A submarine sends out a sonar signal (sound waves) in a direction directly downward it take 2.3 s for the sound waves to travel from the submarine to the ocean bottom and back to the submarine how high (approx) up from the ocean floor is the submarine?speed of sound in water is 1490 m/s

Answers

Answer 1

Explanation:

Using the formula;

2x = vt

x is the distance up from the ocean floor the submarine is

v is the speed of sound in water

t is the time

Given

t = 2.3s

v = 1490m/s

Required

how high (approx) up from the ocean floor is the submarine x

From the formula;

x = vt/2

x = 1490(2.3)/2

x = 745(2.3)

x = 1,713.5m

Hence the submarine is 1713.5m high up from the ocean floor


Related Questions

Can someone please answer how to convert mass into weight?

Answers

Answer:

To find the weight of something, simply multiply its mass by the value of the local gravitational field, and you get a result in newtons (N). For example, if your mass is 50 kg (about 110 pounds), then your weight is (50) (9.8). The point that must be overwhelmingly emphasized is that weight is a force.

Explanation:

A ball thrown vertically upward returns to the thrower 25s later determine the speed with which it was thrown​

Answers

25/2=12.5s

12.5*9.8=122.5 m/s

you drive your car at a speed of 40 km/hr then slow down to 30km/hr

Answers

Answer:

3

Explanation:

3

answer? physics Q for 3rd secondary..

Answers

Answer:

The 300 Ohm resistor

Explanation:

They are in series, so the current has to be the same in both.  P = I^2R, so with the same current, the larger resistor dissipates more power.

The Magnetic Dipole Moment of a Coil Problem A rectangular coil of dimensions 5.40 cm ✕ 8.50 cm consists of 25 turns of wire and carries a current of 20.0 mA. A 0.350 T magnetic field is applied parallel to the plane of the loop. A. What is the magnitude of its magnetic dipole moment?B. What is the magnitude of the torque acting on the loop?

Answers

Answer:

a

 [tex]\mu = 0.0023 \ A\cdot m^2[/tex]

b

[tex]\tau = 0.00080 \ N \cdot m[/tex]

Explanation:

From the question we are told that

  The dimensions of the  rectangular coil is 5.40 cm ✕ 8.50 cm  = 0.054 m X 0.085 m

   The  number of turns is  [tex]N = 25 \ turns[/tex]

   The current it  is carrying is  [tex]I = 20 \ mA = 0.02 \ A[/tex]

    The magnetic field is [tex]B = 0.350 \ T[/tex]

Generally the magnitude of the magnetic dipole moment is mathematically represented as

       [tex]\mu = N * I * A[/tex]

Here  A  is the area which is mathematically represented as

     [tex]A = 0.054 * 0.085[/tex]

=>  [tex]A = 0.00459 \ m^2[/tex]

So  

     [tex]\mu = 25 * 0.02 * 0.00459[/tex]

=>  [tex]\mu = 0.0023 \ A\cdot m^2[/tex]

Generally the  magnitude of the torque acting on the loop is mathematically represented as

    [tex]\tau = \mu * B[/tex]

=>[tex]\tau = 0.0023 * 0.350[/tex]

=>[tex]\tau = 0.00080 \ N \cdot m[/tex]

What average net force is required to accelerate a car with a mass of 1200 kg from rest to 27.0 m/s2 in 10.0 s ?

1620 N
4360 N
444 N
11800 N
3240 N

Answers

Use formula F = MA
Acceleration= (27 m/s2)/10s = 2.7 m/s/s
F = 1200kg * 2.7 m/s/s = 3240 N

The average net force required to accelerate the car is 3240 N

What is acceleration?

This is defined as the rate of change of velocity which time. It is expressed as

a = (v – u) / t

Where

a is the acceleration v is the final velocity u is the initial velocity t is the time

How to determine the acceleration Initial velocity (u) = 0 m/sFinal velocity (v) = 27 m/sTime (t) = 10 sAcceleration (a) =?

a = (v – u) / t

a = (27 – 0) / 10

a = 2.7 m/s²

How to determine the forceMass (m) 1200 KgAcceleration (a) = 2.7 m/s²Force (F) = ?

