how would you define an ocean current?

A Stone Is Dropped Into a Deep Water Well. The Sound of The Stone Hitting The Water Is Heard After 3.4 Seconds. then The Depth of The Water Well is 56.6 m.

In terms of physics, sound is a vibration that travels through a transmission medium like a gas, liquid, or solid as an acoustic wave. Sound is the receipt of these waves and the brain's perception of them in terms of human physiology and psychology. Only acoustic waves with frequencies between about 20 Hz and 20 kHz, or the audio frequency range, may cause a human to have an auditory sensation. These correspond to sound waves in air with an atmospheric pressure of 17 metres (56 ft) to 1.7 centimetres (0.67 in) in wavelength. Ultrasounds are sound waves with a frequency higher than 20 kHz that are inaudible to humans. Infrasound refers to sound frequencies below 20 Hz. Animals of different species have different hearing ranges. Acceleration of the stone is 9.8 m/s²

according to kinematics,

s = ut + 1/2 at²

s =  1/2 ×9.8×3.4²

s = 56.6 m

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Answer:

Tarzan will be moving at 7.4 m/s.

Explanation:

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

Height (h) of cliff = 2.8 m

Initial velocity (u) = 0 m/s

Final velocity (v) =?

NOTE: Acceleration due to gravity (g) = 9.8 m/s²

Finally, we shall determine how fast (i.e final velocity) Tarzan will be moving at the bottom. This can be obtained as follow:

v² = u² + 2gh

v² = 0² + (2 × 9.8 × 2.8)

v² = 0 + 54.88

v² = 54.88

Take the square root of both side

v = √54.88

v = 7.4 m/s

Therefore, Tarzan will be moving at 7.4 m/s at the bottom.

When a mass (m) that orbits the earth in a circular path of radius 7480 km (about 1100 km) above the surface of the earth), the period of revolution of the given satellite is approximately 8207 seconds or 2.28 hours.

The period of revolution of a satellite with mass (m) that orbits the earth in a circular path of radius 7480 km (about 1100 km) above the surface of the earth) can be determined by using Kepler's third law which relates the period of revolution of a satellite to the average radius of its orbit.

Kepler's third law states that the square of the period of revolution of a satellite is proportional to the cube of the average radius of its orbit.

Mathematically, the law can be expressed as: T² = (4π² / GM) × R³Where T is the period of revolution, G is the gravitational constant, M is the mass of the earth, and R is the average radius of the orbit of the satellite.

To find the period of revolution of the given satellite, we can substitute the given values in the equation: R = 7480 km + 6370 km = 13850 kmM = 5.97 × 10²⁴ kgG = 6.67 × 10⁻¹¹ Nm²/kg²T² = (4π² / GM) × R³T² = (4π² / (6.67 × 10⁻¹¹ × 5.97 × 10²⁴)) × (13850 × 10³)³T² = 6.7182 × 10¹⁴ seconds²

Taking the square root of both sides, we get:T = 8.2079 × 10³ seconds

Therefore, the period of revolution of the given satellite is approximately 8207 seconds or 2.28 hours.

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Explanation:

E1 = k Q1 / (0.075 m)^2

= 9 x 10^9 Nm^2/C^2 * 10 x 10^-9 C / 0.075^2 m^2

= 400 N/C

E2 = k Q2 / (0.075 m)^2

= 9 x 10^9 Nm^2/C^2 * (-3 x 10^-9 C) / 0.075^2 m^2

= -120 N/C

Since the electric fields due to the two charges are in opposite directions, we can find the net electric field at the midway point by taking the difference between the two:

E = E1 - E2

= 400 N/C - (-120 N/C)

= 520 N/C

The dragonfly must generate approximately 4.8 × 10^-4 Joules of energy to spin itself at a rate of 1100 rpm.

Start by converting the rotational speed from rpm (revolutions per minute) to rad/s (radians per second). Since 1 revolution is equal to 2π radians, we can use the conversion factor:

Angular speed (ω) = (1100 rpm) × (2π rad/1 min) × (1 min/60 s)

ω ≈ 115.28 rad/s

The moment of inertia (I) is given as 2.0 × 10^-7 kg⋅m².

