a ship sonar sends message and recives the eschoes from the ocean bottom 0.6 second after the sound is sent down the ship how deep is the water beneath the ship
​

Answer:

it traveled 1350 kilometers or 839 miles

Explanation:

You just have to do 450 times three because its 3 HOURS and 450 kilometers per HOUR.

Complete Question

A 8.0 A current is charging a 1.0 -cm-diameter parallel-plate capacitor. What is the magnetic field strength at a point 2.5 mm radially from the center of the wire leading to the capacitor?

Answer:

The magnetic field is  [tex]B = 6.4*10^{-4} \ T[/tex]

Explanation:

From the question we are told that

    The current is  [tex]I = 8.0 \ A[/tex]

    The diameter is  [tex]d = \ cm = \frac{1}{100} = 0.01 m[/tex]

      The position considered is  [tex]d = 2.5 \ mm = 0.0025 \ m[/tex]

 Generally the magnetic field is mathematically represented as

           [tex]B = \frac{\mu_o * I }{ 2 \pi d}[/tex]

Here [tex]\mu_o[/tex] is permeability of free space with value  

[tex]\mu_o = 4\pi * 10^{-7} \ N/A^2[/tex]

      So

                 [tex]B = \frac{ 4\pi * 10^{-7} * 8 }{ 2 \pi * 0.0025 }[/tex]

=>               [tex]B = \frac{ 2 * 10^{-7} * 8 }{ * 0.0025 }[/tex]

=>               [tex]B = 6.4*10^{-4} \ T[/tex]

Answer:

heathy, people, apple, motivational, together  

Explanation:

Answer:

Well, this teacher is pretty dumb if you ask me.

Explanation:

5 N, i think this is right.

Answer:

The orbital velocity [tex]v = 16.4 \ km/s[/tex]

The period is  [tex]T = 14.8 \ hours[/tex]

Explanation:

Generally centripetal force acting ring particle is equal to the gravitational  force between the ring particle and the planet , this is mathematically represented as

       [tex]\frac{GM_s * m }{r^2 } = m w^2 r[/tex]

=>    [tex]w = \sqrt{ \frac{GM}{r^3} }[/tex]

Here G is the gravitational constant with value  [tex]G = 6.67*10^{-11}[/tex]

        [tex]M_s[/tex]  is the mass of with value  [tex]M_s =5.683*10^{26} \ kg[/tex]

        r is the is distance from the center of the  to the  outer edge of the  A ring

i.e r = R  + D  

Here R  is the radius of the planet   with value  [tex]R = 60300 \ km[/tex]

         D  is the distance from the  equator to the outer edge of the  A ring with value  [tex]D = 80000 \ kg[/tex]

So  

       [tex]r =80000 + 60300[/tex]

=>    [tex]r =140300 \ km = 1.4*10^{8} \ m[/tex]

So

    =>    [tex]w = \sqrt{ \frac{ 6.67*10^{-11}* 5.683*10^{26}}{[1.4*10^{8}]^3} }[/tex]

    =>    [tex]w = 1.175*10^{-4} \ rad/s[/tex]

Generally the orbital velocity is mathematically represented as

       [tex]v = w * r[/tex]

=>     [tex]v = 1.175*10^{-4} * 1.4*10^{8}[/tex]

=>     [tex]v = 1.64*10^{4} \ m /s = 16.4 \ km/s[/tex]

Generally the period is mathematically represented as

     [tex]T = \frac{2 \pi }{w }[/tex]

=> [tex]T = \frac{2 * 3.142 }{ 1.175 *10^{-4} }[/tex]

=> [tex]T = 53473 \ second = 14.8 \ hours[/tex]

Answer:

The orbital velocity [tex]v = 16.4 \ km/s[/tex]

The period is  [tex]T = 14.8 \ hours[/tex]

Explanation:

Generally centripetal force acting ring particle is equal to the gravitational  force between the ring particle and the , this is mathematically represented as

       [tex]\frac{GM_s * m }{r^2 } = m w^2 r[/tex]

Answer:

electron shell

Explanation:

the proton and nucleus are inside the electron shell so the arrow is point on the outer shell which is the electron shell.

Answer:

in the downward movement of the movement when the constant is lost

Explanation:

When the coin is on the piston it has a relationship given by

       a = d²x / dt²

the piston position is

      x = A cos wt

      a = - A w² cos wt

the maximum acceleration is

     a = - A w²

When the piston raises the acceleration of gravity and that of the piston go in the same direction, when the piston descends they relate it is contrary to gravity, therefore when the frequency increases, the point where the acceleration of the piston is greater than gravity arrives and the coin loses contact.

