The greatest ocean depths on the Earth are found in the Marianas Trench near the Philippines. Calculate the pressure (in atm) due to the ocean at the bottom of this trench, given its depth is 10.3 km and assuming sea water density is constant all the way down. atm

Answers

Answer 1

Answer:

[tex]P = 103867260[/tex] atm

Explanation:

The pressure at the bottom of any liquid column is equal to product of density of the liquid , gravitational acceleration constant (g) and height of the water column

Thus, [tex]P = \rho*g*h[/tex]

Substituting the given values, we get -

[tex]P = 1029[/tex] kg/m3 [tex]* 9.8[/tex] m/s^2  [tex]*10.3 *1000[/tex] meters

[tex]P = 103867260[/tex] atm


Related Questions

An open organ pipe emits E (330 Hz ) when the temperature is 26 ∘C. The speed of sound in air is v≈(331+0.60T)m/s, where T is the temperature in ∘C. Determine the length of the pipe.

Answers

Answer:

0.525 m

Explanation:

From the question,

In an open pipe,

v = λf................... Equation 1

Where v = speed of sound in the pipe, f = frequency, λ = wave length.

But,

λ = 2l................ Equation 2

Where l = length of the pipe

Substitute equation 2 into equation 1

v = 2lf.............. Equation 3

make l the subject of the equation

l = v/2f............. Equation 4

Given:

v ≈ (331+0.60T) m/s at T = 26°C

v ≈ (331+0.60×26) ≈ 346.6 m/s

f = 330 Hz.

Substitute these values into equation 4

l = 346.6/(2×330)

l = 346.6/660

l = 0.525 m

Hence the length of the pipe is 0.525 m

What is the rate of flow of electric charge around a circuit

Answers

Answer:

Current

Explanation:

hope it helps and your day will be full of happiness

An air-filled pipe is found to have successive harmonics at 980 Hz , 1260 Hz , and 1540 Hz . It is unknown whether harmonics below 980 Hz and above 1540 Hz exist in the pipe. What is the length of the pipe

Answers

Answer:

  L = 0.7 m

Explanation:

This is a resonance exercise, in this case the air-filled pipe is open at both ends, therefore we have bellies at these points.

          λ / 2 = L                       1st harmonic

          λ = L                            2nd harmonic

          λ = 2L / 3                    3rd harmonic

         λ =  2L / n                    n -th harmonic

the speed of sound is related to wavelength and frequencies

           v =λ f

            f = v /λ

           

we substitute

            f = v n / 2L

the speed of sound in air is v = 343 m / s

suppose that the frequency of f = 980Hz occurs in harmonic n

            f₁ = v n / 2L

            f₂ = v (n + 1) / 2L

            f₃ = v (n + 2) / 2L

we substitute the values

             2 980/343  =  n / L

              2 1260/343 = (n + 1) / L

               2 1540/343 = (n + 2) / L

             

we have three equations, let's use the first two

              5.714 = n / L

              7.347 = (n + 1) / L

we solve for L and match the expressions

              n / 5,714 = (n + 1) / 7,347

              7,347 n = 5,714 (n + 1)

              n (7,347 -5,714) = 5,714

              n = 5,714 / 1,633

              n = 3.5

as the number n must be integers n = 4 we substitute in the first equation

              L = n / 5,714

              L = 0.7 m

Using a Boltzmann distribution, find the fraction of atoms in the excited state versus the ground state (i.e. the relative population) in a plasma source and a flame source. Assume that the lowest energy of a sodium atom lies at 3.371x10-19 J above the ground state, the degeneracy of the excited state is 2, whereas that of the ground state is 1, and the temperature of the flame is 3000 K and 10,000 K for plasma.

Answers

Answer:

0.174 plasma

[tex]$5.85 \times 10^{-4}$[/tex] flame

Explanation:

Given :

Energy :

[tex]$\Delta E=3.371 \times 10^{-19} $[/tex] J per atom

[tex]$g^*=2$[/tex] (degenraci of excited state)

g = 1  (degenraci of excited state)

Boltzmann Distribution

[tex]$\frac{N^*}{N}=\frac{g^*}{g}e^{-\frac{\Delta E}{kT}}$[/tex]

where,

[tex]$N^*$[/tex] = atoms in excited state

N = atoms in lower energy level

k = [tex]$1.38 \times 10^{-23}$[/tex]  J/K

Therefore,

Relative population in plasma

T = 10,000 K

[tex]$\frac{N^*}{N}=\frac{g^*}{g}e^{-\frac{\Delta E}{kT}}$[/tex]