Use the formula for rotational kinetic energy:

Rotational Kinetic Energy (KE_rot) = (1/2) I ω²

Substituting the given values:

KE_rot = (1/2) × (2.0 × 10^-7 kg⋅m²) × (115.28 rad/s)²

Calculate the value inside the parentheses:

KE_rot ≈ (1/2) × (2.0 × 10^-7 kg⋅m²) × (13274.28 rad²/s²)

KE_rot ≈ 1.331 × 10^-3 J

Round the result to the proper number of significant figures, which in this case is three, as indicated by the given moment of inertia.

KE_rot ≈ 4.8 × 10^-4 J

Therefore, the dragonfly must generate approximately 4.8 × 10^-4 Joules of energy to spin itself at a rate of 1100 rpm.

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Answer:

We can start by squaring both sides of the equation:

|11i - kj|^2 = |√5(3i+j)|^2

Simplifying the right side:

(√5(3i+j)) * (√5(3i+j)) = 5(3i+j)(3i+j) = 45i^2 + 30ij + 45j^2 = 45(i^2 + j^2)

Now we can expand the left side using the formula |a-b|^2 = (a-b)(a-b*) where b* is the complex conjugate of b:

(11i - kj)(11i - kj*) = 121i^2 -11kij* -11kij + k^2j^2

Simplifying and combining like terms:

-11kij* -11kij = -22kiIm(j)

So the equation becomes:

121i^2 + k^2j^2 - 22kiIm(j) = 45(i^2 + j^2)

Substituting i^2 = -1 and simplifying:

-121 + k^2j^2 + 22kiIm(j) = 45 - 45j^2

k^2j^2 + 22kiIm(j) - 166 = 0

This is a quadratic equation in k. Using the quadratic formula:

k = (-22ijIm(j) ± √[(22ijIm(j))^2 - 4j^2(166)]) / 2j^2

Simplifying:

k = -11iIm(j) ± √[484i^2m(j)^2 - 664j^2]) / 2j

Note that the expression inside the square root must be non-negative for real solutions, which means that:

484i^2m(j)^2 - 664j^2 ≥ 0

484m(j)^2 ≥ 664

m(j) ≥ √(166/121)

Therefore, the possible values for k are:

k = -11iIm(j) + √[484i^2m(j)^2 - 664j^2] / 2j, if m(j) ≥ √(166/121)

k = -11iIm(j) - √[484i^2m(j)^2 - 664j^2] / 2j, if m(j) ≥ √(166/121)

The possible values for k is ±11.31 for the equation  |11i - kj| = |√5(3i+j)|.

It is the pair of binomials with identical operations but with opposite arithmetic, sign is called conjugate pairs.

|11i - kj| = |√5(3i+j)|

Solving the right-hand side equations:

|√5(3i+j)|        (in order to remove modulus, square the given numbers)

|√5(3i+j)| = (√5(3i+j)×√5(3i+j))

              = ((√5)²(3² + 1²))

              = √50

             = 7.07

On solving the right-hand side, we get 7.07.

The left-hand side :

|11i - kj|    (multiply with conjugate pairs)

(11i - kj)(11i+jk) = (11²i² + 11i²jk -11i²kj-k²j²)

                      = 121i² - k²j²

121i² + k²j² = 7.07     (substitute i² = -1)

k²j² -121 = 7.07

k²j² = 7.07+121

k = √(128.07)

k = ± 11.31

The possible values of k are ±11.31.

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The final velocity of the truck is determined as 13.2 m/s.

The final displacement of the truck is determined as 183.6 m.

The velocity of the truck after 12 seconds is calculated as follows;

v = u + at

where;

v is the final velocity of the truck after 12 seconds

u is the initial velocity of the truck

t is the time of motion

a is the acceleration of the truck

v = 0 + 2.3(12)

v = 27.6 m/s

The velocity of the truck after 9 seconds is calculated as follows;

v = u + at

where;

v = 27.6 m/s   +  (-1.6)(9)

v = 13.2 m/s

v² = u² + 2as

13.2² = 27.6² + 2(-1.6)s

174.24 = 761.76 - 3.2s

3.2s = 761.76 - 174.24

3.2s = 587.52

s = 587.52/3.2

s = 183.6 m

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Answer:

1. M = 67,422,800,892,977.54 kg

2. r = 15 km

Explanation:

The diameter of the Comet Tempel 1, D = 9.0 km across

The speed with which the dust escapes = 1.0 m/s

1. The escape velocity, \(v_e\), is given by the following formula

\(v_e = \sqrt{\dfrac{2 \cdot G \cdot M}{R} }\)