F = ma

F = 1200 × 2.7

F = 3240 N

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A tungsten filament at 25-degree celsius has a resistance of 0.350 ohms. What would be the resistance if the temperature is increased to 2850°C? The temperature coefficient of resistance for tungsten is 0.5*10^-3 /°C

Answers

Answer:

R₂₈₅₀ =  0.838 ohms

Explanation:

The resistance of an element at some other temperature than 0°C can be given by the following formula:

Rₓ = R₀(1 + ∝ΔT)

where,

Rₓ = Resistance at temperature x

R₀ = Resistance at 0°C

∝ = Coefficient of linear expansion of tungsten = 0.5 x 10⁻³ /°C

ΔT = Change in Temperature

So, for the resistance of tungsten wire at 25°C:

Rₓ = R₂₅ = 0.35 ohms

ΔT = 25°C

Therefore,

0.35 ohms = R₀[1 + (0.5 x 10⁻³ /°C)(25°C)]

R₀ = (0.35 ohms)/1.0125

R₀ = 0.345 ohms

Now, we find the resistance at 2850°C

R₂₈₅₀ = (0.345 ohms)[1 + (0.5 x 10⁻³ /°C)(2850°C)]

R₂₈₅₀ =  0.838 ohms

A car is traveling at 60 mph. If the rate of speed increases 4 mph each hour,
how long will it be before the car is traveling at a rate of 80 mph?
A) 4 hours
B) 5 hours
C) 6 hours
D 7 hours​

Answers

Answer:

5 hours ................

Answer:

B=5 hrs

explanation:

becoz if you add 4 mph/hr

it will graually be 80 in 5 hrs

Consider a person standing in an elevator that is moving at a constant velocity down. The upward normal force N exerted by the elevator floor on the person is Select one: a. smaller than the downward force of gravity on the person. b. identical to the downward force of gravity on the person. c. larger than the downward force of gravity on the person.

Answers

Answer:

b. identical to the downward force of gravity on the person.

Explanation:

For an object in an elevator,

F = mg - ma       (g > a)

But since the velocity is uniform, a = 0.

Then,

F = mg - 0

F = mg

This is the actual weight of the object.

The object does not feel weightless, so that its actual weight can be measured during the downward motion of the elevator with uniform velocity.

Thus, the upward normal force, N, exerted by the elevator floor on the person is identical to the downward force of gravity on the person.

Jessica pulls on a washing machine with 200 N of force. Mark helps out and pulls on the washing machine in the same direction with 400 N of force. What is the net force on the washing machine

Answers

Answer:

600N

Explanation:

The x component of vector A is -25.0m and the y component id +40.0m (a) what is the magnitude of A?(b) What is the angle between the direction of A and the positive direction of x?

Answers

Answer:

θ = 122°

Explanation:

Components of a Vector

A vector in the plane can be defined by its rectangular components:

[tex]\vec A =<x,y>[/tex]

Or also can be given by its polar components:

[tex]\vec A =<r,\theta>[/tex]

Where r is the magnitude of the vector and θ is the angle it forms with the positive direction of x.

The relation between them is:

[tex]r=\sqrt{x^2+y^2}[/tex]

[tex]\displaystyle \theta=\arctan\frac{y}{x}[/tex]

It's given the x-component of vector A is x=-25 m and the y-component is y=40 m

(a)

The magnitude of the vector is:

[tex]r=\sqrt{(-25)^2+40^2}[/tex]

[tex]r=\sqrt{625+1600}[/tex]

[tex]r=\sqrt{2225}[/tex]

[tex]r\approx 47.2\ m[/tex]

(b)

[tex]\displaystyle \theta=\arctan\frac{40}{-25}[/tex]

[tex]\displaystyle \theta=\arctan (-1.6)[/tex]

The calculator gives us the value

θ = -58°

But the real angle lies on the second quadrant since x is negative and y is positive, thus:

θ = -58° + 180° = 122°

θ = 122°

what is the speed of rocket that travels 9km in 10 seconds​

Answers

Answer:

900

Explanation:

v = s / t = 9000m / 10 s = 900m/s

Explanation:

The speed of a rocket that travel 9km in 10 second is 900

Coach ulcer paces the sidelines. Sarting at the 30 yd. line (A), he moves to the 10 yd. line (B), back to the 50 yd. line (C) and finally to the 20yd. Line (D) in 200 seconds. Determine his average speed and velocity.