[tex]$\frac{N^*}{N}=\frac{2}{1}e^{-\frac{-3.37\times 10^{-19}}{1.38 \times 10^{-23} \times 10000}}$[/tex]

[tex]$\frac{N^*}{N}=2 \times e^{-2.44275}$[/tex]

[tex]$\frac{N^*}{N}=2 \times 0.8692$[/tex]

[tex]$\frac{N^*}{N}=0.1738$[/tex]

Relative population in flame    

T = 3000

[tex]$\frac{N^*}{N}=\frac{g^*}{g}e^{-\frac{\Delta E}{kT}}$[/tex]

[tex]$\frac{N^*}{N}=\frac{2}{1}e^{-\frac{-3.371\times 10^{-19}}{1.38 \times 10^{-23} \times 3000}}$[/tex]

[tex]$\frac{N^*}{N}=2 \times e^{-8.1425}$[/tex]

[tex]$\frac{N^*}{N}=2 \times 2.9090 \times 10^{-4}$[/tex]

[tex]$\frac{N^*}{N}=5.85 \times 10^{-4}$[/tex]

Find the current passing through each of the 3 resistors connected parallel to each other as shown in the figure (i1, i2, I3). Show your actions clearly.

Answers

Answer:

I1 = ε/R1

I2 = ε/R2

I3 = ε/R3

Explanation:

From the image, we see that the resistors are connected in parallel. This means that the voltage passing through them is the same.

Now, formula for current is; I = V/R

In this case, V which is voltage is denoted by ε.

Thus;

I1 = ε/R1

I2 = ε/R2

I3 = ε/R3

To decrease the angle between the anterior surface of the foot and anterior surface of the lower leg is described as:

Answers

Answer:

dorsiflexion

Explanation:

To decrease the angle between the anterior surface of the foot and anterior surface of the lower leg is described as: dorsiflexion

Answer:

dorsiflexion

Explanation:

Hope this helps

A 1.0-kg object moving 9.0 m/s collides with a 2.0-kg object moving 6.0 m/s in a direction that is perpendicular to the initial direction of motion of the 1.0-kg object. The two masses remain together after the collision, and this composite object then collides with and sticks to a 3.0-kg object. After these collisions, the final composite (6.0-kg) object remains at rest. What was the speed of the 3.0-kg object before the collisions

Answers

Answer:

         v₃ = - (3 i ^ + 4 j ^) m / s

v₃ = 5 m / s,  θ = 233º

Explanation:

This is a momentum problem. Let us form a system formed by the three objects so that the forces during the collisions have been internal and the moment is conserved.

Let's start working with the first two objects.  As each object moves in a different direction let's work with the components in an xy coordinate system

X axis

initial instant. Before the shock

       p₀ₓ = m₁ v₁₀ + 0

final instant. After the crash

      p_{fx} = (m1 + m2) vₓ

the moment is preserved

       p₀ₓ = p_{fx}

       m₁ v₀₁ = (m₁ + m₂) vₓ

       vₓ = [tex]\frac{m_1}{m_1+m_2} \ v_{o1}[/tex]

Y axis  

initial instant

       p_{oy} = 0 + m₂ v₀₂

final moment

       p_{fy} = (m₁ + m₂) v_y

the moment is preserved

      p_{oy} = p_{fy}

      m₂ v₀₂ = (m₁ + m₂) v_y

      v_y = [tex]\frac{m_2}{m_1 +m_2 } \ v_{o2}[/tex]

We already have the speed of the set of these two cars, now let's work on this set and vehicle 3

X axis    

initial instant

      p₀ₓ = (m₁ + m₂) vₓ + m₃ v₃ₓ

final instant

      p_{fx} = 0

      p₀ₓ = p_{fx}

      (m₁ + m₂) vₓ + m₃ v₃ₓ = 0

       v₃ₓ = [tex]- \frac{m_1+m_2 }{m_3} \ v_x[/tex]

