A frequency generator sends a 550Hz sound wave through both water and ice. What is the difference in wavelength between the wave produced in ice and the wave produced in water?

Answer:

OXYGEN

Explanation:brainlyist me

Answer:

Nitrogen

Explanation:

Oxygen is second

Answer:

Object's weight = 6,839.42 N

Explanation:

Given

Above waterline = 30%

Volume of object = 1m^3

Required

Determine the weight of the object

First, we need to calculate its Mass

Mass = Density of Water * Volume of object in water

Density of water = 997kg/m³

If 30% is above waterline, then 70% is in water.

So:

Mass = Density of Water * Volume of object in water

Mass = 997kg/m³ * 70%m³

Mass = 997kg * 70%

Mass = 697.9 kg

The object weight sis then calculated as thus:

Weight = Mass * Acceleration of gravity

Weight = 697.9 kg * 9.8m/s²

Weight = 6 839.42 N

An easy way to encourage students to stay hydrated is to let them know hydration improves performance.

Drinking water is essential for our body system to function well. It helps in the regulation of the body temperature, the prevention of infections, as well as the delivery of nutrients to different parts of the body.

Staying hydrated is also vital in getting enough sleep and keeps the body in shape.

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Answer:

pt 1: [tex]m=1.66698*10^{21} kg[/tex]

Pt 2: [tex]KE=1212.23531 J[/tex]

Explanation:

Information Given: (p = density)

l = 5200km  d = 35km p = 2700kg/[tex]m^{2}[/tex]

Part 1: Mass

Part 2: Kinetic Energy

[tex]KE=\frac{1}{2}mv^2[/tex]

[tex]KE=\frac{1}{2} (1.66698*10^{21})(1.20497*10^{-9})^2[/tex]

[tex]KE=1212.23531 J[/tex]

Part 3: Jogger Speed

set up, because I don't have the mass :(

Information given:

[tex]KE_{jogger}[/tex]

Hope it helps :)

Black I think

this is due to the fact that blue lights only reflect blue things so if yellow is shone on it it will reflect black appearance.

how it useful

Answer: Plates shifting

Explanation: After years and years of plates colliding into solid rock, they slowly become closer together. As recent studies have shown, Africa is currently moving closer to Europe one centimeter every year (one inch every 2.5 years).

Answer:

nvudbwasivnjlscv bwbfvsz

Explanation:

Answer:

222/88 Ra

Explanation:

We have to wait 5.57 half lives for a radioactive sample to drop to 2.10 % of its original activity.

To find the tike taken for the activity, we need to know about radioactivity and half-life.

                R=R₀e^(-λ×t)

                 t= 1/λ ln(R₀/R)

    t=(half-life/ln2)×ln(R₀/R)

Here R=0.021R₀, so t= (half-life/ln2)×ln(R₀/0.021R₀)=5.57 half-lives

Thus, we can conclude that we have to wait 5.57 half lives for a radioactive sample to drop to 2.10 % of its original activity.

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Answer:

The fish is experiencing a water pressure of 502.8 kPa.

Explanation:

The water pressure the fish is experiencing can be found as follows:

[tex]P = \rho gh[/tex]  (1)

Where:

g: is the gravity = 9.81 m/s²

h: is the height (depth) = 50.0 m

ρ: is the seawater's density = 1.025 g/cm³  

By replacing the above values into equation (1) we have:

[tex] P = \rho gh = 1.025 \frac{g}{cm^{3}}*\frac{1 kg}{1000 g}*\frac{(100cm)^{3}}{1 m^{3}}*9.81 m/s^{2}*50.0 m = 502.8 kPa [/tex]        

Therefore, the fish is experiencing a water pressure of 502.8 kPa.

I hope it helps you!        

Answer:

a) [tex]H_{in}=39.34 kJ[/tex]

b) Efficiency=76.77%

Explanation:

a)

In order to solve this problem, we can use the following formula:

[tex]H_{in}=H_{out}+W[/tex]

the problem provides us with all the necessary information so we can directly use the formula:

[tex]H_{in}=9.14kJ+30.2kJ[/tex]

[tex]H_{in}=39.34 kJ[/tex]

b) In order to find the efficiency, we can use the following formula:

[tex]Efficiency=\frac{W}{H_{in}}*100\%[/tex]

so we get:

[tex]Efficiency=\frac{30.2kJ}{39.34kJ}*100\%[/tex]

Efficiency=76.77%

Answer: b is probly correct  beacuse all the others make no sence

The statement that is true about international space station is The force of gravity produces the same acceleration for the earth and the space station.

International space station is a space station that is located at the lower part of the Earth's orbit . It is the third largest space object in the space and it is a multinational project that involves five different space agencies are NASA, Roscosmos, JAXA, ESA, and CSA.