Answers

Answer:

See the answer below

Explanation:

Average speed = total distance traveled/total time taken

In order to determine the total distance traveled by the coach, consider the attached image.

Distance covered:

30 yd. line to 10 yd. line (A to B)= 20 yds

10 yd. line to 50 yd. line (B to C) = 40 yds

50 yd. line to 20 yd. line (C to D) = 30 yds

Total distance covered = 20 + 40 + 30 = 90 yds

Time taken = 200 seconds

Average Speed = 90/200 = 0.45 yd/s

Velocity = speed with direction

Hence,

His Velocity = 0.45 yd/s to the left of his starting point.

Why evaporation takes place from the Surface?​

Answers

Answer:

in the water cycle evaporation occurs when the sunlight warms the surface of the water the heat from the sun makes the water molecules move faster and faster until they move so fast they can escape as a gas once evaporated a molecule from water that vapor spends about 10 days in the air

4. A substance has a density of 0.79 g/cm'. It is soluble in water. List all the possibilities of what it might be How could you determine the actual identity?​

Answers

Answer:

See explanation

Explanation:

Given that the density of the unknown substance is 0.79 g/cm3 and is soluble in water, the possible substances it could be are;

i) t-butanol

ii) ethanol

iii) 2-propanol

iv) acetone

However, the actual identity of the unknown substance can be obtained by carrying out a boiling point test. The four substances listed above have different boiling points. Hence the boiling point of the unknown substance ultimately discloses its identity.

Table 1
Ball Na
dom
dom
n. Pas
1). Pas
1
4.3
5.86-10
5.91-10
5.89-10
8.09.10
4.91
3
6.02-10
6.32-10
5.24 10
14
5.91-10
6.98 to
8.07.10
6.41 10
5,8910
5.99-10
638-10
5.38.10
4.16
5
5.43
At 20 °C, the density of the ball substance is p = 2,2 10kg/m: fluid is p = 0,91-10kg/m
Distance between two marks is? - 1 m

Answers

Answer:

the answer is c.

Explanation:

What is the answer to this question

Answers

Answer:

0.11 m/s

Explanation:

From the question given above, the following data were obtained:

Initial displacement (d1) = 1.09 m

Final displacement (d2) = 2.55 m

Time (t) = 12.8 s

Average velocity =?

Next, we shall determine the total displacement (i.e change in displacement). This can be obtained as follow:

Initial displacement (d1) = 1.09 m

Final displacement (d2) = 2.55 m

Total displacement = d2 – d1

Total displacement = 2.55 – 1.09

Total displacement = 1.46 m

Finally, we shall determine the average velocity of the beetle. This can be obtained as follow:

Total Displacement = 1.46 m

Total time (t) = 12.8 s

Average velocity =?

Average velocity = Total Displacement / Total time

Average velocity = 1.46/12.8

Average velocity = 0.11 m/s

Thus, the average velocity of the beetle is 0.11 m/s

What is the de Broglie wavelength of an electron accelerated from rest through a potential difference of 50.0 V?
(a) 0.100 nm
(b) 0.139 nm
(c) 0.174 nm
(d) 0.834 nm
(e) none of those answers

Answers

Answer:

(c) 0.174 nm

Explanation:

According to de Broglie hypothesis, the wavelength of the wave associated with electron is given by:

[tex] \boxed{ \bf{\lambda = \sqrt{\dfrac{150}{V \ (in \ Volt)}} \: \text{\AA}}}[/tex]

V → Potential Difference (50.0 V)

By substituting value of potential difference in the equation, we get:

[tex] \rm \longrightarrow \lambda = \sqrt{\dfrac{150}{50}} \: \text{\AA} \\ \\ \rm \longrightarrow \lambda = \sqrt{3} \: \text{\AA} \\ \\ \rm \longrightarrow \lambda = 1.74 \: \text{\AA} \\ \\ \rm \longrightarrow \lambda = 0.174 \: nm[/tex]

A 5 kg bowling ball with a velocity of +10 m/s collides with a stationary 2 kg bowling pin. If the ball's final velocity is +8 m/s, what is the pin's final velocity?

a 5 m/s
b 2.5 m/s
c 10 m/s
d 5.2 m/s

Answers

Answer:

The pin's final velocity is 5m/s

Explanation:

Step one:

given data

mass of ball m1=5kg

initial velocity of ball u1=10m/s

mass of pin m2=2kg

initial velocity of pin u2= 0m/s

final velocity of ball v2=8m/s

final velocity of pin v2=?

