The tank is 51m long and 20m wide.The sea water in the tank is 11m deep and has a density of 1030 kilograms per cubic meters.
Calculate the mass of water in the tank.

Taking into account the definition of Scientific notation, the correct representation of 5,970,000 in scientific notation is 5.97×10⁶.

Scientific notation is a quick way to represent a number using powers of base 10.

The numbers are written as a product:

a×10ⁿ

where:

In this case, to write the number 5,970,000 in scientific notation, the following steps are performed:

Finally, the correct representation of 5,970,000 in scientific notation is 5.97×10⁶.

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The average horizontal force applied on the gate from the lake water. will be 882.9 kN

Force is defined as the external agent applied to any object which tries to displace the body from one place to another or resist the body movement.

Given,

width of gate, W = 5 m

depth of the lake, h = 6 m

density of water = 1000 kg/m³

Atmospheric pressure = 101 kPa

Pressure at the bottom of the gate will be calculated as below:-

P = ρ g h

P = 1000 x 9.81 x 6

P = 58860 Pa

Now, Force acting on the plate is

F = ( 1 / 2)pwh

F = ( 1 / 2 ) x 58860 x 5 x 6

F = 882900 N

F = 882.9 k N

The average horizontal force applied on the gate from the lake water is equal to F = 882.9 kN

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The acceleration of the bullet leaving the barrel is 2.25 x 10⁵ m/s².

Given,

Velocity of the bullet leaving the barrel, v = 600 m/s

Distance travelled by the bullet = Size of the rifle = s = 0.800 m

According the Third Equation of Motion which relates velocity with the distance and acceleration says that,

v² = u² + 2as ;

where 'v' and 'u' are final and initial velocity of the bullet respectively

'a' is the acceleration of the bullet

's' is the distance travelled by the bullet

here u = 0 m/s

(600)² = 0² + 2 * (0.800) * (a)

360000 = (1.6) * (a)

a = [tex]\frac{360000}{1.6}[/tex]

a = 225000 m/s²

a = 2.25 x 10⁵ m/s²

Therefore the rate at which bullet accelerated while being in the barrel is 2.25 x 10⁵ m/s².

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E. The magnitude of the average emf induced in the loop is 0.20 mV.

The magnitude of the average emf induced in the loop is calculated as follows;

emf = NA(B₁ - B₂)/t

where;

A = πr² = π(0.04)² = 5.027 x 10⁻³ m²

emf = (1)(5.027 x 10⁻³)(0.06 - 0.04)/0.5

emf = 2.01 x 10⁻⁴ V

emf = 0.201 mV

Thus, the magnitude of the average emf induced in the loop is 0.20 mV.

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The situation is not possible here because "the total sound now has twice the loudness it had when only one neighbor was mowing" - this sentence can't be happened.

The scope of sounds estimated on the decibel scale is from 0 dB (the calmest sound) to 140 dB (the edge of agony). Sounds over 85 dB are viewed as by particular associations like NIOSH (the US Public Establishment for Word related Wellbeing and Wellbeing) to be risky to human hearing. Sound level alludes to different logarithmic estimations of discernible vibrations and may allude to. Sound openness level, proportion of the sound openness of a sound comparative with a reference esteem Sound power level, proportion of the rate at which sound energy is radiated, reflected, communicated or got, per unit time.

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Hi! Scientists use models for a lot of reasons. Primarily to study objects in detail that they wouldn't be able to normally. For example, to study something small like a cell, making a model could be easier. Also, things that are too dangerous to study in real life can be turned into models as well.

To put it shortly, scientists use models to observe things in better detail.

I can include a link to a website if that will help :)

https://study.com/academy/lesson/why-scientists-use-models-simulations.html

Good luck!

The value of y for the line in the plane of the two wires along which the total magnetic field is zero is 0.1667 m below the upper wire .

It is given that force per unit length is 320μ N/m and one wire carries a current [tex](I_1)[/tex] of 20.0 A and is located along the line y= 0.500 m .

Putting above values in equation (1) , we get

[tex]320\times10^{-6}=\frac{4\pi \times10^{-7}\times20.0\times I_2}{2\pi \times0.500}\\\ \\320\times10^{-6}=\frac{4 \times10^{-5}\times I_2}{5}\\\\I_2=\frac{320\times10^{-6}\times5}{4 \times10^{-5}} \\\\I_2=40A[/tex]

Now the magnitude of the magnetic field at a distance "a" due to long current carrying is given by

           [tex]B=\frac{u_0I}{2\pi a}[/tex]

The direction of this magnetic field is given by the right-hand rule of thumb of holding the wire in the right hand, with the outstretched thumb pointing in  the  direction of current flow and the direction of curled fingers pointing in the direction of magnetic field circulation. of wire.

