When a power plant sends out high voltage electricity for long distance transmission,
what is needed to make the electricity suable in houses?

•a step down transformer
•a step up transformer
•a generator
•a commutator

a) To find the speed of a blade 10 seconds after the fan is turned off, we need to calculate the angular velocity of the fan and then convert it to linear velocity.

Given:

- Diameter of the fan blades = 80 cm = 0.8 m

- Fan is turning at 60 rpm (revolutions per minute)

First, we find the angular velocity:

Angular velocity (ω) = 2π × (Revolutions per minute) / 60

                   = 2π × 60 / 60

                   = 2π rad/s

Next, we find the linear velocity of the tip of the blade:

Linear velocity (v) = (Angular velocity) × (Radius)

                   = (2π rad/s) × (0.4 m)

                   = 0.8π m/s

After 10 seconds, the blade will travel a distance equal to its linear velocity multiplied by the time:

Distance = (Linear velocity) × (Time)

        = (0.8π m/s) × (10 s)

        = 8π m

Therefore, the speed of the tip of the blade 10 seconds after the fan is turned off is 8π m/s.

b) To calculate the number of revolutions the fan makes while stopping, we need to find the time it takes for the fan to come to a stop and then convert it to the number of revolutions.

Given:

- Fan coasts to a stop in 25 seconds

The time for one revolution is the inverse of the angular velocity:

Time for one revolution = 1 / (Angular velocity)

                      = 1 / (2π rad/s)

                      = 1 / (2π) s

The number of revolutions can be calculated by dividing the total stopping time by the time for one revolution:

Number of revolutions = (Total stopping time) / (Time for one revolution)

                    = 25 s / (1 / (2π) s)

                    = 25 × (2π)

                    = 50π

Therefore, the fan makes approximately 50π revolutions while stopping.

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Answer:

There are three types of magnets: permanent magnets, temporary magnets, and electromagnets. Permanent magnets emit a magnetic field without the need for any external source of magnetism or electrical power.

The Earth's magnetic field is believed to be generated by electric currents in the conductive iron alloys of its core, created by convection currents due to heat escaping from the core.

Explanation:

Answer:

Q = N e     where N is number of electrons and Q is total charge

I = Q / t       where I is current and t = sec

I = Q / t = 12.55E-6 Coul / Sec

Q = 12.55E-6 Coul/sec * 2.39 sec = 3.00E-5 Coul    total charge

N = 3.00E-5 coul / 1.60E-19 coul = 1.87E14 electrons

(electronic charge = 1.60E-19 Coul)

Answer:

split traction

Explanation:

Answer:

insulin and glucagon, to regulate blood-glucose levels

Explanation:

When a rope of length L and mass m is suspended from the ceiling, the tension in the rope at position y can be found using the following expression:

T(y) = mg + λy where g is the acceleration due to gravity, λ is the linear mass density of the rope, and y is the distance measured upward from the free end of the rope.

Here's how to derive the expression: Let's consider an element of length dy of the rope at a distance y from the free end of the rope. The weight of the element is dm = λdy and acts downward. The tension in the rope on the element can be resolved into two components - one acting downward and another acting upward. Let T be the tension in the rope at point y and T + dT be the tension in the rope at point (y + dy).The upward component of tension on the element is given by Tsinθ, where θ is the angle between the element and the vertical. As the rope is assumed to be in equilibrium, the horizontal components of tension balance each other and the net vertical force on the element is zero. Therefore, we have,

Tsinθ - dm g = 0 ⇒ Tsinθ = dm g ⇒ Tsinθ = λdyg

The angle θ can be found using the equation tanθ = dy/dx ≈ dy/dy = 1. Therefore, sinθ = dy/√(dy²+dx²) ≈ dy and we have,T dy = λdyg ⇒ T = λgThis expression gives the tension in the rope at the free end of the rope. The tension in the rope at position y, measured upward from the free end of the rope is given by,T(y) = mg + λy

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Answer:

2 seconds lol im just a kid but I know these lol who has tiktok

Answer:

1200

Explanation:

a = 1.5 m/s2

F = 1800N

M =?

F = Ma

M = F/a = 1800/1.5 = 1200Kg

A halo around the sun or moon is often an indication that cirrostratus clouds are present.

Cirrostratus clouds are high-level clouds that form above 20,000 feet (6,000 meters). They are composed of ice crystals and have a thin, often transparent appearance. When these clouds are present, they can create a halo effect around the sun or moon.

The halo is formed when light from the sun or moon is refracted, or bent, as it passes through the ice crystals in the cirrostratus clouds. This refraction causes the light to spread out and create a ring or halo-like appearance around the sun or moon.

The presence of a halo is often an indication of thin, high-level cloud cover, typically preceding an approaching warm front or an area of moisture. It can also be a sign of changing weather conditions.