Y Axis

initial instant

        p_{oy} = (m₁ + m₂) v_y + m₃ v_{3y}

final moment

        p_{fy} = 0

        p_{oy} = p_{fy}

        (m₁ + m₂) v_y + m₃ v_{3y} = 0

        v_{3y} = [tex]- \frac{m_1+m_2}{m_3} \ v_y[/tex]

now we substitute the values ​​of the speeds

       v₃ₓ = [tex]- \frac{m_1+m_2}{m_3} \ \frac{m_1}{m_1+m_2} \ v_{o1}[/tex]  

       v₃ₓ = [tex]- \frac{m_1}{m_3} \ v_{o1}[/tex]

       v_{3y} = [tex]- \frac{m_1+m_2}{m_3} \ \frac{m_2}{m_1+m_2} \ v_{o2}[/tex]

       v_{3y} = [tex]- \frac{m_2}{m_3} \ v_{o2}[/tex]

let's calculate

         v₃ₓ = - ⅓ 9

         v₃ₓ = - 3 m / s

         v_{3y} = - ⅔ 6

         v_{3y} = - 4 m / s

therefore the speed of vehicle 3 is

         v₃ = - (3 i ^ + 4 j ^) m / s

It can also be given in the form of modulus and angles using the Pythagorean theorem

         v₃ = [tex]\sqrt{v_{3x}^2 + v_{3y}^2}[/tex]

          v₃ = [tex]\sqrt{3^2+4^2}[/tex]

          v₃ = 5 m / s

let's use trigonometry for the angle

          tan θ' = [tex]\frac{v_{3y}}{v_{3x}}[/tex]

          θ' = tan⁻¹ (\frac{v_{3y}}{v_{3x}})

          θ' = tan⁻¹ (4/3)

          θ'  = 53º

That the two speeds are negative so this angle is in the third quadrant, measured from the positive side of the x axis

          θ = 180 + θ'

          θ = 180 +53

          θ = 233º

An object that weighs 20 N encounters 20 N of air resistance. What is the acceleration in the y direction?

Answers

Answer:

Explanation:

F=ma.

a=m÷fcos 0

In the process of changing a flat tire, a motorist uses a hydraulic jack. She begins by applying a force of 58 N to the input piston, which has a radius r1. As a result, the output plunger, which has a radius r2, applies a force to the car. The ratio r2/r1 has a value of 8.2. Ignore the height difference between the input piston and output plunger and determine the force that the output plunger applies to the car.

Answers

Answer:

The correct answer is "3899.92 N".

Explanation:

The given values are:

Force,

[tex]F_{app}=58 N[/tex]

Ratio,

[tex]\frac{R_2}{R_1}=8.2[/tex]

As we know,

Area, [tex]A=\pi r^2[/tex]

or,

⇒  [tex]\frac{F_2}{F_1} =\frac{A_2}{A_1}[/tex]

On substituting the value of "A", we get

⇒  [tex]\frac{F_2}{F_1} =\frac{\pi r_2^2}{\pi r_1^2}[/tex]

⇒  [tex]\frac{F_2}{F_1} =\frac{r_2^2}{r_1^2}[/tex]

On applying cross-multiplication, we get

⇒  [tex]F_2=F_1\times (\frac{r_2}{r_1} )^2[/tex]

On substituting the given values, we get

⇒       [tex]=58\times (8.2)^2[/tex]

⇒       [tex]=58\times 67.2[/tex]

⇒       [tex]=3899.92 \ N[/tex]

After graduation, you face a choice. You can work for a multinational consulting firm and earn a starting salary (including benefits) of $40,000, or you can start your own consulting firm using $5,000 of your own savings. If you keep your money in a savings account, you can earn an interest rate of 7 percent. You choose to start your own consulting firm. At the end of the first year, you add up all of your expenses and revenues. Your expenses include $14,000 for rent, $1,000 for office supplies, $24,000 for labor, and $4,500 for telephone expenses. After operating your consulting firm for a year, your total revenues are $88,000.

Required:
What are your total explicit cost and total implicit costs?

Answers

Answer:

Explanation:

Money is used to cover the explicit cost. They're what we're used to seeing, and they're easy to identify. There are several classifications for explicit costs; we can distinguish between fixed and variable costs, as well as direct and indirect costs. Raw materials, manpower, indirect production costs, and so on are all factors that contribute to its expense.

The opportunity cost of using a resource is the amount of money paying for it that might have been used on something else.

Alternative benefit options or money that we miss earning by doing such business acts are referred to as opportunity costs.