Hence, field due the wire has an outward direction from the top side of the wire to the bottom side of the wire. Thus, the field is added, above the upper wire comes out of the paper for two wires  and enters the paper below the lower wire.

Between the two wires, due to top wire magnetic field is in the paper and due to the bottom wire magnetic field is outside so it is subtracted. The magnetic field is zero here.

Let the resultant magnetic field is zero at a distance x below the upper wire and then magnitude of the field at this distance due to both field must be equal and opposite and is given by

[tex]B=\frac{u_0I_1}{2\pi x} -\frac{u_0I_1}{2\pi (r-x)} \\\\0=\frac{u_0I_1}{2\pi x} -\frac{u_0I_1}{2\pi (r-x)} \\\\\frac{u_0I_1}{2\pi x} =\frac{u_0I_1}{2\pi (r-x)} \\\\\frac{I_1}{x} =\frac{I_2}{r-x} \\\\\frac{20}{x} =\frac{40}{0.500-x} \\\\20(0.500-x)=40x\\\\10-20x=40x\\\\10=60x\\\\\x=\frac{10}{60} \\\\x=0.1667m[/tex]

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(a) The block does not move a bit.

(b) The force on the block from the wall is −(12N)i+(5.0N)j

(a)

To know that the block is going to move or not we are using the formula,

We compare f and μF₁

If

f<μF₁,

the block does not slide on the wall

but if

f>μF₁,

the block does slide.

We are given,

F= This is the applied force

F₁= This is the normal force of the wall on the block.

f=This is the force of friction.

mg= This is the force of gravity.

The horizontal component of Newton’s second law is,

F-F₁=0

So, F= F₁ = 12N

and

μ F₁ =(0.60)(12N)=7.2N.

The vertical component is,

f-mg=0

so, f=mg=5.0N.

From the above calculation we can get that,

f<μF₁,

So The block does not move a bit

(b)Since the block does not move,

f=5.0N.

F₁ = 12N

So, The force of the wall on the block is

F(wall) =−F₁ i + f j=−(12N)i+(5.0N)j

Therefore from the above calculation we can conclude that, The force on the block from the wall is −(12N)i+(5.0N)j

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Disclaimer: The question is incomplete in the portal here is the total question.

Question:

A 12N horizontal force pushes a block weighing 5.0 N against a vertical wall (Fig.). The coefficient of static friction between the wall and the block is 0.60, and the coefficient of kinetic friction is 0.40. Assume that the block is not moving initially. (a) Will the block move? (b) In unit-vector notation, what is the force on the block from the wall?

A) 42.7% is the energy efficiency of the device.

Energy efficiency is a measure of how much energy is wasted. There are several ways to save energy, especially thermal energy, and the rate at which it is lost needs to be slowed down. Energy efficiency is defined as requiring less energy to complete an activity or reach a goal. Homes, structures, and production facilities that use less energy to create their products also consume less energy to heat, cool, and run gadgets and appliances.

Energy efficiency is one of the most straightforward and cost-effective ways to halt global warming, reduce consumer energy costs, and increase the competitiveness of American businesses. Energy efficiency is also a key component in the decarbonization process to attain net-zero emissions of carbon dioxide.

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3.56 km/s is the speed of a one-dimensional compressional wave moving along a thin copper rod.

At one end of the rod, a hard hammer strike is delivered. With a time interval of Δt between the two pulses, a listener at the other end of the rod hears the sound twice as it travels through the metal and the air.

The time interval is given by t = L/v.

The delay between pulses arrivals is:

Δt = L [(1/v(air)) - (1/v(copper))]

Now,

When the copper rod is swapped out for a different substance and the sound speed is measured as v(r).

The speed of air, v(air) = 343 m/s

Then,

L = [  Δt / (1/v(air)) - (1/v(r)) ]

L = [  Δt / (1/343) - (1/v(r)) ]

Here L is the length of the rod, Δt is in seconds and v(r) is the speed of sound in the rod.

The amount of time travel in the rod is now minimal as v(r) approaches infinity. As a result, the rod's length will be close to 343Δt, which is the distance that sound travels through the air during the delay.