It is worth noting that halos can also be formed by other cloud types, such as altostratus or thin, wispy cirrus clouds, depending on the specific atmospheric conditions.

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The distance of Uranus from the Sun in astronomical units is 9.13 AU.

What is astronomical unit?

In our solar system, distances are really great. Therefore, astronomers frequently avoid using miles or kilometers when describing the distances to planets, asteroids, comets, or spacecraft. They substitute astronomical units, or AU, which represent the typical separation of the Earth from the sun. That translates to around 93 million miles, 150 million kilometers, or 8 light-minutes.

By above definition, we can say that

1 astronomical unit (AU) = 150 million kilometers (km)

In order to find the distance in astronomical units multiply the given distance expressed in kilometers of Uranus, which is 2870 million kilometers, by the unit ratio, 1 AU per 150 million kilometers, as shown on the expression:

\((2870 million Km)(\frac{1Au}{150millionKm})\)

Therefore, the distance of Uranus from the Sun is 19.13 AU.

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​

As an astronaut meal planner, here is a menu for a day's worth of meals in space includes Breakfast, Lunch and Dinner.

Breakfast:

Scrambled Eggs with Vegetables

Special Packaging: Individual vacuum-sealed pouches

Preparation: The scrambled eggs with vegetables will be precooked and dehydrated. Astronauts will rehydrate the meal with hot water, stir, and allow it to sit for a few minutes before consuming. This ensures the eggs and vegetables rehydrate properly in the zero-gravity environment.

Lunch:

Teriyaki Chicken Stir-Fry with Rice

Special Packaging: Sealed pouches for each component (chicken, vegetables, rice, and sauce)

Preparation: The chicken, vegetables, and rice will be precooked and dehydrated separately. The teriyaki sauce will be in a separate pouch. Astronauts will rehydrate each component individually with hot water, then mix them together in a resealable pouch. After a few minutes, the meal will be ready to eat.

Dinner:

Grilled Salmon with Quinoa and Roasted Vegetables

Special Packaging: Vacuum-sealed pouches for each component

Preparation: The salmon will be cooked and then freeze-dried for preservation. The quinoa and roasted vegetables will be precooked and dehydrated separately. Astronauts will rehydrate the quinoa and vegetables with hot water, then warm up the salmon using a food warmer. The components will be combined on a tray and secured with fasteners to prevent them from floating away.

Modifications for space include using dehydration and freeze-drying techniques to remove water content from the food while preserving nutrients. Vacuum-sealed and sealed pouches are used to prevent spoilage and maintain freshness. Rehydration with hot water is necessary to restore the food's texture and taste.

Additionally, it's important to consider portion control and individual packaging to prevent cross-contamination and maintain food safety. The use of condiments and seasonings may be limited to avoid loose particles floating in the spacecraft.

By designing meals that are easy to prepare, consume, and clean up, astronauts can enjoy nutritious and satisfying meals while adapting to the unique challenges of the zero-gravity and non-refrigerated environment in space.

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Results are never more exact than the least precise measurement. The average lap time surpasses clock precision.  I will have a fair amount of confidence in the results of the report

This is further explained below.

Generally, A result can never be more accurate than the measurement that was the least accurate.

The calculated average lap time is more accurate than the clock can measure.

A conclusion can never be more accurate than the measurements that were used to get at it. The accuracy that may be achieved with the clock is exceeded by the average lap time that was determined.

Hence, for a clock that bears a precision of 0.2 seconds, I will be confident in its results bearing precision in mind, Because knowing the precision you can easily manipulate the results to get the desired outcome

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Answer:

6.25 second

Explanation:

we know that

s= (u+v)/2*T

125=20t

t=6.25 second

plz mark as brainalist

The majority of electromagnetic energy from space cannot reach the Earth's surface. The only light that reaches sea level is radio waves, visible light, and some ultraviolet light.

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1a) The total energy stored when all three weights are raised by 70 cm is 80.5 J.

1b) The height of the tank is 2 m.

2b) The potential energy is converted into kinetic energy when the water flows out of the taps 6.32 m/s.

3a) The mass of water that goes over the falls every second is 1 x 10⁶ kg.

3b) The gravitational potential energy of 1 kg of water at the top of the falls 1000 J.

3c) The speed of the water as it reaches the bottom if all the GPE stored in 1 kg of water is transferred to kinetic energy is 44.72 m/s.

3d) The water will not be falling as fast as the speed calculated in part c suggests because of the presence of air resistance and the fact that the water falls through a medium (air) which offers resistance to its motion.

3e) The energy transferred when the GPE stored in the water falling in one second is transferred to other energy stores is finally stored in thermal energy stores due to the heat generated by the water as it hits the bottom.