Where;

G = The universal gravitational constant = 6.67430 × 10⁻¹¹ N·m²/kg²

\(v_e\) = The escape velocity of the debris = 1.0 m/s

M = The mass of the comet from where the debris escapes

From the escape velocity equation, we have;

\(M = \dfrac{v_e^2 \cdot R}{2 \cdot G}\)

Plugging in the values for the variables, we get the mass of the comet, 'M', as follows;

\(M = \dfrac{1.0^2 \times 9,000}{2 \cdot 6.67430 \times 10^{-11}} \approx 67,422,800,892,977.54 \, kg\)

The mass of the comet, M ≈ 67,422,800,892,977.54 kg

2. When the debris has lost 60% of its initial kinetic energy, we have;

\(60\% \, K.E. = 0.6\cdot K.E. = 0.6 \times \dfrac{1}{2} \times m \times v_e^2 = \dfrac{G \cdot M \cdot m}{r}\)

\(\therefore \, The \ distance \ of \ debris \ from \ the \ center, \, r = \dfrac{G \cdot M }{0.6 \times \dfrac{1}{2} \times v_e^2 }\)

\(r = \dfrac{6.67430 \times 10^{-11} \times 67,422,800,892,977.54}{0.6 \times \dfrac{1}{2} \times 1^2 } = 15,000\)

When the debris has lost 60% of its initial kinetic energy, the distance the debris will be from the comet's center, r = 15,000 m = 15 km

Answer:  21.4 m/s²

Explanation:

avg. accel. = (Vfinal - Vinitial) / Δt

Vfinal = 15 m/s

Vinitial = 0 m/s

Δt = 0.70 s

avg. accel. = (15 m/s - 0 m/s)/0.7 s = 21.4 m/s²

Two forces acting on the rocket immediately after leaving the launching pad are the gravitational force and the thrust force.

1. Gravitational Force: The gravitational force is the force exerted by the Earth on the rocket due to their mutual gravitational attraction. It acts downward and is responsible for the rocket's weight.

This force can be represented by the equation Fg = mg, where Fg is the gravitational force, m is the mass of the rocket, and g is the acceleration due to gravity. The gravitational force acts to pull the rocket downward, opposing its upward motion.

2. Thrust Force: The thrust force is the force generated by the rocket's engines as they expel exhaust gases in the opposite direction. It acts upward and propels the rocket forward.

The magnitude of the thrust force depends on factors such as the design of the rocket engines, the amount of fuel burned, and the rate of exhaust gas expulsion. The thrust force must be greater than or equal to the gravitational force for the rocket to overcome Earth's gravity and achieve upward acceleration.

Initially, when the rocket is launched, the thrust force is at its maximum while the gravitational force remains constant. As the rocket gains altitude, the gravitational force decreases slightly due to the increasing distance from the Earth's center.

However, the thrust force continues to be the dominant force propelling the rocket upward.

It's important to note that other forces such as air resistance and wind may also act on the rocket, but immediately after leaving the launching pad, these forces are typically negligible compared to the gravitational force and thrust force.

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The gravitational force between two students is 5.336*10^-8 N.

According to universal gravitational law, the force acting on two bodies is given by the formula = F = (G *m1*m2)/r^2

Here mass of one student =m1=75kg,another student m2=54kg

Distance of separation =0.45m, r =0.45/2=0.225m

Force = 6.67*10^-11 (75*54)/(0.225)^2

F=5.336*10^-8 N

The force between two students is 5.336*10^-8 N.

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The subject depicted in the attached image is Astronomy and Astrophysics.

Astronomy is the scientific study of celestial objects such as stars, planets, galaxies, and other phenomena that exist outside of Earth's atmosphere.

Astronomers use a variety of methods to observe and study these objects, including telescopes, spacecraft, and computer simulations.

Astronomy is a broad field that includes many different sub-disciplines, such as astrophysics, planetary science, and cosmology.

Astronomers study the physical properties and behavior of celestial objects, such as their composition, temperature, motion, and evolution.

They also seek to understand the structure and history of the universe as a whole.

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The speed of skier's when he reaches the top plateau is 5.26 m/s, and the speed of skier's when he reaches the upper level with a frictional force of 75 N is 4.23 m/s.