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321 kPa is the calculated pressure.

The measurement of pressure in relation to the pressure of a complete vacuum, or absolute zero pressure, is known as absolute pressure. In the same way as temperature must be expressed by its absolute unit, the Kelvin, the ideal gas law also requires the measurement of absolute pressure. Absolute pressure sensors have a permanently sealed absolute vacuum chamber that is exposed to the side of the sensor that is not in contact with the pressure media. Since the diaphragm employs the sealed vacuum as its reference and zero point, ambient pressure has no effect on its deformation.

Gauge pressure plus atmospheric pressure equals absolute pressure (220 + 101 = 321 kPa).

The pressure as compared to atmospheric pressure is known as gauge pressure.

Gauge pressure and ambient pressure are added to get absolute pressure.

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The period of motion is given as  ​π/2(√md³/KeqQ)

What is Coulombs law?

Coulomb's law states that the electrical force between two charged objects is directly proportional to the product of the quantity of charge on the objects and inversely proportional to the square of the separation distance between the two objects.

The period of motion from the instance given  ​π/2(√md³/KeqQ).

w²= (2π/T)²

T=2π/w

where w is gotten as 16KeqQ/md³

T=  ​π/2(√md³/KeqQ).

m is the mass of the body with charge-Q

In conclusion, the Coulomb's law equation provides an accurate description of the force between two objects whenever the objects act as point charges.

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Since Sandi's observation is a biased one, we should be most concerned about the observer bias.

When a researcher's expectations, viewpoints, or prejudices affect what they see or record in a study, this is known as observer bias. It typically has an impact on studies when observers are aware of the research objectives or hypotheses.

Similarly in Sandi's case, she has already drawn conclusions and formed biases on MacDonald's customers. She already anticipates that they will be overweight. Thus, this will have an impact on her observations.

Therefore, in Sandi's case, observer bias should be our main concern.

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The time spent in air by the tuna before it hits the ground is determined as 1.375 seconds.

The time spent in air by an object or a projectile is the total time of motion of the object or projectile.

The time spent in air by the tuna can be determined by finding equation of the final velocity of the tuna when it hits the ground as follows;

h = 16t² + ut

where;

Velocity of the tuna when it hits the ground,

v = dh/dt

dh/dt = 32t + u

when the tuna hits the ground, final velocity = 0

0 = 32t + u

0 = 32t + 44

t = -44/32

|t| = 1.375 seconds

Thus, the time spent in air by the tuna before it hits the ground is determined as 1.375 seconds.

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Rxpression for the electric field E at the center of the semicircle is

Q/[tex]\frac{}{}[/tex](2π² ∈₀ R²) (-j) , where j is the unit vector.

Since the total charge on the semicircle is Q and the Radius of the semicircle is  R.

if we consider that an elemental charge makes an angle theta with the horizontal on the semicircle and along the y axis the charge distribution is symmetrical, which means there is a symmetrical charge on the other side too.

So, the vertical component will get added and the horizontal component will cancel out each other .

No let consider  due to  elemental charge dE  be the electric Field and λ    be the charge per unit length .

So, we can write λ =Q/πR.

k = 1 /(4π∈₀)

Now the total Electric Field at the center is

= 2dE sinθ

= 2kλ/r [tex]\int\limits^_ {} \,[/tex] Sinθ dθ

After integrating the above term from 0 to π/2, we get

E= 2kλ / R (-j)

E=Q/(2π²∈₀R²) (-j)

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Option (B) many

According to the lesson, it can be said that many adults have some of the same problems with sleep that Bailey does.

Sleeplessness is one of the most common issues, every second adult or a teenager is facing now a days. Due to a lot of responsibilities, we end up having little time for sleep and this in turn affects our body functions.

Good sleep hygiene is often recommended to improve sleep quality, which  also helps in alleviating anxiety due to work load and responsibilities.

Answer is Option (B) many

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According to the question, the truck will pass now with a frequency of 10.2×10⁻³Hz.

The number of waves that pass a fixed point in a unit of time is known as the frequency in physics. It is also the number of cycles or vibrations that a body in periodic motion experiences in a unit of time. When a body in periodic motion moves through a series of events or locations before returning to its initial state, it is said to have experienced one cycle or one vibration.

Explanation:

To find the frequency of the cyclist being passed by the truck,

f' = {(19.7+(-1.56))/19.7}×1/90

  = 10.2×10⁻³Hz.