Therefore, The statement that is true about international space station is The force of gravity produces the same acceleration for the earth and the space station.

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Answer:

A. ) F₁ₓ = -45.0 N F₁y = 53.6 N

B.)  F₂ₓ = 3.48 N F₂y = -19.7 N

C.)  Fₓ = -41.5 N Fy = 33.9 N

D)  F = 53.6 N

E)  θ = -39. 2º (320.8º)

Explanation:

A)

       [tex]F_{x1} = F_{1} * cos (130) = 70 N * cos (130) = -45 N (1)\\F_{y1} = F_{1} * sin (130) = 70 N * sin (130) = 53.6 N (2)[/tex]

B)

       [tex]F_{x2} = F_{2} * cos (280) = 20 N * cos (280) = 3.48 N (3)\\F_{y2} = F_{2} * sin (280) = 20 N * sin (280) = -19.7 N (4)[/tex]

C)  

        sum of the x- and - y components of F₁ and F₂:

D)

       [tex]F_{t} =\sqrt{F_{x} ^{2} + F_{y} ^{2} } = \sqrt{(-41.5N)^{2} +(33.9N)^{2}} = 53.6 N (7)[/tex]

E)

       [tex]\theta = arc tg \frac{33.9N}{(-41.5N)} = arc tg (-0.817) = -39. 2 \deg[/tex]

Answer:

C

Explanation:

Sled A has more potential energy because it's mass is 100 kg, and it is higher up than Sled B. The more high up the sled is and the lighter it is, the faster it gets, it creates more and more potential energy.

Hope this helps!

Answer:

Explanation:

Where,

[tex]\Rightarrow\:\:\rm{s = 10(60) + \dfrac{1}{\cancel{2}} \times \cancel{2} \times (60)^{2}}[/tex]

[tex]\Rightarrow\:\:\rm{s = 600+60\times 60}[/tex]

[tex]\Rightarrow\:\:\rm{s = 600 + 3600}[/tex]

[tex]\Rightarrow\:\:\rm{s = {\bf{\red{4200\:m}}}}[/tex]

Answer:

The input energy to  engine 2  is [tex]E_2 = 3807.7\ J[/tex]

Explanation:

From the question we are told that

     The efficiency of engine one is  [tex]\eta_1 = 0.18[/tex]

     The input energy required is [tex]E = 5500 \ J[/tex]

       The efficiency of engine 2 is [tex]\eta_2 = 0.26[/tex]

Generally the workdone by the engine 1  is mathematically represented as

       [tex]W_1 = E * \eta_1[/tex]

=>     [tex]W_1 = 5500 * 0.18[/tex]

=>     [tex]W _ 1= 990 \ J[/tex]

Generally the workdone by engine 2  is mathematically represented as  

        [tex]W _2= E_1 * \eta_2[/tex]

=>     [tex]W _2 = E_1 * 0.26[/tex]

=>     [tex]W_2 =0.26E_1[/tex]

From the question we are told that [tex]W_1 = W_2[/tex]

So  

       [tex]990 =W_2 =0.26E_1[/tex]

=>     [tex]E_2 = 3807.7\ J[/tex]

Answer:

In 1869 Russian chemist Dimitri Mendeleev started the development of the periodic table, arranging chemical elements by atomic mass. He predicted the discovery of other elements, and left spaces open in his periodic table for them.

Explanation:

Answer: Mendeleev first published a table of elements arranged according to increasing atomic masses. He noticed that some elements near each other had differing properties, but elements in vertical columns had similar properties. Moseley then rearranged the table according to atomic numbers and this eliminated the discrepancies found in Mendeleev’s attempt. Today’s version of the periodic table displays elements in order based on their atomic number; the atomic number indicates the number of protons within the atoms of a particular element. Rows are called periods and columns are called groups. Elements in the same group have similar properties. Elements are grouped into nine categories: noble gases, halogens, nonmetals, alkali metals, alkaline earth metals, transition metals, other metals, metalloids, and rare earth elements.

Answer:

Velocity (m/s) over time (s) graph

Velocity (m/s) over time (s) graph

We could write out our average acceleration as:

a = Δv/ Δta=Δv/Δta, equals, Δ, v, slash, Δ, t

a = (15 m/s - 0 m/s) / 0.2 seconds

a = 15 m/s / 0.2 seconds

a = 75 m/s / second

Explanation:

What this formula is telling us is that if we know the acceleration of an object, and the ... we can plug in our acceleration of 12.5 m/s2 for a, and 4 seconds for t.

