all uninterruptible power supplies are measured in ________ and in volt-amps (va).

The number of teeth on the driven gear is 14 and the theoretical center-to-center distance is 3.34 inches.

Speed Ratio: Speed Ratio = (Number of Teeth on Driven Gear)/(Number of Teeth on Driving Gear). The Speed Ratio = 1400 rev/min/2100 rev/min = 0.6667.

Therefore, the number of teeth on the driven gear = (Number of Teeth on Driving Gear) x (Speed Ratio) = 20 x 0.6667 = 13.33. Rounding up, we can conclude that the number of teeth on the driven gear is 14.

The next step is to find the theoretical center-to-center distance. To do this, we need to use the formula for calculating Pitch Diameter: Pitch Diameter = (Number of Teeth)/(Diametral Pitch).

In this case, the Pitch Diameter of the driving gear is (20 teeth)/(12 teeth/in) = 1.67 inches. Therefore, the center-to-center distance = Pitch Diameter x 2 = 1.67 inches x 2 = 3.34 inches.

Hence the number of teeth on the driven gear is 14 and the theoretical center-to-center distance is 3.34 inches.

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Answer:

C. Light does not need a medium to travel through, but since sound waves must have a medium to vibrate, sound is not created where no air is present.

Explanation:

The reason the bell is seen to vibrate but not heard is that unlike light, which does not require a medium to pass through, sound waves require an air-filled space in order to vibrate. Option C is correct.

A sound wave is produced when a medium begins to vibrate. When an entity vibrates, a pressure wave is formed, which causes sound.

Light does not need a medium to travel through, but since sound waves must have a medium to vibrate, the sound is not created where no air is present best describes why the bell is seen vibrating but not heard.

The complete question is attached.

Hence option C is correct.

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Answer: The answer is B) False

Explanation: There are only three muscle types in the exercise physiology.

Skeletal Muscle

Smooth Muscle

Cardiac Muscle

If the ice extended right around the earth, forming a loop that the puck could go around in circles, the puck would not keep going forever. The laws of physics would eventually cause the puck to come to a stop. This is because of two factors: air resistance and gravity.

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The same property of Uranus's rings that is the same as Saturn's rings is the composition. Both rings are composed of countless small particles of ice, dust, and rocks that range in size from a few centimeters to several meters.

These particles are believed to be remnants of a broken moon or other celestial body that has been destroyed by a collision with another object in the Solar System. Uranus and Saturn's rings also share similar colors, which range from blue to gray, depending on the light they receive from the Sun.

However, the other properties of Uranus's rings make them distinct from Saturn's. The most noticeable difference is the number of rings; Uranus has 13 rings compared to Saturn's 7. Additionally, Uranus's rings are much thinner than Saturn's, and their orbital inclinations are much larger. Finally, Uranus's rings contain more dust particles than Saturn's rings.

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Answer:

Cool.

Explanation:

What's the question..? :|

Answer:

Everything is made up of tiny particles called Matter

Answer:

exposing it to a flame to see if it catches on fire

Explanation:

A chemical property is when the matter changes into a different substance. This is true when you expose it to a flame because if a substance is burned, it will change into a new substance. All the other options are examples of physical changes.

-I also took the quiz and got 100%

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Answer:

51.54 m/s

Explanation:

Applying,

Law of conservation of momentum,

Total momentum before collision = Total momentum after collision

mu+m'u' = mv+m'v'.................... Equation 1

Where m = mass of the head of the golf club, m' = mass of the gulf ball, u = initial velocity of the head of the gulf club, u' = initial velocity of the gulf ball, v = final velocity of the head of a gulf club, v' = final velocity of the gulf ball

From the question,

Given: m = 151 g = 0.151 kg, u = 43.9 m/s, m' = 46 g = 0.046 kg, u' = 0 m/s (at rest), v = 28.2 m/s

Substitute these values into equation 1

0.151(43.9)+0.046(0) = 0.151(28.2)+0.046(v')

solve for v'

6.6289+0 = 4.2582+0.046v'

0.046v' = 6.6289-4.2582

0.046v' = 2.3707

v' = 2.3707/0.046

v' = 51.54 m/s

Answer:

1,110,114.6 J

Explanation:

The height is 1857m, the mass is 61 kg, and the accelaration is 9.8 m/s^2. Plug in the equation GPE=h*m*a and you get 1,110,114.6.

h= height

m= mass

a= accelaration

GPE= gravitational potential energy

If you are at the grand canyon standing at the edge of a ledge 1857 meters high you have a mass of 61 kilograms, then your gravitational potential energy at the edge of the cliff would be 1111247.37 Joules.