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It makes sense for the valence electrons to stay the same within a group because these elements have same chemical properties.

We have a group from Periodic table.

We have to investigate the reason behind the fact that why elements with same number of valence electrons share the same group.

Periodic Table is a table of the chemical elements arranged in order of atomic number, usually in rows, so that elements with similar atomic structure (similar chemical properties) appear in vertical columns.

According to the question -

It is true that the number of valence electrons in the outermost shell of the elements of same group are equal. The number of valence electrons in the outermost orbit of the elements of the same group is same as the group number to which they belong. The elements in the same group have same number of valence electrons in their outermost shell. Due to this the elements have almost same Physical properties, Chemical properties and Reactivities of the elements.

Hence, it makes sense for the valence electrons to stay the same within a group because the elements in the same group have same properties.

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Answer:

380 to 700 nanometers

Brainliest please

Explanation:

If a cook puts 9.00g of water in a 2.00 -L pressure cooker that is then warmed to 500°C, the pressure inside the container is 15.8 atm.

From the ideal gas law;

PV = nRT

Where P is pressure, V is volume, n is amount of substance, T is temperature and R is the ideal gas constant (0.08206Latm/molK).

Given the data in the question;

First we determine the amount of water n.

We know that the molar mass of water is 18.01528 g/mol.

Hence

n = mass / molar mass

n = 9.00g / 18.01528 g/mol

n = 0.4995mol

Now, plug the values into the equation above and solve for P

PV = nRT

P = nRT / V

P = ( 0.4995mol × 0.08206Latm/molK × 773.15K ) / 2.00L

P = ( 31.6906Latm ) / 2.00L

P = 15.8 atm

If a cook puts 9.00g of water in a 2.00 -L pressure cooker that is then warmed to 500°C, the pressure inside the container is 15.8 atm.

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T=2ᴫ√(m/k) , T be the time period, m be the mass and k be the force per unit displacement.

To check the correctness of the above equation we are using DIMENSIONAL ANALYSIS THEOREM. After using the theorem we can say that the formula is dimensionally correct.

The dimension of a physical quantity is defined as the power to which the fundamental quantities are raised to express the physical quantity.  . Dimensions are physical quantities that can be measured. A dimension is a combination of length, breadth and height as [M], [L] and [T] respectively.

We shall use DIMENSIONAL ANALYSIS to check the correctness of the above equation. In this analysis , we shall try to check the dimensions on the LHS and RHS :

LHS:

{T}≡{time}

RHS:

{2π√(m/k)}≡{√(m/k)}

Or,{2π√(m/k)}≡{√(m/(force/x))}

Or,{2π√(m/k)}≡{√(M/(MLT⁻²/L))}

Or,{2π√(m/k)}≡{√(1/T⁻²)}

Or,{2π√(m/k)}≡{√T²}

Or,{2π√(m/k)}≡{T}≡{time}

So, LHS = RHS .

Hence, from the above prove we can conclude that, the equation is dimensionally correct.

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Check the correctness of formula, T=2ᴫ√m/k , T be the time period, m be the mass and k be the force per unit displacement.

When approaching the front of an idling jet engine, the hazard area extends forward of the engine approximately 25 feet.

Tetraethyl lead, which is present in gasoline, deposits itself on the turbine blades. Because jet fuel has a higher viscosity than aviation gasoline, it may retain impurities with greater ease.

Once the gasoline charge has been cleared, start the engine manually or with an electric starter while cutting the ignition and using the maximum throttle.

On the final approach, the aeroplane needs to be re-trimmed to account for the altered aerodynamic forces. A substantial nose-down tendency results from the airflow producing less lift on the wings and less downward force on the horizontal stabiliser due to the reduced power and slower velocity.

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λ=0.303m is behind the police car.

The frequency of the emission is [tex]f=1000\, \text{Hz}[/tex]

The police car is traveling at a speed of [tex]$v_s=+40.0 \frac{\mathrm{m}}{\mathrm{s}}$[/tex]

When the air is calm, the sound moves at a speed of [tex]$v=343 \frac{\mathrm{m}}{\mathrm{s}}$[/tex]

When travelling away from the source, the observer's (another car) speed is

[tex]$v_o=-30.0 \frac{\mathrm{m}}{\mathrm{s}}$[/tex]

(c) The sound's frequency is (for the observer at rest): in front of the police car.