1a) To calculate the amount of energy stored in the three weights, we use the formula given below:

E = mgh

Where,E = Energy (Joules)

m = Mass (kg)

g = Gravity (10 N/kg)

h = Height (m)

For the 4.5 kg weight:

E = 4.5 x 10 x 0.7 = 31.5 J

For each of the 3.5 kg weight:

E = 3.5 x 10 x 0.7 = 24.5 J

Thus, the total energy stored when all three weights are raised by 70 cm is:

31.5 J + 24.5 J + 24.5 J = 80.5 J

1b) To calculate how far the 4.5 kg weight must be lifted to store 45 J of energy, we use the formula:

E = mghh = E/mg = 45 / (4.5 x 10) = 1 m2a)

To calculate the gravitational potential energy stored in the water in the tank when it is full, we use the formula given below:

E = mgh

Where,E = Energy (Joules)

m = Mass (kg)

g = Gravity (10 N/kg)

h = Height (m)

The mass of 1 litre of water is 1 kg and the tank can hold 200 litres of water. Therefore, the total mass of water in the tank is:

Mass = 200 kg

The height of the tank is 2 m.

Therefore, the gravitational potential energy stored in the water in the tank is:

E = mgh = 200 x 10 x 2 = 4000 J

Assumptions made in the answer:

We have assumed that the tank is full.

2b) To calculate the speed at which the water would come out of the bathroom and kitchen taps, we use the formula given below:

PE = KEPE = mghKE = 1/2mv²

Where,PE = Potential Energy (Joules)

KE = Kinetic Energy (Joules)

m = Mass (kg)

g = Gravity (10 N/kg)

h = Height (m)

v = Velocity (m/s)

Assuming that the potential energy of the water in the tank is converted into kinetic energy when the water flows out of the taps, the potential energy stored in the water in the tank is given by:

PE = mgh = 200 x 10 x 2 = 4000 J

The potential energy is converted into kinetic energy when the water flows out of the taps.

Therefore, KE = 1/2mv²v² = 2KE/mv² = 2(4000)/200 = 40 m²/s²v = √(40) = 6.32 m/s (speed of the water coming out of the taps)

3a) To calculate the mass of water that goes over the falls every second, we use the formula given below:

Mass = Volume x Density

Where,Volume = 1000 m³/s, Density = 1000 kg/m³, Mass = 1000 x 1000 = 1000000 kg = 1 x 10⁶ kg

3b) To calculate the gravitational potential energy of 1 kg of water at the top of the falls, we use the formula:

E = mgh

Where,m = 1 kg, g = 10 N/kg, h = 100 m, E = 1 x 10 x 100 = 1000 J

3c) To calculate the speed of the water as it reaches the bottom if all the GPE stored in 1 kg of water is transferred to kinetic energy, we use the formula given below:

PE = KEP

E = mgh

KE = 1/2mv²

Where,PE = Potential Energy (Joules)

KE = Kinetic Energy (Joules)

m = Mass (kg)

g = Gravity (10 N/kg)

h = Height (m)

v = Velocity (m/s)

Assuming that all the potential energy is converted into kinetic energy when the water reaches the bottom,

PE = KEKE = mghv² = 2mghv² = 2(1)(10)(100)v² = 2000v = √(2000) = 44.72 m/s

3d) The water will not be falling as fast as the speed calculated in part c suggests because of the presence of air resistance and the fact that the water falls through a medium (air) which offers resistance to its motion.

3e) To calculate the total energy transferred per second as the GPE stored in the water falling in one second is transferred to other energy stores, we use the formula given below:

Power = Energy / Time

Where,Power = 1 x 10⁶ x 10 x 100 = 1 x 10⁹ W = 1 GW (assuming that 1 m³ of water falls every second)3f)

The energy transferred when the GPE stored in the water falling in one second is transferred to other energy stores is finally stored in thermal energy stores due to the heat generated by the water as it hits the bottom.

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The acceleration due to gravity is given as:

                             g = GM/r²

According to Newton's second law of motion,

F = ma

where,

F = force

m = mass

a = acceleration

According to Newton's law of gravity,

Fg = GMm/(r + h)²

Fg = gravitational force

From Newton's second law of motion,

Fg = ma

a = Fg/m

We can refer to "a" as "g"

a = g = GMm/(m)(r + h)²

g = GM/(r + h)²

When the object is on or close to the surface, the value of g is constant and height has no considerable impact. Hence, it can be written as,

g = GM/r²

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Answer:

\(45m\)

Explanation:

Let \(V_0\) be the horizontal speed of the keys and D be the horizontal distance covered by the keys.

\(V_0= 8.0 \frac ms\)

\(D= 28 m\)

The initial velocity in the vertical direction is \(0\).