Mechanical energy is the combination of all the energy in motion represented by total kinetic energy and the total potential energy stored energy in the system which is represented by total potential energy.

As total mechanical energy is the sum of all the kinetic as well as potential energy stored in the system.

ME = KE + PE

The total gravitational potential energy at the edge of the cliff would be

PE = m × g × h

PE =61 × 9.81 × 1857

PE =1111247.37 Joules

Thus, the total gravitational potential energy at the edge of the cliff would be  1111247.37 Joules.

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The surface temperature of Betelgeuse is 3,500 K.

Betelgeuse is a red supergiant and one of the brightest stars in the night sky. It is a relatively young star, estimated to be about 10 million years old and is located about 640 light years away from Earth. It is one of the largest stars known and is about 1,000 times larger in radius than our Sun.

Its outer atmosphere has a temperature of about 3,500 K, which is much hotter than the Sun's surface temperature of 5,778 K. The high temperature of the outer atmosphere is due to the fact that Betelgeuse is a red supergiant. In comparison to other stars, red supergiants are among the hottest due to their large surface areas. Betelgeuse is an important star in astronomical studies because of its size and brightness.

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(a) The image distance is approximately 5.10 cm.

(b) The object distance is approximately 1.9659 cm.

To solve the problem, we can use the lens formula, which relates the object distance (u), image distance (v), and focal length (f) of a lens:

1/f = 1/v - 1/u

Given:

Focal length (f) = -3.20 cm (negative sign indicates a diverging lens)

Image distance (v) = 5.10 cm

(a) Finding the image distance (v):

We know that the focal length (f) and image distance (v) are given. Plugging these values into the lens formula, we can solve for the object distance (u).

1/f = 1/v - 1/u

Substituting the given values:

1/(-3.20 cm) = 1/(5.10 cm) - 1/u

Simplifying:

-0.3125 cm^(-1) = 0.1961 cm^(-1) - 1/u

Rearranging the equation:

1/u = 0.1961 cm^(-1) + 0.3125 cm^(-1)

1/u = 0.5086 cm^(-1)

Taking the reciprocal:

u = 1 / (0.5086 cm^(-1))

u = 1.9659 cm

Therefore, the object distance is approximately 1.9659 cm.

(b) Finding the object distance (u):

We have already found the object distance (u) in the previous step.

Object distance (u) ≈ 1.9659 cm.

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The experimental evidence led to the development of have gathered new evidence. One of the atomic models this atomic model from the one before it is the colors of light emitted from heated atoms had very specific energies.

As scientific knowledge about the structure of the atom has changed, so have models of the atom.

The Rutherford planetary model, which depicts electrons travelling around the nucleus like planets move around the sun in the solar system, is the direct precursor of the model illustrated in the figure attached.

This model was created as a result of the study of light hues released by heated atoms with highly particular energies.

The Bohr model of the atom is so named because Niels Bohr had a major role in developing it. The passage of electrons between energy levels caused the light hues released by heated atoms. Electrons are grouped in shells or energy levels.

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The heat load for Weapons-Grade Plutonium is approximately 0.295 Watts/gram, while the heat load for Reactor-Grade Plutonium is approximately 0.441 Watts/gram.

Weapons-grade (WG) and reactor-grade (RG) plutonium differ in their isotopic composition, which affects their characteristics, including the heat load they generate. Let's compare the two compositions and calculate the heat load for each.