When a corporation foregoes an alternative action but does not make a bill, it incurs implicit costs. These are a company's hidden costs:

1. The use of the firm's own funds (money or assets).

2. The owner's capital, savings, and financial services are used.

From the given information:

The total explicit cost = Rent + office supplies + office staff + telephone expense

=$( 14000 + 1000 + 24000 + 4500)

= $43500

The total implicit cost = forgone salary + forgone interest

= $40000 + 7% of $5000

= $40000 + 350

=$40350

A wave pulse travels along a stretched string at a speed of 100 cm/s. What will be the speed in cm/s if the string's tension is quadrupled, the length halved and its mass is doubled

Answers

Answer:

The new velocity of the string is 100 centimeters per second (1 meter per second).

Explanation:

The speed of a wave through a string ([tex]v[/tex]), in meters per second, is defined by the following formula:

[tex]v = \sqrt{\frac{T\cdot L}{m} }[/tex] (1)

Where:

[tex]T[/tex] - Tension, in newtons.

[tex]L[/tex] - Length of the string, in meters.

[tex]m[/tex] - Mass of the string, in kilograms.

The expression for initial and final speeds of the wave are:

Initial speed

[tex]v_{o} = \sqrt{\frac{T_{o}\cdot L_{o}}{m_{o}} }[/tex] (2)

Final speed

[tex]v = \sqrt{\frac{(4\cdot T_{o})\cdot (0.5\cdot L_{o})}{2\cdot m_{o}} }[/tex]

[tex]v = \sqrt{\frac{T_{o}\cdot L_{o}}{m_{o}} }[/tex] (3)

By (2), we conclude that:

[tex]v =v_{o}[/tex]

If we know that [tex]v_{o} = 1\,\frac{m}{s}[/tex], then the new speed of the wave in the string is [tex]v = 1\,\frac{m}{s}[/tex].

What is the wavelength associated with 0.113kg ball traveling with velocity of 43 m/s?

Answers

Answer:

Wavelength = 1.36 * 10^{-34} meters

Explanation:

Given the following data;

Mass = 0.113 kg

Velocity = 43 m/s

To find the wavelength, we would use the De Broglie's wave equation.

Mathematically, it is given by the formula;

[tex] Wavelength = \frac {h}{mv} [/tex]

Where;

h represents Planck’s constant.

m represents the mass of the particle.

v represents the velocity of the particle.

We know that Planck’s constant = 6.6262 * 10^{-34} Js

Substituting into the formula, we have;

[tex] Wavelength = \frac {6.6262 * 10^{-34}}{0.113*43} [/tex]

[tex] Wavelength = \frac {6.6262 * 10^{-34}}{4.859} [/tex]

Wavelength = 1.36 * 10^{-34} meters

A radial saw has a blade with a 9-in. radius. Suppose that the blade spins at 1000 rpm. (a) Find the angular speed of the blade in rad/min. Incorrect: Your answer is incorrect. webMathematica generated answer key rad/min (b) Find the linear speed of the sawteeth in ft/s. Incorrect: Your answer is incorrect. webMathematica generated answer key ft/

Answers

Answer:

The angular speed of the blade is of 166.67 rad/min

If a radial saw has a blade with a 9-in. radius. Suppose that the blade spins at 1000 rpm then the angular speed of the blade would be 104.72 rad/seconds.

What is speed?

The total distance covered by any object per unit of time is known as speed. It depends only on the magnitude of the moving object. The unit of speed is a meter/second. The generally considered unit for speed is a meter per second.

The mathematical expression for speed is given by

speed = total distance /Total time

As given in the problem If a radial saw has a blade with a 9-in. radius. Suppose that the blade spins at 1000 rpm then we have to find the angular speed of the blade,

the angular speed of the blade (ω) = 2πN/60

                                                          =2×π×1000/60

                                                          = 104.72 rad/seconds

Thus, the angular speed of the radial saw would be 104.72 rad/seconds

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what is a magnetic flux thru closed surface

Answers

Gauss’s Law for magnetism

why is it important to know all about recreational activities?​

Answers

Answer:

It's important

Explanation:

Mental wellness is critical to overall physical health. Participating in recreational activities helps manage stress. Taking time to nurture oneself provides a sense of balance and self-esteem, which can directly reduce anxiety and depression.

Answer:

Mental wellness is critical to overall physical health. Participating in recreational activities helps manage stress. Taking time to nurture oneself provides a sense of balance and self-esteem, which can directly reduce anxiety and depression.

