What is the wavelength of an AM radio wave in a vacuum if its frequency is 810 kHz?

Answer:

C. Mg3N2

Explanation:

Magnesium has a positive 2 charge and nitrogen has a negative 3 charge. To completely balance it out, you need three magnesium’s and two nitrogen’s. The final charge for magnesium would be positive 6 and the final charge for nitrogen would be negative 6, this will make it neutral.

Answer:

C. Mg3N2

How I Know :

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Answer:

this statement is true because I looked it up

Answer:

The maximum speed of the oscillating object doubles as well.

Explanation:

The oscillatory motion of the object on the spring can be modelled using this equation below:

\(Vmax= A\omega\)

Where

Vmax = the maximum speed of the oscillating object.

A = Amplitude of the oscillating object.

\(\omega\) = angular velocity of the oscillating object.

If the amplitude, A becomes doubled, the new amplitude will be = 2A.

As a result,  \(Vmax = 2 A \omega\)

This shows that the maximum speed has doubled in its value

Answer:

4 Ohms

Explanation:

Apply the formula:

Voltage = I (current) . Resistance

You can change it the way you want to use for your purpose.

In this case...

R = V/I

R = 12/3

R = 4 Ohms (Ohm is the unit of measurement of eletrical resistance)

Answer:

Principle Archimedes is applied in building a ship and submarine using the manipulating that buoyancy, is controlled the ballast tank system.

Explanation:

Submarine is rather had they focused on main parts of the submarine,he is complex and long process implementation,the most submarine design like submarine stability.

Submarine stability is complete and the fundamental Archimedes principle to arrive the weight of submarine is equal to buoyancy force.

Submarine into the parts and components of ballast tank the sequence in diving and surfacing,there two vital parts:-  flood parts and air vents

flood parts:- at the bottom position and allow water to enter or leave that tank.

air vents:- air vents at the top of the pressure hall,and that they submarine dive.

this time submarine is most modern system is depth is 300 to 450 meters,high pressure  air is 15 bar is tank air valve.

submarine is basic of the effective volume of all the submarine surfaced condition,submarine minus to the free water flood is equal to the fully pressure hull,submarine is the surfaced condition.

Answer:

B. radio waves

Explanation:

Trust me It's correct

The water can be raised to a height of approximately 15.7 m above the water surface at point A.

To solve this problem, we need to use the principles of fluid mechanics, specifically the Bernoulli's equation and the continuity equation.

A1v1 = A2v2

We know that the diameter of the 8.0 in. pipe is 8 in. = 0.6667 ft, so its cross-sectional area is:

A1 = \(π/4 x (0.6667 ft)^2 = 0.349 ft^2\)

Similarly,

A2 = \(π/4 x (0.3333 ft)^2 = 0.0873 ft^2\)

v1 = (A2v2)/A1 = \((0.0873 ft^2 x 12 ft/s)/0.349 ft^2\) = 2.39 ft/s

Next, we can use the Bernoulli's equation between points A and C to relate the pressure difference to the velocity difference:

P/ρ +\(v^2/2g\) + h = constant

h = (P/ρ - Patm)/g + z

We are given that the pump delivers 25 hp with 60% efficiency, so the power input to the pump is:

Pin = 25 hp/0.6 = 41.7 hp = 31.1 kW

Pin = QP/η

The volumetric flow rate can be calculated from the continuity equation:

Q = A1v1 = \(0.349 ft^2 x 2.39 ft/s = 0.834 ft^3/s\)

Substituting this into the above equation, we get:

P = Pinη/Q = \((31.1 kW x 0.6)/0.834 ft^3/s\) = 1773 Pa

Finally, we can substitute the known values into the Bernoulli's equation:

h = \((P/ρ - Patm)/g + z = (1773 Pa/1000 kg/m^3 - 101325 Pa)/9.81 m/s^2 + 0 m\)

h ≈ 15.7 m

Therefore, the water can be raised to a height of approximately 15.7 m above the water surface at point A.

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Just before a lightning strike, a negative charge within the cloud creates a stepped leader, which moves towards the ground. As it approaches, the ground's positive charge forms an upward streamer. When the stepped leader and upward streamer connect, a powerful electrical discharge occurs, creating the lightning strike from the cloud to the ground.

Just before a lightning strike, the negative charge within the cloud separates from the positive charge, creating an electric field. This electric field becomes strong enough to ionize the air molecules between the cloud and the ground, creating a conductive path for the negative charge to travel to the positively charged ground. This is what causes the lightning bolt to shoot down from the cloud to the ground, resulting in a lightning strike.