[tex]f^{\prime}=\frac{v}{v-v_s} f[/tex]

consequently, the needed wavelength is:

[tex]\lambda=\frac{v}{f^{\prime}}=\frac{v}{\frac{v}{v-v_s} f}[/tex]

Rearranging and canceling v gives us:

[tex]\lambda=\frac{v-v_s}{f}=\frac{343 \frac{\mathrm{m}}{\mathrm{s}}-40 \frac{\mathrm{m}}{\mathrm{s}}}{1000 \mathrm{~Hz}}=0.303 \mathrm{~m}[/tex]

Hence, λ=0.303m  is behind the police car.

A waveform signal's wavelength is defined as the separation between two identical points (adjacent crests) in adjacent cycles as the signal travels through space or along a wire.

This length is typically specified in meters (m), centimeters (cm), or millimeters (mm) in wireless systems (mm).

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So long as the distance from the wall is small compared to the width and height of the wall, the distance does not affect the field.

Density, the mass of a unit volume of a material substance. The formula for density is d = M/V, where d is density, M is mass, and V is volume. Density is commonly expressed in units of grams per cubic centimetre. For example, the density of water is 1 gram per cubic centimetre, and Earth’s density is 5.51 grams per cubic centimetre. Density can also be expressed as kilograms per cubic metre (in metre-kilogram-second or SI units).

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Answer:

Charges B and C are negative

Explanation:

• We are certain of a law of magnetism that states "Field lines move from positive charge to negative charge"

[tex]{}[/tex]

Answer:

B and C are negative.

Explanation:

This is because according to the law of charges in electric field line it says that the lines forming the electric field move from the positive terminal to the negative terminal,hence making B and C be negative as the lines are moving into them.

Answer:

*I think you meant seconds

45 min = 2,700 sec

Explanation:

Hope it helps!!!

:}

The initial velocity will be zero since the car started from rest, and final velocity is 40 m/s.

acceleration: the rate at which the speed and direction of a moving object vary over time. A point or object going straight ahead is accelerated when it accelerates or decelerates. The car began at zero miles per hour, or at rest. It then continued to speed quickly. A vector measurement of the rate and direction of motion is what is meant by the term "velocity." Simply said, velocity is the rate of movement in a single direction. Velocity may be used to gauge both the speed of a rocket launching into space and the speed of a car traveling north o6n a busy freeway.

T= 12s

D=480 m

velocity= 480/12

velocity=40 m/s.

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The marble travel from the top of the ramp is  56.25 m.

Ramp is can be understand in physics, such as the inclined planes, also referred to as ramps . That means,ramp is a type of simple machine which manipulate the direction and magnitude of a force. It is very useful machine by which we can understand the problems in mechenics .

We are given that,

The marble rolls down from the ramp in time = t = 5s

Acceleration of the marble due to gravity = a = 4.5 m/s²

Therefore , for getting the distance (s) of the marble which travelled from the top of the ramp by using the equation of motion i.e. given as,

s = ut + (1/2) at²

Here the marble released from thew rest at the top of the ramp so that initial velocity u = 0 , therefore after putting the value above equation can be written as,

s = (1/2)× 4.5 m/s²×5s × 5s

s = 56.25 m

Hence, the marble travel from the top of the ramp 56.25 m after 5 seconds.

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The specific brain mass: decreases if body mass increases.

Due to the specific brain mass is the ratio of brain mass to body mass is inversely proportional when the body mass increases, the specific brain mass decreases.

If we calculate the specific brain mass of a person with a brain mass of 2 kg and a body mass of 50 kg and then with a body mass of 60 kg we have:

specific brain mass (50 kg) = 2 kg/50 kg = 0.04

specific brain mass (60 kg) = 2 kg/60 kg = 0.03

We can state that the specific brain mass decreases if the body mass increases.

It is the amount of mass associated with a person's body

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The calculated distance is 20 km.

Speed indicates how quickly something, or someone is moving. If you know how far something has traveled and how long it took to get there, you can calculate its average speed. Speed is calculated as follows: speed = distance * time. For instance, if a car drives 120 miles in 2 hours, we may calculate the speed as 120 x 2 = 60 miles per hour. The units for speed are determined by the units of distance and time.

Recognizing that the gap between the trains is closing at a constant rate of 20 km/h,

the total time that elapses before they crash is

t=(20km)/(20km/h)

=1.0h.

During this time, the bird travels a distance of

x=vt

=(20km/1.0h)

=20km.

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