Explanation:

What is the resistance of a 15 ampere current with 8 volts of potential difference?

Answers

Answer:

Resistance in circuit = 0.53 ohm (Approx.)

Explanation:

Given:

Flow of current in circuit = 15 amp

Potential difference = 8 Volts

Find:

Resistance in circuit

Computation:

In an electrical system, resistance is a stopper of a material to electric current.

Resistance in circuit = Potential difference / Flow of current in circuit

Resistance in circuit = 8 / 15

Resistance in circuit = 0.53 ohm (Approx.)

Your car gets a flat! You go from 90 kilometers per hour to a stop in 6 seconds. What is your rate of deceleration?

Answers

Answer:

The rate of deceleration is 15 km/h^2.

Explanation:

Since the car comes in 90km/h, it will need deceleration for 15 km/h for 6 seconds to finally stop the car. 15*6=90

a horizontal 4-cm-diameter water jet with a velocity of 18 m/s. He impinges the jet normally upon a vertical plate of mass 750 kg. The plate rides on a nearly frictionless track and is initially stationary. When the jet strikes the plate, the plate begins to move in the direction of the jet. The water always splatters in the plane of the retreating plate. Determine (a)the acceleration of the plate when the jet first strikes it (time

Answers

Answer:

0.5429 m/s^2

Explanation:

velocity of waterjet = 18 m/s

diameter of water jet ( d ) = 4 cm = 0.04 m

mass of vertical plate(m) = 750 kg

Determine the acceleration of plate when the jet first strikes ( i.e. t = 0 )

first we will determine the impact force

F = β*A*V^2 ----- ( 1 )

where ; β = 1000 kg/m^3  ,  A = π/4 * d^2 , V = 18 m/s

input values into equation 1

F = 407.15 N

finally determine the acceleration at t = 0

F = m*a

a = F / m =  407.15 / 750 = 0.5429 m/s^2

Two loudspeakers placed 6.0 m apart are driven in phase by an audio oscillator whose frequency range is 2193 Hz to 2967 Hz. A point P is located 4.4 m from one loudspeaker and 3.6 m from the other. The speed of sound is 344 m/s. The frequency produced by the oscillator, for which constructive interference of sound occurs at point P, is closest to

Answers

Answer:

"2580 Hz" is the correct solution.

Explanation:

According to the question,

The path difference,

= [tex]4.4 - 3.6[/tex]

= [tex]0.80 \ m[/tex]

Speed,

= 344 m/s

For constructive interference,

⇒  [tex]Path \ difference =n\times \frac{Speed}{frequency}[/tex]

On substituting the values, we get

⇒  [tex]0.80=n\times \frac{344}{frequency}[/tex]

⇒  [tex]frequency=n\times 430[/tex]

⇒                [tex]n=\frac{frequency}{430}[/tex]

If the frequency range is,

f = 2193,

⇒  [tex]n=\frac{2193}{430}[/tex]

        [tex]=5.1[/tex]

If the frequency range is,

f = 2967,

⇒  [tex]n=\frac{2967}{430}[/tex]

        [tex]=6.9[/tex]

hence,

For n = 6, the frequency will be:

⇒  [tex]Frequency=n\times 430[/tex]

                       [tex]=6\times 430[/tex]

                       [tex]=2580 \ Hz[/tex]

An old shade-tree mechanic trick for removing a stubborn bolt is to slip a long pipe over the handle of the wrench, then apply a force to the end of the pipe opposite the bolt.

Why is this effective? What problems might this technique cause?

Answers

Answer: long pipe is used to increase torque and reduce force needed.

Explanation: A torque is needed to open bolt. torque = F·r, if

R increases, Force F needed to open bold is smaller.

Problem is a worn bolt may break down. It sometimes I send useful to heat bolt instead using too much power.

assume the femur has a diameter of 3.0cm and a hollow center of 0.8cm diameter and a length of 50cm.if it is supporting a load of 600N/,what is the stress in the femur? How much will it be shortened by this load?Take the young's modulus for the femur bone as 16 × 10^9N/m²

Answers

Answer:

tbh i think hes right

Explanation:

The femur would be shortened by 0.0434 cm by a load of 600N.

What is pressure?

The total applied force per unit of area is known as the pressure.

The pressure depends both on externally applied force as well the area on which it is applied.