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Answer:Mutual respect. Mutual trust. Mutual affection.

Explanation:

Answer:

a body continues its state of rest or uniform motion in a straight line provided no net force acts on it.

it says that if no net force acts on a body the body will continue its initial state

for example  a chair lying in a room will remain stationary as long as no net force acts on it this is the first part .In second part it says that a body will continue its motion as long as no net force acts on it however this cannot be assumed on earth because if a body starts moving it will come to rest after sometime because of friction and air resistance. T his is only possible in space where there are no resistive forces that can stop the body.

i hope this will be helpful for you.

To operate a central air conditioner, an electric clothes dryer, and an electric range, the ampacity and voltage of the service to the house should be about 240 volts.

The operation of these three appliances requires a sufficient electrical service to the house in terms of ampacity and voltage. The central air conditioner typically requires a dedicated circuit with a higher ampacity due to its significant power consumption. It is commonly connected to a 240-volt circuit and can draw a substantial amount of current depending on its size and efficiency rating.

Similarly, an electric clothes dryer typically operates on a 240-volt circuit and demands a significant amount of power. It requires a dedicated circuit with an appropriate ampacity to handle the load generated by heating elements and the motor.

The electric range, which includes the oven and stovetop, also requires a 240-volt circuit for proper operation. The ampacity of the circuit should be sufficient to support the range's power requirements, including heating elements and other electrical components.

To ensure the safe and efficient operation of these appliances, it is essential to have a service panel with an adequate ampacity rating, typically determined by a qualified electrician based on the specific power requirements of the appliances. Additionally, the voltage supplied to the house should be compatible with the appliances' voltage specifications, commonly 240 volts in this case.

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Answer:   PbBr4 is covalent

Answer: A

Explanation:

Answer:

375 hz

Explanation:

Trust me

Answer:

\(11.3\ \text{g/cm}^3\)

\(7.92\ \text{g/cm}^3\)

\(238.1\ \text{cm}^3\)

Explanation:

8) Volume of lead

\(V=4.5\times 5.2\times 6\\\Rightarrow V=140.4\ \text{cm}^3\)

m = Mass of block = 1587 g

Density is given by

\(\rho=\dfrac{m}{V}\\\Rightarrow \rho=\dfrac{1587}{140.4}\\\Rightarrow \rho=11.3\ \text{g/cm}^3\)

Density of lead is \(11.3\ \text{g/cm}^3\)

9) Volume of lead = Volume of water displaced = (49.1-45.5) = 3.6 mL = \(3.6\ \text{cm}^3\)

m = Mass of iron = 28.5 g

Density is given by

\(\rho=\dfrac{m}{V}\\\Rightarrow \rho=\dfrac{28.5}{3.6}\\\Rightarrow \rho=7.92\ \text{g/cm}^3\)

The density of iron is \(7.92\ \text{g/cm}^3\)

10) m = Mass of silver = 2500 g

\(\rho\) = Density of silver = \(10.5\ \text{g/cm}^3\)

Volume is given by

\(V=\dfrac{m}{\rho}\\\Rightarrow V=\dfrac{2500}{10.5}\\\Rightarrow V=238.1\ \text{cm}^3\)

The volume of silver is \(238.1\ \text{cm}^3\)

(a) The theoretical pressure head is 9.90 m, and the actual pressure head is 7.70 m. (b) The blade outlet angle Bzy is approximately 22.40°.

(a) For the first scenario:

Given:

Rotating speed (n): 1450 rpm

Flow rate (Q): 0.0833 m^3/s

Outer diameter of impeller (D2): 360 mm

Inner diameter of impeller (D1): 138 mm

Blade outlet angle (B2y): 30°

Flow cross-sectional area at impeller outlet (A2): 0.023 m^2

Circulation coefficient (K): 0.7

Assuming Cu (blade outlet velocity coefficient): 0

To calculate the theoretical pressure head, we can use the following equation:

Ht = (Q * K) / (g * A2)

where Ht is the theoretical pressure head, Q is the flow rate, K is the circulation coefficient, g is the acceleration due to gravity (approximately 9.81 m/s^2), and A2 is the flow cross-sectional area at impeller outlet.

Plugging in the given values, we have:

Ht = (0.0833 * 0.7) / (9.81 * 0.023) = 9.90 m

To calculate the actual pressure head, we can use the following equation:

Ha = (Q * K) / (g * A2) - (Cu^2 / (2 * g))

Since Cu is assumed to be 0, the second term in the equation becomes 0.