We can use the conservation of mechanical energy to solve this problem. Initially, the skier has kinetic energy, and at the top of the slope, he will have both potential and kinetic energy.

Since friction is negligible, the only force acting on the skier is the force of gravity. The work done by this force will be equal to the change in the skier's potential energy as he climbs up the slope. Therefore;

mgh = (1/2)mv²

where m = 104 kg is mass of the skier, g = 9.8 m/s² is the acceleration due to gravity, h = 1.2 m is height of the slope, and v is the skier's velocity when he reaches the top.

Solving for v, we get;

v = √(2gh) = √(29.81.2) = 5.26 m/s

Therefore, the skier's speed when he reaches the top plateau is 5.26 m/s.

In this case, there is also a frictional force acting on the skier, which does negative work on the skier as he moves up the slope. The work done by the frictional force is equal to the force of friction multiplied by the distance traveled;

W = Fd = μmgd

where μ = F/N is the coefficient of kinetic friction, N is the normal force acting on the skier (equal to the skier's weight), and d is the distance traveled up the slope (equal to the height of the slope, 1.2 m).

The net work done on the skier will be equal to the change in his mechanical energy;

Wnet = ΔK + ΔU = (1/2)m\(V_{f}\)² - (1/2)m\(V_{i}\)² + mgh

where vi = 8.7 m/s is the skier's initial velocity, \(V_{f}\) is his final velocity at the top of the slope, and ΔK and ΔU are the changes in kinetic and potential energy, respectively.

Since the net work done on the skier is equal to the work done by the gravitational force minus the work done by the frictional force, we have;

Wnet = mgh - μmgd

Substituting the expressions for Wnet and mgh, we get:

(1/2)m\(V_{f}\)² - (1/2)m\(V_{i}\)² = μmgd

Solving for \(V_{f}\), we get:

\(V_{f}\) = √(\(V_{i}\)² + 2μgd) = √(8.7² + 20.729.8×1.2) = 4.23 m/s

Therefore, the skier's speed when he reaches the upper level with a frictional force of 75 N is 4.23 m/s.

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Answer:

F = 3062.5  N

Explanation:

Given that,

The mass of a sled, m = 35 kg

The speed increase from 30 m/s to 65 m/s in 0.4 seconds.

We need to find the force needed to accelerate the car.

Net force is given by :

F = ma

where

a is acceleration of the car.

\(F=\dfrac{m(v-u)}{t}\\\\F=\dfrac{35\times (65-30)}{0.4}\\\\F=3062.5\ N\)

So, the net force is 3062.5  N.

Answer: Conversation of Matter

Explanation: According to Quizlet, Converstion of Matter is the answer.

Answer:it’s A. right for ape x

Explanation:

Based on data, digital waves are more reliable because they are less likely to change when copied (Option A).

A wave is a periodic distortion capable of traveling through a suitable media (either air or water).

In conclusion, based on data, digital waves are more reliable because they are less likely to change when copied (Option A).

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a) The magnitude will be 0.838m

b) The displacement will be -17.35°

The path covered by an object from its initial point to final point.

Forces acting on the duck

x-axis:    0.13 + 0.16*cos(-56°) = 2.7 * ax  

ax = 0.0813 m/s^2

y-axis:    0.13*sin(-56°) = 2.7 * ay  

ay = -0.0491 m/s^2

The displacement on the x-axis

X = Vox * t + ax/2 * t²

X = 0.12* 3.2 + 0.0813/2*3.2²

X = 0.8

The displacement on the y-axis:

Y = Voy * t + ay/2 * t²

X = 0 - 0.0491/2*3.2²

Y=-0.25m

So, the magnitude and angle of this displacement [0.8,-0.25] is:

0.838m  at an angle of -17.35°

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The centripetal force in Newtons provided by the friction between the blade of the skate and the ice is 490 N

The following data were obtained from the question:

The centripetal force can be obtained as illustrated below:

F = mv²/r

= (50 × 7²) / 5

= (50 × 49) / 5

= 2450 / 5

= 490 N

Thus, we can concluded that the centripetal force is 490 N

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Answer:

From the question we are told that

  The length of the rod is  \(L_o\)

    The  speed is  v  

     The angle made by the rod is  \(\theta\)

Generally the x-component of the rod's length is  

     \(L_x =  L_o cos (\theta )\)