Let \(g\) be the acceleration due to gravity, which always acts vertically downwards, so, it will not change the horizontal direction of the speed, i.e. \(V_0\) will remain constant throughout the projectile motion.

Let \(t\) be the total time of flight in seconds. So,

\(D= V_0t\)

\(\Rightarrow 28=8t\)

\(\Rightarrow t=3 s \; \cdots (i)\)

Let \(H\) be the hight of the cliff.

Now, from the equation of motion

\(s=ut+\frac 1 2 at^2\;\cdots(ii)\)

Where is the displacement is the direction of force, is the initial velocity in the direction of force,   is the constant acceleration due to force and is the time of flight.

Here, , and  (negative sign is for taking the sigh convention positive in the direction as shown in the figure.)

From equation (ii)

\(-H=0\times t + \frac 1 2 (-g)t^2\)

\(\Rightarrow H= \frac 1 2 gt^2\)

Take \(g= 10 \frac m{s^2}\) and put the value of \(t\) from equation (i), we have

\(H=\frac 1 2 \times 10\times 3^2\)

\(\Rightarrow H= 45 m\).

Hence, the height of the cliff was \(45 m\).

The correct answer is A) bent toward the normal. When a light ray moves from air into glass, which has a higher index of refraction, its path will be bent toward the normal. This is due to the fact that light travels more slowly in glass than in air and thus is bent at the interface between the two materials.

When a light ray passes from one medium to another, it can be refracted, or bent. This is due to the fact that the speed of light varies depending on the medium it is travelling through. When a light ray moves from a medium with a lower index of refraction, such as air, to a medium with a higher index of refraction, such as glass, its path will be bent toward the normal. This phenomenon is known as refraction, and it is an important part of optics and optical systems.

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Answer:

Explanation:

That’s dangerous who would do that.

Me

The maximum height reached by the ball is 38.42 meters

The maximum height attained by the projectile is determined from the time taken for the projectile ton reach the maximum height and it is calculated as follows.

The time taken for the projectile to reach maximum height is calculated as follows;

t = ¹/₂ x 5.6 seconds = 2.8 seconds

Maximum height reached by the projectile is calculated as follows;

h = ¹/₂gt²

h = 0.5(9.8)(2.8²)

h = 38.42 m

Thus, the maximum height reached by the ball is 38.42 meters.

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The forces that came into play in this act are the following:
Force of friction: This is the force that opposes the motion between two surfaces in contact. In the case of the car, the force of friction between the tires and the road helps to slow down the car.

Air resistance: This force is also known as drag and is the resistance encountered by an object moving through the air. As the car moves forward, it encounters air resistance, which opposes its motion and causes it to slow down. Gravity: This force pulls the car towards the earth. Even though the car is moving forward, gravity is constantly acting on it and trying to slow it down.

When the car starts from rest, the only force acting on it is the force of the engine, which propels it forward. As the car moves, it encounters resistance from the forces of friction and air resistance. These forces act in the opposite direction to the car's motion and cause it to slow down. Eventually, the car comes to rest due to the combined effect of these forces and gravity.

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The main answer is that the person makes it to the safe building.From the information given in the question, we have;Initial Velocity, u = 8.10 m/sTime, t = ?Distance, d = 6.50 m (horizontal) and 3.00 m (vertical)It's evident that the motion of the person can be described by two independent equations.

The horizontal motion equation can be given as;{d = u*t} --------(1)The vertical motion equation can be given as;{d = (1/2)*g*t²} -------(2)Where g = 9.81 m/s² is the acceleration due to gravity. Since the person is falling, it's negative. By substituting equation (1) into equation (2), we get;{(1/2)*g*t²} = {(u*t)}² - {6.50 m}²This equation can be solved for time. To solve for the time, use the quadratic formula;{t = (-b ± sqrt(b² - 4ac))/2a}

Where; a = 1/2g, b = -8.1 and c = -(6.5)² + (3.0)Thus;{t = (-(-8.1) ± sqrt((-8.1)² - 4(1/2*-9.81)*(-(6.5)² + (3.0))))/2(1/2*-9.81)}Using the calculator, we get;{t = 1.36 s} (time is positive)From equation (1);{d = u*t} = (8.10 m/s) x (1.36 s) = 11.016 mThis distance is greater than the horizontal distance between the two buildings (6.50 m). Therefore, the person makes it across to the safe building.

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Answer:

a = (V2 - V1) / t     average acceleration between points 1 & 2

V2 = speed at bottom = -4.8 m/s

V1 = 4.8 m/s     speed when thrown, choosing upwards as positive

a = (-4.8 - 4.8) / .98       time upwards = time downwards

a = -9.6 m/s / .98 s

a = -9.8 m/s^2

Answer:

there is no image to go off of. If you could share that I would be happy to help.

Explanation:

Please share the image!