Plugging in the values, we have:

Ha = (0.0833 * 0.7) / (9.81 * 0.023) = 7.70 m

(b) For the second scenario:

Given:

Outer diameter of impeller (Dz): 350 mm

Blade outlet width (b2): 12.7 mm

Rotating speed (n): 1200 rpm

Flow rate (Q): 1.27 m^3/min

Pressure difference at inlet and outlet: 272 kPa

Circulation coefficient (K): 0.9

Hydraulic efficiency (mn): 70%

Assuming Ciu (blade inlet velocity coefficient): 0

To calculate the blade outlet angle (Bzy), we can use the following equation:

Bzy = arcsin(2 * mn * Q) / (π * Dz * b2 * sqrt(2 * g * (p2 - p1)))

where Bzy is the blade outlet angle, mn is the hydraulic efficiency, Q is the flow rate, Dz is the outer diameter of the impeller, b2 is the blade outlet width, g is the acceleration due to gravity, and p2 - p1 is the pressure difference at the inlet and outlet.

Plugging in the given values, we have:

Bzy = arcsin((2 * 0.70 * 1.27) / (π * 350 * 12.7 * sqrt(2 * 9.81 * 272))) ≈ 22.40°

(a) The theoretical pressure head is 9.90 m, and the actual pressure head is 7.70 m in the first scenario.

(b) The blade outlet angle Bzy is approximately 22.40° in the second scenario.

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No number one cutting-edge is required to saturate the class C800 Current Transformer in this scenario.

To determine the number one present-day required to saturate the class C800 Current Transformer (CT), we need to take into account its saturation traits. Typically, the saturation of a CT happens while the primary contemporary reaches a positive stage that causes the magnetic middle to saturate, ensuing in distorted output waveforms.

In this situation, the CT has a ratio of 100/5, which means that for every 1000Amps of number one contemporary, there could be 5Amps of secondary modern. The secondary burden is given as a hundred ohms.

To calculate the primary current required to saturate the CT, we want to consider the burden impedance, which is contemplated to the primary facet of the CT. The meditated burden impedance is calculated with the use of the turns ratio squared:

Reflected Burden Impedance = (Turns Ratio)² * Burden Impedance

In this example, the contemplated burden impedance is:

Reflected Burden Impedance = (1000/5)² * 100 = 400,000 ohms

Now, the number one modern required to saturate the CT can be determined by applying Ohm's Law:

Primary Current = Voltage / Impedance

Since the CT is connected to a present-day supply, the voltage can be taken into consideration as negligible. Therefore, the number one present-day required to saturate the CT is:

Primary Current = 0/ 400,000 = 0 Amps

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Answer:

i.e. velocity of boy is 2.82 m/s towards west

Explanation:

Given

mass of boy m_b=70.9\ kgm

b

=70.9 kg

mass of girl m_g=43.2\ kgm

g

=43.2 kg

speed of girl after push v_g=4.64\ m/sv

g

=4.64 m/s

Suppose speed of boy after push is v_bv

b

initially momentum of system is zero so final momentum is also zero because momentum is conserved

P_i=P=fP

i

=P=f

0=m_b\cdot v_b+m_g\cdot v_g0=m

b

⋅v

b

+m

g

⋅v

g

v_b=-\frac{m_g}{m_b}\times v_gv

b

=−

m

b

m

g

×v

g

v_b=-\frac{43.2}{70.9}\times 4.64v

b

=−

70.9

43.2

×4.64

v_b=-2.82\ m/sv

b

=−2.82 m/s

i.e. velocity of boy is 2.82 m/s towards west

The angular diameter of an object is the size of the angle it subtends at a given distance. One arc second is equal to 1/3600th of a degree. The angular diameter of the sun is approximately 0.5 degrees or 1800 arc seconds.

To calculate the distance the sun would have to be moved so that its angular diameter would be 1 arc second, we can use the formula:

distance = (diameter/2) / tan(angle/2)

Plugging in the values, we get:

distance = (1392000/2) / tan(1/2)

distance = 694400 / tan(0.5)

distance = 694400 / 0.00873

distance = 79456880 km

Therefore, the sun would have to be moved approximately 79.5 million kilometers(approximately 79,578,500,000 km) away from Earth to have an angular diameter of 1 arc second.

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Answer: Universities

Business and industry

Medical Institutes

Explanation: I hope that helped