1. Weapons-Grade Plutonium (WG):

Weapons-grade plutonium is highly enriched in the fissile isotope plutonium-239 (Pu-239). It typically contains more than 93% Pu-239. The remaining composition consists of other plutonium isotopes such as Pu-240, Pu-241, and trace amounts of Pu-238.

2. Reactor-Grade Plutonium (RG):

Reactor-grade plutonium is obtained from spent nuclear fuel in nuclear reactors. It contains a lower percentage of Pu-239 and a higher percentage of other isotopes, particularly Pu-240. Reactor-grade plutonium generally has a Pu-239 content of less than 93%, typically around 70-85%. The remaining composition includes Pu-240, Pu-241, Pu-242, and other isotopes.

Now, let's calculate and compare the heat load generated by each composition.

The heat load (heat generated per unit mass) for plutonium isotopes can be estimated based on their specific decay heat values. Decay heat is the heat generated by the radioactive decay of isotopes. Here are the approximate decay heat values for the commonly occurring plutonium isotopes:

- Pu-238: 0.56 Watts/gram

- Pu-239: 0.56 Watts/gram

- Pu-240: 0.05 Watts/gram

- Pu-241: 0.19 Watts/gram

- Pu-242: 0.02 Watts/gram

To calculate the heat load for each composition, we need the isotopic composition in weight percent. Let's assume the following compositions:

- Weapons-Grade Plutonium (WG):

   - Pu-238: 0.5%

   - Pu-239: 93.5%

   - Pu-240: 6%

   - Pu-241: 0.9%

   - Pu-242: 0.1%

- Reactor-Grade Plutonium (RG):

   - Pu-238: 1%

   - Pu-239: 70%

   - Pu-240: 25%

   - Pu-241: 3%

   - Pu-242: 1%

Now, let's calculate the heat load for each composition:

For Weapons-Grade Plutonium (WG):

Heat load = (0.005 × 0.56) + (0.935 × 0.56) + (0.06 × 0.05) + (0.009 × 0.19) + (0.001 × 0.02) = 0.295 Watts/gram

For Reactor-Grade Plutonium (RG):

Heat load = (0.01 × 0.56) + (0.7 × 0.56) + (0.25 × 0.05) + (0.03 × 0.19) + (0.01 × 0.02) = 0.441 Watts/gram

Therefore, the heat load for Weapons-Grade Plutonium is approximately 0.295 Watts/gram, while the heat load for Reactor-Grade Plutonium is approximately 0.441 Watts/gram.

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We are given two vectors and we are asked to determine the direction of their sum. To do that we will rewrite them on coordinates form. For vector A, since it is a horizontal vector it only has an x-coordinate equivalent to its magnitude, like this:

For vector B we will use the following formula:

Replacing the values:

Solving the operations

(a)

The work done by the spring force can be calculated with the formula below:

Using k = 320 N/m and x = 0.075 m, we have:

(b)

The increase in thermal energy is given by the work done by the friction force.

To calculate this work, first let's find the friction force:

Now, calculating the work, we have:

(c)

The block speed can be found by converting the potential energy from the spring (same value of the calculated work in item a) into kinetic energy for the block:

The pellet has a kinetic energy of 4.9 10-2 J as it exits the spring.

Kinetic energy is the term for the energy present in an object in motion. Examples of kinetic energy include you walking down the street, the globe rotating around the sun, and motion in space. The relationship between kinetic energy and an object's mass and square of its velocity is direct: K.E. = 1/2 m v2.

To calculate the kinetic energy of the pellet, solve the energy conservation equation using the necessary variables as substitutes.

KE = EPE

KE = 1/2 kx²

KE = 1/2(20N/m) (7cm ×1/100cm)²

KE = 4.9 × 10⁻²J.

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Answer:

650N

Explanation:

f = ma

130kg x 5 m/s^2 = 650N

Answer:

12 pounds mine does

Explanation:

Answer: 10 pnds

Explanation:

The mechanical advantage of the nutcracker is 3.5.

The Mechanical Advantage (MA) is a measure of how much a machine, such as a nutcracker, amplifies the input force. It is calculated by dividing the output force by the input force.