Velocity (m/s) over time (s) graph

Velocity (m/s) over time (s) graph

We could write out our average acceleration as:

a = Δv/ Δta=Δv/Δta, equals, Δ, v, slash, Δ, t

a = (15 m/s - 0 m/s) / 0.2 seconds

a = 15 m/s / 0.2 seconds

a = 75 m/s / second

Answer:

That is false. Take a look at this way. You can push a ball with your own breath, you just need to blow it. And you can pull something from afar with a magnet. It is possible to do both.

Explanation:

Not all physical things can be done only physically. Like I just said, it is possible to use other forces (no, not the dark side one),  such as a magnetic force, displayed by a magnet or anything with a force like so.

Answer:

67000N

Explanation:

We solve for the acceleration using the the 3rd constant-acceleration equation.

(Vx)f² = (Vx)i² + 2ax∆x

We have the displacement to be

∆x = Xf - Xi = 940m

Vx = 70m/s

The acceleration = (70m/s)²/2(940m)

= 4900/1880

= 2.61m/s²

From isaac newton's second law,

51000kg x 2.61m/s²

= 133,000N

The engines thrust is half of this value

Therefore thrust = 67000N or 67kN

Answer:

e) 67°

Explanation:

the force on the wire can be calculated using the expression below

F = BILsinФ

But we are looking for the angle between the wire segment and the magnetic field, then we can make Ф the subject of the formula from the above expresion, then we have,

Ф =sin⁻¹ (F/BIL)

The parameters is defined as

I =current that is been carried by the wire= 0.35 A

Ф = angle between the wire segment and the magnetic field, which is the unknown?

L = length of the wire=65 cm

B = magnetic field = 1.24

F= force on the wire = 0.26 N,

Ф =sin⁻¹ (F/BIL)

Ф =sin⁻¹ X .....................eqn(#)

Where X= (F/BIL)

We can calculate for X= (F/BIL), from eqn(#) by substituting value of Force, Lenght and

magnetic field

X=(F/BIL)= 0.26/(1.24×0.35×0.65)

= 0.26/0.2821

=0.922

Then substitute X into eqn (Ф =sin⁻¹ X)

Then

Ф =sin⁻¹ (0.922)

Ф=66.42°

Ф=67° approximately

Therefore, the angle between the wire segment and the magnetic field is 67°

Answer:

Examples of Radiant Energy All Around You

The term radiant energy refers to energy that travels by waves or particles, particularly electromagnetic radiation such as heat or x-rays. Radiant energy is created through electromagnetic waves and was discovered in 1885 by Sir William Crookes. Fields in which this terminology is most often used are telecommunications, heating, radiometry, lighting, and in terms of energy created from the sun. Radiant energy is measured in joules.

Everyday Examples of Radiant Energy

Virtually anything that has a temperature gives off radiant energy. Some examples of radiant energy include:

•The heat emitted from a campfire

•Emission of heat from a hot sidewalk

•X-rays give off radiant energy

•Microwaves utilize radiant energy

•Space heaters produce radiant energy

•Heat created by the body can be radiant energy

•Lighting fixtures

√Home heating units

•Fixtures that convert solar energy to heat

•Visible light

•Gamma rays

•Radio waves

•Electricity

•A surface heated by the sun converts the energy of the light into infrared energy which is a form of radiant energy

•Cell phones utilize radiant energy to function

•Magnetic motor generators that utilize •neodymium magnets create radiant energy

•Audio signals that come to home or cars via radio waves

•Ultraviolet light

√Infrared radiation

•The light emitted from a campfire

•The light generated from a light bulb

•A heated brake disc giving off heat

•The heat from a grill used for cooking

•Water can reflect or absorb radiant energy

•Soil can absorb radiant energy

•Light from the sun

•Heat emitted from a bunsen burner

•Heat from an overused computer

•Heat caused by friction

•Heat emitted from a dryer

•The heat generated by a light bulb

•Heat generated through reflection of visible light

•A window reflects radiant energy

•Heat created from a stove or oven

•Heat emitted from a washing machine

Answer: Reflection is the only process in which the wave does not continue moving forward.

Explanation:

Reflection is a process in which the direction of the wave changes when it is exposed to a bounce off barrier. Refraction can be defined as the change in the direction of the wave when the wave passes through one medium to another. Diffraction is a process in which the direction of the wave changes when the wave passes through a particular opening near the barrier.

Answer:

Reflection is the only process in which the wave does not continue moving forward.