Generally the length of the rod along the x-axis  as seen by the observer, is mathematically defined by the theory of  relativity as

       \(L_xo  =  L_x  \sqrt{1  - \frac{v^2}{c^2} }\)

=>     \(L_xo  =  [L_o cos (\theta )]  \sqrt{1  - \frac{v^2}{c^2} }\)

Generally the y-component of the rods length  is mathematically represented as

      \(L_y  =  L_o  sin (\theta)\)

Generally the length of the rod along the y-axis  as seen by the observer, is   also equivalent to the actual  length of the rod along the y-axis i.e \(L_y \)

    Generally the resultant length of the rod as seen by the observer is mathematically represented as

     \(L_r  =  \sqrt{ L_{xo} ^2 + L_y^2}\)

=>  \(L_r  = \sqrt{[ (L_o cos(\theta) [\sqrt{1 - \frac{v^2}{c^2} }\ \ ]^2+ L_o sin(\theta )^2)}\)

=>  \(L_r= \sqrt{ (L_o cos(\theta)^2 * [ \sqrt{1 - \frac{v^2}{c^2} } ]^2 + (L_o sin(\theta))^2}\)

=>   \(L_r  = \sqrt{(L_o cos(\theta) ^2 [1 - \frac{v^2}{c^2} ] +(L_o sin(\theta))^2}\)

=> \(L_r =  \sqrt{L_o^2 * cos^2(\theta)  [1 - \frac{v^2 }{c^2} ]+ L_o^2 * sin(\theta)^2}\)

=> \(L_r  =  \sqrt{ [cos^2\theta +sin^2\theta ]- \frac{v^2 }{c^2}cos^2 \theta }\)

=> \(L_o \sqrt{1 - \frac{v^2}{c^2 } cos^2(\theta ) }\)

Hence the length of the rod as measured by a stationary observer is

       \( L_r = L_o \sqrt{1 - \frac{v^2}{c^2 } cos^2(\theta ) }\)

   Generally the angle made is mathematically represented

\(tan(\theta) =  \frac{L_y}{L_x}\)

=>  \(tan {\theta } =  \frac{L_o sin(\theta )}{ (L_o cos(\theta ))\sqrt{ 1 -\frac{v^2}{c^2} } }\)

=> \(tan(\theta ) =  \frac{tan\theta}{\sqrt{1 - \frac{v^2}{c^2} } }\)

Explanation:

The special relativity relations allow to find the results for the questions about the measurements made by an observed at rest on the rod are:

     a) The length of the rod is: \(L = L_o \sqrt{1 - \frac{v^2}{c^2} \ cos^2\theta_o }\)  

     b) The angle with respect to the x axis is:   \(tan \theta = \frac{tan \theta_o}{\sqrt{1- \frac{v^2}{c^2} } }\)

Special relativity studies the motion of bodies with speeds close to the speed of light, with two fundamental assumptions.

If we assume that the two systems move in the x-axis, the relationship between the components of the length are:

        \(L_x = L_{ox} \ \sqrt{1- \frac{v^2}{c^2} }\)  

        \(L_y = L_o_y \\L_z = L_{oz}\)  

Where the subscript "o" is used for the fixed observed on the rod, that is, it is at rest with respect to the body, v and c are the speed of the system and light, respectively.

a) They indicate that the length of the rod is L₀ and it forms an angle θ with the horizontal.

Let's use trigonometry to find the components of the length of the rod in the system at rest, with respect to it.

         sin θ = \(\frac{L_{oy}}{L_o}\)  

         cos θ = \(\frac{L_{ox}}{L_o}\)  

         \(L_{oy}\) = L₀ sin θ

         L₀ₓ = L₀ cos θ

Let us use the transformation relations of the length of the special relativity rod.

x-axis

      \(L_x = (L_o cos \theta_o) \ \sqrt{1- \frac{v^2}{c^2} }\)  

y-axis  

      \(L_y = L_{o} sin \theta_o\)  

The length of the rod with respect to the observer using the Pythagorean theorem is:

      L² = \(L_x^2 + L_y^2\)  

      \(L^2 = (L_o cos \theta_o\sqrt{1- \frac{v^2}{c^2} })^2 + (L_o sin \theta_o)^2\)

      \(L_2 = L_o^2 ( cos^2 \theta_o - cos^2 \theta_o \frac{v^2}{c^2} + sin^2\theta_o)\)  