The mathematical expression for the pressure

Pressure = Force /Area

As given in the problem assume the femur has a diameter of 3.0cm and a hollow center of 0.8cm diameter and a length of 50cm.if it is supporting a load of 600N,

The stress in the femur = force /area

                                       =600 / π(.004²-.0015²)

                                        = 13.89 ×10⁶ Pa

The length reduced by the load = PL/AE

                                                      = 13.89 ×10⁶ ×50/ 16 × 10⁹ cm

                                                       =0.0434 cm

The femur would be shortened by 0.0434 cm by a load of 600N.

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BRAINLIEST PLEASE HELP ASAP DO NOT ANSWER UNLESS YOU KNOW AND DON'T GIVE ME A LINK OR I WILL REPORT YOU
1. Explain what type of lever is a baseball bat and why the bat is that type of lever in relation to the input force, output force, and fulcrum.
2. Explain for which class of lever is the output force always greater than the input force.
3. Explain for which class of lever is the output force always less than the input force.
4. Use the law of conservation of energy to explain why in a second class lever the distance over which the input force is applied is always greater than the distance over which the output force is applied.
5. A lever has an IMA of 4. If the length of the length of the input arm is 1.0 meter, what is the length of the output arm.

Answers

Answer:

In a third class lever, the effort is located between the load and the fulcrum. ... If the fulcrum is closer to the effort, then the load will move a greater distance. A pair of tweezers, swinging a baseball bat or using your arm to lift something are examples of third class levers.

Explanation:

what can i use to curl my eyelashes other den mascara and eyelash curler​

Answers

Explanation:

a hot toothbrush. ......lol

Spoon, heat a spoon in a mug of warm water then press the curved side over your lid and gently press lashes against the curved edge of the spoon for 10 secs.

Charge of uniform density (40 pC/m2) is distributed on a spherical surface (radius = 1.0 cm), and a second concentric spherical surface (radius = 3.0 cm) carries a uniform charge density of 60 pC/m2. What is the magnitude of the electric field at a point 2.0 cm from the center of the two surfaces?

Answers

Answer:

E = K Q / R^2     by Gauss' Law where Q is the  charge enclosed by the surface of revolution and R is the distance from the enclosed charge

Since Q = d 4 pi r^2   where d is the charge density and r the radius of the inner sphere

E = K / R^2 * (4 d pi r^2) = 4 K pi d (r / R)^2 = 4 K pi d * 1/4

E =  9 * 10E9 * 3.14 * 40 * 10E-12 = 1.13 N / C

In what way would a digital thermometer be preferable to those from a liquid-based thermometer?

Answers

Reading a digital thermometer is more preferable because you can just easily read the measurement from a screen, a number will just show in a screen as compared to a liquid-based thermometer where you still have to read the measurement by counting the markers in the thermometer. Hope this helps! Mark brainly please!

A uniform metre rule of mass 100g balance the 40cm mark when a mass x is placed at the 20cm mark
what is the value of x​

Answers

Answer:

X = 50 g

Explanation:

Please see attached photo for explanation.

From the attached photo,

Anticlock–wise moment = X × 20

Clockwise moment = 100 × 10

Anticlock–wise moment = clockwise moment

X × 20 = 100 × 10

X × 20 = 1000

Divide both side by 20

X = 1000 / 20

X = 50 g

Therefore, the value of X is 50 g

For your answer to this problem, just type in the numerical magnitude of the momentum - no units.

An object with a mass M and a velocity v has a momentum of 15 kg•m/s. An object with a mass of 2M and 4v would have a momentum of kg•m/s

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Answer:

120 kg•m/s.

Explanation:

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

Case 1

Mass of object = M

Velocity of object = V

Momentum = 15 kg•m/s

Case 2

Mass of object = 2M

Velocity of object = 4V

Momentum = ?

Momentum is defined as follow:

Momentum = mass × velocity

The momentum of object in case 2 can be obtained as follow:

From case 1

Momentum = mass × velocity

15 = M × V

15 = MV ....... (1)

From case 2:

Momentum = mass × velocity

Momentum = 2M × 4V

Momentum = 8MV ....... (2)

Finally , substitute the value of MV in equation 1 into equation 2.

Momentum = 8MV

MV = 15

Momentum = 8 × 15

Momentum = 120 kg•m/s

Therefore, an object with a mass of 2M and 4V would have a momentum of 120 kg•m/s

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