The point where you lose contact is

      a = g

      g = A w²

In short, in the downward movement of the movement when the constant is lost

Answer:

B ( 1.1 s )

Explanation:

Given that a thin cylindrical shell is released from rest and rolls without slipping down an inclined ramp that makes an angle of 30° with the horizontal. 

Let the height = h

Sin 30 = h / 3.1

h = 3.1 Sin 30

The P.E at the top = mgh

P.E = 9.8 × 3.1 Sin 30 × m

P.E = 15.19m

The K.E = P.E

K.E = 1/2mv^2

1/2 × V^2 × m = 15.19m

The mass m will cancel out.

V^2 = 30.38

V = 5.51 m/s

Using third equation of motion to find acceleration

V^2 = U^2 + 2as

Since it starts from rest, initial velocity u = 0

30.38 = 2 × 3.1 × a

30.38 = 6.2a

a = 30.38 / 6.2

a = 4.9 m/s^2

Using equation one of linear motion to calculate time t

V = U + at

5.51 = 4.9t

t = 5.51 / 4.9

t = 1.12 s

Therefore, it will take approximately 1.1 s to travel the first 3.1 m

The correct option is B

Answer:total displacement is 550m.

Explanation:

Answer:

A. the same speed as the first trip

Explanation:

there's no friction

pls mark as brainliest

Answer:

734.16 kg m/[tex]s^{2}[/tex]

Explanation:

The problem is asking for the Force of pushing off the ground.

Given = Mass: 600 newtons (N)

             Acceleration: 12 m/[tex]s^{2}[/tex]

We have to convert the mass into kg first. Remember that 1 kg is equal to 9.80665 newtons.

Let x be the mass in newtons.

Let's convert: [tex]\frac{1 kg}{9.80665 N}[/tex] x [tex]\frac{x}{600 N}[/tex] = [tex]\frac{600}{9.80665}[/tex] = 61.18 kg

Phil's weight is 61.18 kg

Let's go back to finding the force.

F = m x a

F = 61.18 kg x 12 m/[tex]s^{2}[/tex]

F = 734.16 kg m/[tex]s^{2}[/tex]

Answer:

Yes

Explanation:

Answer:

The distance is  [tex]R = 79.5 \ m[/tex]

Explanation:

From the question we are told that

     The angle at which the ball is thrown is [tex]\theta = 60^o[/tex]

     The initial speed is [tex]u = 30 \ m/s[/tex]

Generally the distance from the original position which the ball covered is mathematically  evaluated as

            [tex]R = \frac{ u^2 * sin 2\theta }{g}[/tex]

=>         [tex]R = \frac{ 30^2 * sin (2* 60)}{9.8}[/tex]  

=>         [tex]R = 79.5 \ m[/tex]

Answer:

a)  k = 3.2 N / m

b)    T = 0.27 s

some problem measuring the time of the oscillation.

Explanation:

a) With the student's observations we can make a graph of elongation against applied weight, see attached, with this graph, Hooke's law must comply

              F = -k x

where force is the applied weight

             F = W = mg

substituting

            W = k x

            x = 1 / k W       (1)

the linear regression of the graph gives

           x = 31.2 w + 0.23

the slope of the graph is

            m = 31.2     cm/N

if we relate this value with equation 1

           1 / k = 131.2

           k = 0.032 N / cm

Let's reduce to the SI system

          k = 0.032 N / cm (100cm / 1m)

          k = 3.2 N / m

This is the best value we can obtain from the spring constant with these experiential data

b) When the student oscillates the system, he has a simple harmonic motion whose angular velocity is

            w =[tex]\sqrt\frac{k}{m} {}[/tex]

for the masses of m = 60 gr = 60 10-3 kg = 0.060 kg

            w = √ (3.2 / 0.060)

            w = 22.82  rad/s

angular velocity is related to frequency and period

           w = 2π f = 2π / T

           T = 2π / w

           T = 2π / 22.82

            T = 0.27 s

A great discrepancy is observed between the theoretical value and the experimental value, it is not common erorr so high it can be due to some problem measuring the time of the oscillation.

Answer:

Water at high temperatures

Explanation:

When heated, molecules move faster which results in the sound waves being transmitted faster.