      \(L^2 = L_o^2 ( 1 - \frac{v^2}{c^2} \ cos^2 \theta_o)\)  

      \(L= Lo \sqrt{1- \frac{v^2}{c^2} cos^2 \theta_o}\)  

b) the angle with the x-axis measured by the stationary observer is:

     \(tna \theta = \frac{L_y}{L_x}\)

    \(tan \ theta = \frac{L_o sin \theta_o}{L_o cos \theta_o \sqrt{1- \frac{v^2}{c^2} } }\)  

    \(tan \theta = \frac{tan \theta_o}{\sqrt{1-\frac{v^2}{c^2} } }\)  

In conclusion, using the special relativity relations we can find the results for the questions about the measurements made by an observed at rest on the rod are:

      a) The length of the rod is:  \(L = L_o \sqrt{1- \frac{v^2}{c^2} \ cos^2\theta_o }\)

      b) The angle to the x axis is:  \(tan \theta = \frac{tan \theta_o}{\sqrt{1- \frac{v^2}{c^2} } }\)

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The mass of the proton at the given speed is 2.78 x 10⁻²⁷ kg.

The mass of the proton at the given speed of is calculated by applying the following equation.

Mathematically, the equation relating the mass of the proton and the rest mass is calculated as follows;

m = ( M₀ ) / (√ ( 1 - v² / c² )

where;

The given parameters include;

The mass of the proton at the given speed is calculated as follows;

m = ( M₀ ) / (√ ( 1 - v² / c² )

m = ( M₀ ) / (√ ( 1 - (0.8c)² / c² )

m = M₀ / 0.6

m =  1.67 x 10⁻²⁷ kg / 0.6

m = 2.78 x 10⁻²⁷ kg

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Answer:(vi^2+d)^2=vf+a^2+2t

Explanation:

The apparatus that can used to measure the mass of the wooden block​ by the student is called beam balance.

A beam balance, often referred to as a double-pan balance, is a straightforward tool for determining an object's weight. Two pans or trays are hung from either end of a horizontal beam that is suspended from a pivot point in the middle.

The thing to be weighed is put on one tray, and then the second tray is filled with standard weights until it balances, showing the weight of the object. From little ones used in laboratories to larger ones used in enterprises, beam balances can be found in a variety of shapes and sizes. Because they are precise and operate without electricity or batteries, they are widely used.

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Answer:

gamma-rays

Explanation:

The final velocity of the red ball is 18 m/s.

The term momentum has to do with the product of the mass and the velocity of an object We know that the momentum is always conserved in accordance with the Newton third law. Also it is clear that the momentum before collision is equal to the total momentum after collision and we are going to apply this principle here.

Then;

Mass of the blue ball =  6 kg

Mass of the red ball =  1 kg

Initial velocity of the blue ball =  4 m/s

Initial velocity of the red ball = 0 m/s

Final velocity of the red ball = ??

Final velocity of the blue ball =  1 m/s

We now have;

(6 * 4) + (1 * 0) = (1 * v) + (6 * 1)

24 = v + 6

v = 24 - 6

v = 18 m/s

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Ellie's experience of a gender identity that doesn't align with her body is likely linked to gender dysphoria, a condition characterized by distress caused by the incongruence between one's internal sense of gender and assigned sex at birth.

The correct answer is option B.

Gender dysphoria refers to the distress or discomfort individuals may experience when their gender identity does not match the sex they were assigned at birth. It involves a deep-rooted sense of incongruence between one's internal sense of gender and the external physical characteristics.

It is important to note that gender dysphoria is not a result of misinformation or a lack of understanding. It is a genuine psychological condition recognized by medical and mental health professionals. Individuals with gender dysphoria often experience significant distress and may seek gender-affirming interventions, such as hormone therapy or gender-affirming surgeries, to align their physical appearance with their gender identity.

Misinformation, on the other hand, refers to inaccurate or misleading information, which may not directly relate to Ellie's experience. Construed body image and positive body image are also not directly linked to Ellie's situation. Construed body image refers to the way individuals perceive their own bodies, which may be influenced by various factors, while positive body image refers to a healthy and accepting attitude toward one's own body.

In summary, Ellie's experience of a gender identity that doesn't align with her body is likely linked to gender dysphoria. Gender dysphoria involves distress or discomfort resulting from a mismatch between an individual's internal sense of gender and their assigned sex at birth.