Explanation:

Answer:

14.17H

Explanation:

Energy stored in an inductor is expressed as;

E = 1/2LI²

L is the inductance of the inductor

I is the current flowing through the inductor

Given

E = 10.2J

I = 1.20

Required

Inductance L

Substitute the given parameters into the formula;

10.2 = 1/2L(1.2)²

10.2 = 1/2*1.44L

10.2 = 0.72L

L = 10.2/0.72

L = 14.17H

Hence the self inductance of the solenoid is 14.17H

Answer:

The fundamental frequency is  [tex]f_1 =128 \ Hz[/tex]

Explanation:

From the question we are told that

   The frequency of one harmonics is  [tex]f_x= 448 \ Hz[/tex]

    The next higher harmonic is  [tex]f_z = 576 \ Hz[/tex]

Generally the frequency of an air column open at both ends is mathematically represented as

              [tex]f_n = \frac{nv }{ 2 L }[/tex]

Here n  is the order of the harmonics (frequency)

        v is the velocity of the sound

        L  is the length of the column

So for one harmonics we have that

        [tex]f_k = \frac{n v }{2L}[/tex]

Then for the next higher harmonics

       [tex]f_x = \frac{n+1 ) v}{2 L }[/tex]

Generally the difference between these frequencies is mathematically represented as  

       [tex]f_z- f_x = \frac{(n+1 )v}{ 2L} - \frac{(n )v}{ 2L}[/tex]

=>    [tex]576 - 448 = \frac{vn + v - nv }{2L}[/tex]

=>    [tex]\frac{ v }{2L} = 128[/tex]

Generally for fundamental  frequency n =  1

So  

       [tex]f_1 = n * \frac{v}{2L}[/tex]

So

       [tex]f_1 =1 * 128[/tex]

=>    [tex]f_1 =128 \ Hz[/tex]

Answer:no

Explanation:

No.....................................

Answer:

a

  [tex]\theta _1 =0.687 ^o[/tex]

  [tex]\theta _2 =0.630 ^o[/tex]

b

 Generally given that the domain arcsine function is  between -1 and 1 then the arcsine of 2.22 will not be valid

Generally given that the domain arcsine function is  between -1 and 1 then the arcsine of 2.1 will not be valid

Explanation:

From the question we are told that

     The slit grating is  [tex]N = 68 \ slits / cm = 6800 \ slits / m[/tex]

      The order of spectrum is [tex]n = 4[/tex]

Generally the width of the slit is mathematically represented as  

               [tex]a = \frac{1}{ 6800}[/tex]      

=>            [tex]a = 0.000147 \ m[/tex]  

Generally the condition for constructive interference is

              [tex]asin\theta = n * \lambda[/tex]

Now for the first wavelength the angle is evaluated as

             [tex]\theta _1 = sin ^{-1} [ \frac{n \lambda_1 }{a} ][/tex]

    =>     [tex]\theta _1 = sin ^{-1} [ \frac{4* 434 *10^{-9} }{ 0.000147 } ][/tex]

    =>    [tex]\theta _1 =0.687 ^o[/tex]

Now for the second wavelength the angle is evaluated as

             [tex]\theta _2 = sin ^{-1} [ \frac{n \lambda_2 }{a} ][/tex]

    =>     [tex]\theta _2 = sin ^{-1} [ \frac{4* 410 *10^{-9} }{ 0.000147 } ][/tex]

    =>    [tex]\theta _2 =0.630 ^o[/tex]

Gnerally if grating is   [tex]N = 12800 \ slits per cm = 1280000 \ slits / m[/tex]

Generally the width of the slit is mathematically represented as  

               [tex]a = \frac{1}{ 1280000}[/tex]      

=>            [tex]a = 7.813 *10^{-7} \ m[/tex]  

Generally the condition for constructive interference is

              [tex]asin\theta = n * \lambda[/tex]

Now for the first wavelength the angle is evaluated as

             [tex]\theta _1 = sin ^{-1} [ \frac{n \lambda_1 }{a} ][/tex]

            [tex]\theta _1 = sin ^{-1} [ \frac{4* 434 *10^{-9} }{ 7.813*10^{-7} } ][/tex]

    =>     [tex]\theta _1 = sin ^{-1} [ 2.22][/tex]

Generally given that the domain arcsine function is  between -1 and 1 then the arcsine of 2.22 will not be valid

    =>    [tex]\theta _1 =0.687 ^o[/tex]

Now for the second wavelength the angle is evaluated as

             [tex]\theta _2 = sin ^{-1} [ \frac{n \lambda_2 }{a} ][/tex]

    =>     [tex]\theta _2 = sin ^{-1} [ \frac{4* 410 *10^{-9} }{ 7.813*10^{-7} } ][/tex]

     =>  [tex]\theta _2 = sin ^{-1} [2.1 ][/tex]

Generally given that the domain arcsine function is  between -1 and 1 then the arcsine of 2.22 will not be valid

Answer:

Mike can travel 80 Km in 4 hours

Answer:

Explanation:

gas, liquid, soild

liquid, Gas, solid

Gas, Solid, liquid