Step two:

The expression for elastic collision is given as

m1u1+m2u2=m1v1+m2v2

substituting we have

5*10+2*0=5*8+2*v2

50+0=40+2v2

50-40=2v2

10=2v2

divide both sides by 2

v2=10/2

v2=5m/s

The pin's final velocity is 5m/s

2. A force of 500 N separates two objects. The distance between their centers is 67 m. The first object has a mass of 20 kg. What is the mass of the second object?​

Answers

Answer:

The Equation of Newton's Law of Gravity. G is called the Gravitational Constant, and has the value 6.67×10-11 N ∙m2kg-2 (N is for Newton, the physicists' unit of force)or 1.5×10-11 lb∙m2kg-2

Explanation:

Gravitational force is an attractive force that exists between all objects with mass . an object with mass attracts another object with mass.

Mass of second object is [tex]1.6825*10^{15}[/tex] kg

If two bodies have mass m₁ and m₂ . Then attractive force between them is calculated by formula show below,

                                  [tex]F=G\frac{m1m2}{r^{2} }[/tex]

Where r is distance between bodies. and G is gravitational constant.

G = [tex]6.67*10^{-11}[/tex]  [tex]Nm^{2}/kg^{2}[/tex]

Here,  F = 500N, m₁=20 kg and r= 67 m

Substituting above values in force equation.

            [tex]500=\frac{6.67*10^{-11}*20*m2 }{(67)^{2} } \\\\m2=1.6825*10^{15} Kg[/tex]

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describe three possible careers in physical science

Answers

Answer:

physical sicence boilagiy sxcience and earth science

Explanation:

Which of the following is a self-fulfilling prophecy?

mutual views that are often held by conflicting people
shared goals that override differences among people and require their cooperation
a perceived incompatibility of actions, goals, or ideas
a belief that leads to its own fulfillment

Answers

Answer:

last one

Explanation:

In the winter sport of curling, players give a 20 kg stone a push across a sheet of ice. The Slone moves approximately 40 m before coming to rest. The final position of the stone, in principle, onlyndepends on the initial speed at which it is launched and the force of friction between the ice and the stone, but team members can use brooms to sweep the ice in front of the stone to adjust its speed and trajectory a bit; they must do this without touching the stone. Judicious sweeping can lengthen the travel of the stone by 3 m.1. A curler pushes a stone to a speed of 3.0 m/s over a time of 2.0 s. Ignoring the force of friction, how much force must the curler apply to the stone to bring it op to speed?A. 3.0 NB. 15 NC. 30 N
D. 150 N2The sweepers in a curling competition adjust the trajectory of the slope byA. Decreasing the coefficient of friction between the stone and the ice.
B. Increasing the coefficient of friction between the stone and the ice.C. Changing friction from kinetic to static.D. Changing friction from static to kinetic.3. Suppose the stone is launched with a speed of 3 m/s and travel s 40 m before coming to rest. What is the approximate magnitude of the friction force on the stone?A. 0 NB. 2 NC. 20 ND. 200 N4. Suppose the stone's mass is increased to 40 kg, but it is launched at the same 3 m/s. Which one of the following is true?A. The stone would now travel a longer distance before coming to rest.B. The stone would now travel a shorter distance before coming to rest.C. The coefficient of friction would now be greater.D. The force of friction would now be greater.

Answers

Answer:82. Since you have a distance and a force, then the easiest principle to use is energy, i.e. work.

The work done by friction is F * d. This work cancels out the kinetic energy of the stone (1/2)mv^2

Fd = (1/2)mv^2

F = (1/2)mv^2/d.

Plug in m = 20 kg, v = 3 m/sec, d = 40 m.