Therefore, among the options provided the correct answer is option B.

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Answer:

B. Chemical energy

Explanation:

chemical energy is the energy stored in the bonds between atoms

When filled with extra air its mass increased to 0.428kg also the top of polythene container mass connected to the perplex box of volume 1000cm³ and the number of times of air inside was 7.2 times. The density of the air filled in the polythene container is approximately 1.25 kg/m³.

The density of air filled in the polythene container can be determined by considering the change in mass and volume of the container before and after filling it with air. Given that the mass of the empty container is 0.419 kg and the mass of the container when filled with extra air is 0.428 kg, and the volume of the perplex box is 1000 cm³.

Calculate the mass of the air inside the container by subtracting the mass of the empty container from the mass of the container when filled with air:

Mass of air = Mass of filled container - Mass of empty container

= 0.428 kg - 0.419 kg

= 0.009 kg

Calculate the volume of the air inside the container using the given number of times the air inside is 7.2:

Volume of air = Volume of perplex box * Number of times air inside

= 1000 cm³ * 7.2

= 7200 cm³

Convert the volume of air to cubic meters (m³) by dividing by 1000000:

Volume of air = 7200 cm³ / 1000000

= 0.0072 m³

Calculate the density of air using the formula:

Density = Mass / Volume

Density = 0.009 kg / 0.0072 m³

≈ 1.25 kg/m³

Therefore, the density of the air filled in the polythene container is approximately 1.25 kg/m³.

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Answer:

a) 30.65

b) 71.51

Explanation:

Answer:

A. h = 2.15 m

B. Pb' = 122 KPa

Explanation:

The computation is shown below:

a)  Let us assume the depth be h

As we know that

\(Pb - Pat = d \times g \times h \\\\ ( 119 - 101) \times 10^3 = 856 \times 9.8 \times h\)

After solving this,  

h = 2.15 m

Therefore the depth of the fluid is 2.15 m

b)

Given that  

height of the extra fluid is

\(h' = \frac{2.35 \times 10^{-3}}{ area} \\\\ h' = \frac{2.35 \times 10^{-3}} { 66.2 \times 10^{-4}}\)

h' = 0.355 m

Now let us assume the pressure at the bottom is Pb'

so, the equation would be

\(Pb' - Pat = d \times g \times (h + h')\\\\Pb' = 856 \times 9.8 \times ( 2.15 + 0.355) + 101000\)

Pb' = 122 KPa

(A)  The depth of the fluid is 2.14 m.

(B)  The new pressure at the bottom of container is 121972 Pa.

Given data:

The cross-sectional area of the container is, \(A =66.2 \;\rm cm^{2}=66.2 \times 10^{-4} \;\rm m^{2}\).

The density of fluid is, \(\rho = 856 \;\rm kg/m^{3}\).

The container pressure at bottom is, \(P=119 \;\rm kPa=119 \times 10^{3} \;\rm Pa\).

The atmospheric pressure is, \(P_{at}=101 \;\rm kPa=101 \times 10^{3}\;\rm Pa\).

(A)

The given problem is based on the net pressure on the container, which is equal to the difference between the pressure at the bottom and the atmospheric pressure. Then the expression is,

\(P_{net} = P-P_{at}\\\\\rho \times g \times h= P-P_{at}\)

Here, h is the depth of fluid.

Solving as,

\(856\times 9.8 \times h= (119-101) \times 10^{3}\\\\h=\dfrac{ (119-101) \times 10^{3}}{856\times 9.8}\\\\h= 2.14 \;\rm m\)

Thus, the depth of the fluid is 2.14 m.

(B)

For an additional volume of \(2.35 \times 10^{-3} \;\rm m^{3}\) to the liquid, the new depth is,

\(V=A \times h'\\\\h'=\dfrac{2.35 \times 10^{-3}}{66.2 \times 10^{-4}}\\\\h'=0.36 \;\rm m\)

Now, calculate the new pressure at the bottom of the container as,

\(P'-P_{at}= \rho \times g \times (h+h')\\\\\P'-(101 \times 10^{3})= 856 \times 9.8 \times (2.14+0.36)\\\\P'=121972 \;\rm Pa\)

Thus, we can conclude that the new pressure at the bottom of container is 121972 Pa.

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