price rises by 20%, the new cost per unit of coal (Coal_new) becomes 1.2 × 100 = 120.

We can use the original isocost equation, but with the new cost per unit value:

TC = 120C + 500G

Rearranging the equation to solve for G:

G = (TC - 120C) / 500

Now we can plot the isocost curve with the increased coal price.

By plotting these isocost curves on a graph with G on the y-axis and C on the x-axis, we can visualize the different combinations of gas and coal that can be purchased at the given expenditures or price changes.

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Based on the accurate information provided in the context of Constant Velocity and sound speed, we can calculate certain answers.

a. The frequency of the sound wave emitted by the motorcycle is 700 Hz, so the period is

b. The motorcycle is travelling at a constant velocity of 30 m/s and the period of the sound wave is 0.0014 s, so the distance travelled by the motorcycle during one period of the sound oscillation is

c. The speed of sound in air is 340 m/s and the period of the sound wave is 0.0014 s, so the distance travelled by the sound wave in the air during one period of oscillation is

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The length of the rubber band that the boy must use is 0.024 m or 24 mm.

To determine the length of the rubber band, we can use the formula for the potential energy stored in a stretched spring, which is also applicable to a stretched rubber band:

where U is the potential energy stored in the rubber band, k is the spring constant (or in this case, the rubber band constant), and x is the displacement of the rubber band from its natural length.

Since the rubber band is stretched by 0.25 of its natural length, the displacement x is 0.25 times the natural length of the rubber band.

We can solve for the rubber band constant k by using the formula for the velocity of a projectile launched by a spring (or in this case, a rubber band):

where v is the velocity of the projectile, m is the mass of the rubber band, M is the mass of the projectile, and k is the spring constant. We can rearrange this equation to solve for k:

k = (v² M) / (2 m)

We can now combine the two equations to solve for the length of the rubber band, L:

U = 1/2 k x²

U = 1/2 ((v² M) / (2 m)) (0.25 L)²

U = (v² M L²) / (32 m)

The potential energy stored in the rubber band must be equal to the kinetic energy of the projectile when it is launched:

U = 1/2 M v²

(v² M L²) / (32 m) = 1/2 M v²

L = ((16 m v²) / (k M))

L = ((16 m v²) / ((v² M) / (2 m) M))

L = √(32 m^2 / M)

L = (0.032 M)

Substituting the given values, we get:

L = √(0.032 * 0.006)

L = 0.024 m

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The object moving the fastest at point C.

We want to assess some of the points in this time-position graph, which has some points marked on it.

The object is moving there at the highest speed since point C has the largest slope on the location versus time graph.A distance-time graph is a graph that displays an object at various positions and times. This graph's slope provides the speed value. is shifting in such a way that the slope of the position-time graph varies at six separate places.

The steepest slope of the x-t graph is found at point C. The object is therefore travelling the fastest at point C.

Your question is incomplete but most probably your full question attached below

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