83. With more mass, the kinetic energy is higher now. The work needed is higher. W = F * d and F is the same.

Explanation:Hope I helped :)

At noon on a clear day, sunlight reaches the earth's surface at Madison, Wisconsin, with an average intensity of approximately 2.00 kJ·s−1·m^−2. If the sunlight consists of photons with an average wavelength of 510.0 nm, how many photons strike a 5.20 cm^2 area per second?

Answers

Answer:

The number of photons per second that strike the given area is 2.668 x 10⁸ photons/second

Explanation:

Given;

intensity of the sunlight, I = 2.00 kJ·s−1·m^−2

area of incident, A = 5.2 cm² = 5.2 x 10⁻⁴ m²

Energy of incident photons per second on the given area;

E = IA

E = (2000)( 5.2 x 10⁻⁴)

E = 1.04 J/s

Energy of a photon is given is by;

[tex]E = \frac{hc}{\lambda} \\\\E = \frac{(6.626*10^{-34})(3*10^8)}{(510*10^{-9})}\\\\E = 3.898*10^{-19} \ J/photon[/tex]

The number of photons per second that strike the given area is;

[tex]n = \frac{1.04 \ J/s}{3.898*10^{-19} \ J/photon} \\\\n = 2.668*10^{18} \ photons/second[/tex]

Therefore, the number of photons per second that strike the given area is 2.668 x 10⁸ photons/second

You serve a tennis ball of mass 60g at a speed of 50
m/s, what is the impulse exerted on the ball? ( ball starts from rest )

Answers

Answer:

[tex]J = 3~Kg.m/s[/tex]

Explanation:

Impulse and Momentum

The impulse-momentum theorem states that the change in momentum of an object equals the impulse applied to it.

The equation can be written as follows:

[tex]J =\Delta p = p_2-p_1[/tex]

Where:

J    = Impulse

p2 = Final Momentum

p2 = Initial Momentum

The momentum can be calculated as:

p = m.v

Where m is the mass of the object and v is the velocity.

The tennis ball with mass m=60 g = 0.06 Kg was served from rest (v1=0) to v2=50 m/s. The change in momentum is:

[tex]\Delta p = 0.06Kg~50~m/s-0[/tex]

[tex]\Delta p = 3~Kg.m/s[/tex]

Thus the impulse is:

[tex]\marhbf{J = 3~Kg.m/s}[/tex]

The CEO, ellen misk, left her martian office but accidentally left a cylindricall can of coke (3.1 inches in diameter, 5.42 inches in height) on her desk. If the can exerts a pressure of 510 Pascals, what is the specific gravity of the can?

Answers

Answer:

Specific Gravity = 0.378

Explanation:

First, we will find the force exerted by the can on the table. This force will be equal to the weight of the can:

Pressure = Force/Area = Weight/Area

Weight = Pressure*Area

where,

Area = πdiameter²/4 = π[(3.1 in)(0.0254 m/1 in)]²/4 = 4.8 x 10⁻³ m²

Weight = (510 N/m²)(4.8 x 10⁻³ m²)

Weight = 2.48 N

Now, the weight is given as:

Weight = mg

2.48 N = m(9.8 m/s²)

m = (2.48 N)/(9.8 m/s²)

m = 0.25 kg

Now, we calculate volume of can:

Volume = (Area)(Height) = (4.8 x 10⁻³ m²)(5.42 in)(0.0254 m/1 in)

Volume = 6.6 x 10⁻⁴ m³

Hence, the density of can will be:

Density of Can = m/Volume = 0.25 kg/6.6 x 10⁻⁴ m³

Density of Can = 378.32 kg/m³

So, the specific gravity of Can will be:

Specific Gravity = Density of Can/Density of Water

Specific Gravity = (378.32 kg/m³)/(1000 kg/m³)

Specific Gravity = 0.378

(ii) In two parts of the graph in Figure 1 the forces are balanced.
State the letters of the two parts of the graph where the horizontal forces
acting on the car are balanced.
121
part
and part

this is the graph:

Answers

Answer:

Hii I m Indian.......♥️

A painter on a scaffold drops a 1.25 kg can of paint from a height of 5.00 m.
a) What is the kinetic energy of the can when the can is at a height of 3.00 m?
b) With what speed will the can hit the ground? (Neglect air resistance)

Answers

Answer:

(a) The kinetic energy of the can when the can is at a height of 3.00 m.