In your scenario, an input force of 75 N (Newtons) is applied to the nutcracker, and the output force exerted on the walnut is 262.5 N. To find the mechanical advantage, we will use the formula:

MA = Output Force / Input Force

Plugging in the values from your question, we have:

MA = 262.5 N / 75 N

MA = 3.5

The mechanical advantage of the nutcracker in this situation is 3.5. This means that the nutcracker amplifies the applied force by 3.5 times, allowing you to crack the walnut with greater ease. A higher mechanical advantage indicates that the machine is more effective at increasing the applied force, making tasks easier to accomplish.

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Answer:

please help

Explanation:

The minimum weight of a bowling ball with a diameter of 8.5 inches that would sink in water is approximately 33.4 pounds. To calculate the minimum weight required for the bowling ball to sink, we can use the principle of buoyancy.

The buoyant force acting on the ball must exceed the weight of the ball for it to sink. The buoyant force is equal to the weight of the water displaced by the ball. The volume of the bowling ball can be calculated using the formula for the volume of a sphere, which is given by \(\(\frac{4}{3}\pi r^3\)\), where r is the radius of the ball. By substituting the given diameter of 8.5 inches, we can find the radius to be 4.25 inches or approximately 0.108 meters.

To determine the weight of the water displaced by the ball, we need to find the volume of water displaced. The volume of the water can be calculated using the formula \(\(V = \pi r^2 h\)\), where h is the height of the submerged portion of the ball. When the ball is completely submerged, the height will be equal to the diameter, which is 8.5 inches or approximately 0.216 meters. By substituting the values into the formula, we can find the volume of the water to be approximately 0.00848 cubic meters.

Since the density of water is given as 1.0 g/mL, we can convert the volume of water into mass by multiplying it by the density. This gives us a mass of 8.48 kilograms. Finally, to determine the minimum weight of the ball, we need to convert the mass from kilograms to pounds. 1 kilogram is approximately equal to 2.20462 pounds, so the minimum weight of the bowling ball required to sink is approximately 18.7 pounds.

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===================================

\( \large \sf \underline{Problem:}\)

===================================

\( \large \sf \underline{Answer:}\)

\( \huge \qquad \quad \sf{48km/h}\)

===================================

\( \large \sf \underline{Solution:}\)

Given:

\( : \implies \sf \quad{s = \frac{d}{t} }\)

\(: \implies \sf \quad{s = \frac{144}{3}} \)

\(: \implies \sf \quad{\underline{\underline{\pmb{48 \: km/h}}}}\)

The variables of speed, distance and time has a correlation on one another. In finding the speed, you should divide the distance to time because speed is inversely proportional to distance and time.

===================================

Answer:

Gamma and x-ray are electromagnetic and highly penetrating,

Beta radiation is due to energetic electrons while alpha radiation is consists of atomic nuclei and much more massive

Alpha radiation requires the least amount of protective clothing because the particles are much more massive and less penetrating

Answer:

7. They arethe meter (m), the kilogram (kg), the second (s), the kelvin (K), the ampere (A), the mole (mol), and the candela (cd)

Explanation:

7. They arethe meter (m), the kilogram (kg), the second (s), the kelvin (K), the ampere (A), the mole (mol), and the candela (cd)

B that is the correct answer

Answer:

Conventional current is the rate at which positive charge flows in a circuit.

Explanation:

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The time of motion of the track star is determined as 0.837 s.

The time of motion of the track star is calculated as follows;

T = (2u sinθ)/g

where;

T = (2 x 12 x sin20)/9.8

T = 0.837 s

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Answer:

The object accelerates downward at 4 m/s² since the tension on the rope is less than weight of the object.

Explanation:

Given;

mass of the object, m = 2 kg

weigh of the object, W = 20 N

tension on the rope, T = 12 N

The acceleration of the object is calculated by applying Newton's second law of motion as follows;

T = F + W

T = ma + W

ma = T - W

\(a = \frac{T-W}{m} \\\\a = \frac{12 - 20}{2} \\\\a = -4 \ m/s^2\) (the negative sign indicates deceleration of the object)

The object accelerates downward at 4 m/s² since the tension on the rope is less than weight of the object.