Also sound waves travel fastest in solids. Fast in liquids, and slowest in the air.

Answer:

Sound waves will move faster in water at high temperature.

Explanation:

The waves in which particles of the medium vibrate in the direction of propagation, is called longitudinal wave.

Hence sound waves will move faster in the water at high temperature.

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Answer:

my essay

i typed my my essay like u said

Answer:

Energy

A medium

Explanation:

A wave is a disturbance that travels through a medium, transferring energy from one point to another in the medium.

Mechanical waves are waves that require a medium for their propagation.

Answer:

THE ANSWER FOR THE FIRST SENTENCE IS ENERGY WHILE THE ANSWER FOR THE SECOND SENTENCE IS MATTER.

Explanation:

HOPE THIS HELPS

Explanation:

Short, Made with medal, and screws into places to hold things together.

Hope This Helps!!!

screws are one kind kd simple machine. They have a corkscrew-shaped ridge, know as a thread, wrapping around a cylinder. The head is specially shaped to allow a screwdriver or wrench to grip the screw driving it in.

Answer:

v₂₃₁ = 304.4 m/s

v₃₅₁ = 355.5 m/s

Explanation:

The formula used to find the speed of sound in air, at different temperatures is given as follows:

v = v₀ √(T/273)

where,

v = speed of sound at given temperature

v₀ = speed of sound at 0°C = 331 m/s

T = temperature in K

For T = 231 K:

v₂₃₁ = (331 m/s)√(231 K/273 K)

v₂₃₁ = 304.4 m/s

For T = 315 K:

v₃₅₁ = (331 m/s)√(315 K/273 K)

v₃₅₁ = 355.5 m/s

Answer:

third trust eh if not then here a cookie for my forgivness

Explanation:

Answer:

The actual frequency produced by the train whistle is 444 Hz.

Explanation:

Given;

speed of the train, Vs = 120 km/h = 33.33 m/s

observed frequency, Fo = 0.4 kHz = 400 Hz

speed of sound in air, V = 340 m/s

Note: as the train moves away, the observed frequency will be smaller than the source frequency. Thus, the source frequency which is frequency of the train whistle will be greater than 400 Hz.

The source frequency is given by Doppler effect formula;

[tex]F_s = F_o [\frac{V}{V-V_s} ]\\\\F_s = 400[\frac{340}{340-33.33} ]\\\\F_s = 400(1.109)\\\\F_s = 443.6 \ Hz[/tex]

Fs = 444 Hz

Therefore, the actual frequency produced by the train whistle is 444 Hz.

Answer:B

Explanation:Just took it.

200 coils of insulated wire around a steel bolt is the strongest electromagnet among them. Hence, option (A) is correct.

A wire coil is used to build electromagnets (wire curled in series). In comparison to a straight wire, this is more efficient in creating a magnetic field. A powerful magnetic core consisting of a magnetic substance, such as iron, can be tightly wound around a wire to enhance this effect.

In electromagnet, the core serves a great role in creating magnetic field.  The core of ferromagnetic materials is far efficiency compared to hollow cardboard core. And  the more number of turns means more amount of current passes through it and the strength of the electromagnet is more. That is why, 200 coils of insulated wire around a steel bolt is the strongest electromagnet among them.

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Answer:

DEVICE E

Explanation:

Answer:

R.E.T.A.R.D.A.T.I.O.N

Explanation:

It won't let me spell it normal

Joffrey talks and moves slowly. When asked a question, he answers slowly in monotone monosyllables, if he answers at all then it means that he is experiencing R.E.T.A.R.D.A.T.I.O.N .

The rate of change in an object's velocity with respect to time is known as acceleration in mechanics. The vector quantity of accelerations. The direction of the net force that is acting on an object determines its acceleration.

Since acceleration has both a magnitude and a direction, it is a vector quantity. Velocity is a vector quantity as well. The definition of acceleration is the change in velocity vector over a time interval divided by the time interval.

R.E.T.A.R.D.A.T.I.O.N is the process of inhibiting something's growth. However, negative acceleration has a very different meaning in physics. The negative acceleration is just the opposite of acceleration. A shift in speed is meant by the phrase "acceleration." In general, acceleration signifies an increase in velocity; while, acceleration in the opposite  denotes a reduction in velocity.

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Answer:

[tex]\mathbf{x = 3te ^{-2 \sqrt{2t}}}[/tex]

Explanation:

From the given information concerning the spring-mass system:

Let us apply Hooke law.