(b) The speed with which the can will hit the ground is 9.9 m/s

Explanation:

(a) Kinetic energy is a form of energy. It is defined as the energy associated with bodies that are in motion and this energy depends on the mass and speed of the body.

Kinetic energy is defined as the amount of work necessary to accelerate a body of a given mass and at rest, until it reaches a given speed.

Kinetic energy is represented by the following formula:

Ec = ½ mv²

Where Ec is kinetic energy, which is measured in Joules (J), m is mass measured in kilograms (kg), and v is velocity measured in meters over seconds (m/s).

On the other hand, potential energy is the energy that measures the ability of a system to perform work based on its position. In other words, this is the energy that a body has at a certain height above the ground.

Gravitational potential energy is the energy associated with the gravitational force. This will depend on the relative height of an object to some reference point, the mass, and the force of gravity. Then for an object with mass m, at height h, the expression applied to the gravitational energy of the object is:

Ep = m g h

Where Ep is the potential energy in joules (J), m is the mass in kilograms (kg) is h the height in meters (m) and g is the acceleration of fall in m/s² (approximately 9.8 m/s²)

Energy is neither created nor destroyed, but only transforms from one state to another. This principle also extends to mechanical energy, so that in an isolated system, the sum of kinetic and potential energies between two instants of time remains constant:

½ m*v1² + m*g*h1=  ½ m*v2² + m*g*h2

 where v1 and h1 are the initial speed and height and v2 and h2 the final speed and height.

In this case:

v1= 0 m/sh1= 5 mv2= ?h2= 3 m

Replacing:

½ m*0² + m*g*5=  ½ m*v2² + m*g*3

Solving:

m*g*5= ½ m*v2² + m*g*3

The mass m appears in all terms, being able to simplify:

g*5=  ½ v2² + g*3

Solving for v2:

[tex]v2=\sqrt{\frac{g*5-g*3}{\frac{1}{2} } }[/tex]

being g= 9.8 m/s²:

v2= 6.26 m/s

The speed at 3 meters height is 6.26 m/s

Then the kinetic energy is calculated as:

Ec = ½ mv²

Ec = ½ 1.25 kg* (6.26 m/s)²

Ec= 24.49 J

The kinetic energy of the can when the can is at a height of 3.00 m.

(b) You know:

½ m*v1² + m*g*h1=  ½ m*v2² + m*g*h2

 where v1 and h1 are the initial speed and height and v2 and h2 the final speed and height.

In this case:

v1= 0 m/sh1= 5 mv2= ?h2= 0 m

Replacing:

½ m*0² + m*g*5=  ½ m*v2² + m*g*0

Solving:

m*g*5= ½ m*v2²

g*5=  ½ v2²

[tex]v2=\sqrt{\frac{g*5}{\frac{1}{2} } }[/tex]

being g= 9.8 m/s²:

v2= 9.9 m/s

The speed with which the can will hit the ground is 9.9 m/s

A student bought a 1.55-ounce chocolate bar and left it in
a car on a hot day.
How many ounces of chocolate are in the melted bar?
A. Exactly 1.55 ounces
B. At least 1.55 ounces
C. An unknown number of ounces
D. Less than 1.55 ounces

Answers

Answer:

a

Explanation:

A student bought a 1.55-ounce chocolate bar and left it in a car on a hot day.  Exactly 1.55 ounces of chocolate are in the melted bar. Hence, option (A) is correct.

What is mass?

In physics, mass is a quantitative measurement of inertia, a basic characteristic of all matter. It essentially refers to a body of matter's resistance to changing its speed or location in response to the application of a force.

The change caused by an applied force is smaller the more mass a body has. The kilogram serves as the unit of mass in the International System of Units (SI).

As mass does not depends on the state of the object, exactly same mass of chocolate are in the melted bar that it was in chocolate bar. So, 1.55 ounces of chocolate are in the melted bar. Hence, option (A) is correct.

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A force of 150 N is applied on an object at 60 degrees above the positive x-axis. Determine its

horizontal and vertical components.

Answers

Answer:

horizontal component=fcostita

=150cos60

use calculator to evaluate it

for vertical=fsintita

=150sin60

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