Then, we have:

mg = ks

8 = k4

k = 8/4

k = 2

Provided that the mass weighing 8 lbs is attached to a spring.

Then, we can divide it by gravity 32 ft/s².

∴

m = 8/32

m = 1/4 slugs

The medium that offers the damping force [tex]\beta = \sqrt{2}[/tex]

Now, let us set up a differential equation that explains the motion of the spring-mass system.

The general equation is:

[tex]mx '' + \beta x' + kx = 0[/tex]

where;

[tex]m = \dfrac{1}{4}[/tex]

k = 2,  and

[tex]\beta = \sqrt{2}[/tex]

Then;

[tex]\dfrac{1}{4}x'' + \sqrt{2} x' + 2x = 0[/tex]

By solving the above equation, the auxiliary equation is:

[tex]m^2 + 4 \sqrt{2} m + 8n = 0[/tex]

Using quadratic formula:

[tex]\dfrac{-b \pm \sqrt{b^2 -4ac}}{2a}[/tex]

[tex]m = \dfrac{-4 \sqrt{2} \pm \sqrt{(4 \sqrt{2})^2 -4(1)(8) }}{2}[/tex]

[tex]m = \dfrac{-4 \sqrt{2} \pm \sqrt{32 -32 }}{2}[/tex]

[tex]m = -2 \sqrt{2}[/tex]

Since this is a repeated root, the solution to their differential equation took the form.

[tex]x_c = c_1 e^{mt} + c_2 t e^{mt}[/tex]

[tex]x_c = c_1 e^{-2\sqrt{2} t} + c_2 t e^{-2\sqrt{2} t}[/tex]

From the initial condition.

At equilibrium position where the mass is being from:

x(0) = 0

Also, at the downward velocity of 3 ft/s

x'(0) = 3

Then, at the first initial condition:

[tex]x_c (0) = c_1 e^{-2\sqrt{2} *0} + c_2 (0) e^{-2\sqrt{2} *0}[/tex]

[tex]0= c_1 e^{0} + 0[/tex]

[tex]0= c_1[/tex]

At the second initial condition;

[tex]x' = -2 \sqrt{2} c_1 e^{-2 \sqrt{2} t } -2 \sqrt{2} c_2 t e^{-2 \sqrt{2} t } + c_2 e^{-2 \sqrt{2} t}[/tex]

where;

x'(0) = 3

[tex]x' (0) = -2 \sqrt{2} c_1 e^{-2 \sqrt{2}* 0 } -2 \sqrt{2} c_2 (0) e^{-2 \sqrt{2} *0 } + c_2 e^{-2 \sqrt{2} *0}[/tex]

[tex]3 = -2 \sqrt{2} * 0 *e^0 - 0 + c_2 e^0[/tex]

[tex]3 = 0 + c_2[/tex]

[tex]3 = c_2[/tex]

Replacing in the constraints, the equation of the motion is:

[tex]\mathbf{x = 3te ^{-2 \sqrt{2t}}}[/tex]

Answer:

D. the same as force. the applied force per cross-sectional area.

Explanation:

Tensile stress of a material is defined as the ratio of the applied force on the material to its cross sectional area. this is expressed mathematically as;

Tensile stress = Force/cross sectional area

Tensile stress = F/A

Force is measured in newton while cross sectional area is measured in m

Hence the unit of Tensile stress is N/m²

The tensile stress is the same as force. the applied force per cross-sectional area. Hence, option (D) is correct .

The given problem is based on the concept of stress and strain. The force applied on a material per unit area is known as Stress. It is denoted by the symbol [tex]\sigma[/tex] .

And due to the applied force, the ratio of change in length to the original length is known as Strain. It is denoted by the symbol [tex]\epsilon[/tex].

And the ratio of Stress and Strain is known as Elastic modulus. That is,

[tex]E = \dfrac{stress}{strain}\\\\E = \dfrac{\sigma }{\epsilon}[/tex]

So, the tensile stress is obtained as,

[tex]E = \dfrac{\sigma}{\epsilon}\\\\\sigma = E \times \epsilon[/tex]

So, clearly stress is nothing but the force applied per unit cross sectional area. That is,

[tex]\sigma = \dfrac{F}{A}[/tex]

Thus, we can conclude that the tensile stress is the same as force. the applied force per cross-sectional area. Hence, option (D) is correct .

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