a chunk of ice ( at = -20°c) is added to a thermally insulated container of cold water ( t = 0°c). what happens in the container?

The gravitation force with which the earth is being pulled can be determined by applying Newton's law of universal gravitation.

According Newton's law of universal gravitation, the force exerted between two objects in the universe is directly proportional to the product of masses of the two objects and inversely proportional to the square of the distance between the two objects.

Mathematically, the formula for gravitation force is given as;

F = GmM/R²

where;

If the mass of the object is know and the distance between earth  and the object is also known, the force with which the earth is being pulled can be calculated by applying Newton's law of universal gravitation as shown in the above equation.

Thus, the force with which the earth is being pulled can calculated as well.

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At a depth of 50 m, the pressure is 512 kPa (kilopascals), a diver breathing air (78% \(N_2\), 21% \(O_2\)) at this depth would have a partial pressure of oxygen in their lungs of 124.338 kPa.

The absolute pressure , is 50 meters x 9.81 \(m/s^2\) x 1 Pa/m = 490.5 kPa. 

Total pressure = atmospheric pressure + water pressure

total pressure = 101.3 kPa + 490.5 kPa

total pressure = 591.8 kPa

partial pressure of oxygen in the air mixture, first the partial pressure of nitrogen is found out, nitrogen makes up 78% of the air mixture, the partial pressure of nitrogen is:

partial pressure of nitrogen = total pressure x 0.78

partial pressure of nitrogen= 591.8 kPa x 0.78

partial pressure of nitrogen = 461.724 kPa

Similarly, the partial pressure of oxygen is:

partial pressure of oxygen = total pressure x 0.21

partial pressure of oxygen = 591.8 kPa x 0.21

partial pressure of oxygen = 124.338 kPa

Hence, a diver breathing air (78% \(N_2\), 21% \(O_2\)) at this depth would have a partial pressure of oxygen in their lungs of 124.338 kPa.

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Answer:

108 KPa

Explanation:

Answer:

a) 3.6 sec

b) 17.7 m/s

c) 17.7 m/s

d) 63.6 m

Explanation:

y = (25sin45)t - 4.9t^2 =17.7t - 4.9t^2

a) to find hang time set y = 0

4.9t^2 = 17.7t or t = 3.6 sec

b) vx = 25cos45 = 17.7 m/s

c) vy = 25sin45 = 17.7 m/s

d) To find the range R, use the results in (a) and (b) into the equation R = vxt:

R = (17.7 m/s)(3.6 sec) = 63.6 m

Answer:

The movement of a 2000-kg truck travelling at 19 m/s is 38,000 kgm/s.

The first reason is that in the past humans didn't necessarily see the connections between the oceans and thus labelled them that way separating them into 5 oceans and the names stuck to this day.

Secondly, while all the oceans of the world constitute one large Ocean, people need some way to identify each part of it since it is easier to think when we have different categories instead of one large category.

Hope that helped!

Answer:

At point A;

GPE = 58800 J

KE = 0 J

At point B;

GPE = 44100 J

KE = 14700 J

At point C;

GPE = 0 J

KE = 58800 J

At point D;

GPE = 29400 J

KE = 29400 J

EPE = 29400 J.

Explanation:

The Gravitational Potential Energy, GPE, is given as follows;

GPE = The mass. m × The acceleration due to gravity, g × The elevation or height, h

∴ GPE = m × g × h

The Kinetic Energy, KE = 1/2  × The mass × The square of the velocity, v²

KE = 1/2 × m × v²

At point A;

The parameters are;

h = 100m, v = 0 (Starting point), Cart mass, m = 60 kg, we have;

GPE = m × g × h = 60 × 9.8 × 100 = 58800

GPE = 58800 J

KE = 1/2 × m × v² = 1/2 × 60 × 0² = 0

KE = 0 J

Mechanical Energy, ME = GPE + KE = 58800 + 0 = 58800 = Constant

ME = 58800 J

At point B;

GPE = 60 × 9.8 × 75 = 44100

GPE = 44100 J

KE = ME - GPE = 58800 - 44100 = 14700

KE = 14700 J

At point C;

GPE = 60 × 9.8 × 0 = 0 J

KE = ME - GPE = 58800 - 0 = 58800

KE = 58800 J

At point D;

v = 0, the Cart is brought to rest by the spring

GPE = 60 × 9.8 × 50 = 29400

GPE = 29400 J

KE = ME - GPE = 58800 - 29400 = 29400

KE = 29400 J

The Elastic Potential Energy EPE gained by the spring = KE = 29400

EPE = 29400 J.

  The centre of mass of the system is the point at which the total

  mass of system could be concentrated without changing the moment

  of the system.

Centre of Mass is the point at which the whole mass of the system

is assumed to be concentrated.

 The general formula for the COM is:

              xₙ =  Σmₐxₐ / Σmₐ         where,  a = 1,2,3.........n

  Here the term Σ mₐ xₐ is called the first moment of the system and the

  denominator expression is called total mass of the system.

    Therefore, from this theory we can say that the moment of the system

    remain unchanged while calculating the COM.

  Hence, The centre of mass of the system is the point at which the total

  mass of system could be concentrated without changing the moment

  of the system.

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The frequency of the second overtone if the fundamental frequency is 308 Hz is equal to 1540 Hz.

An overtone can be described as any resonant frequency above the fundamental frequency of a sound. Overtones can be defined as all pitches higher than the lowest pitch within an individual sound. While the fundamental can be described as usually heard most prominently, overtones are present in any pitch except a true sine wave.

The lowest normal frequency is called the fundamental frequency, while the higher frequencies are known as overtones.

Given, the fundamental frequency, f₀ = 30 8 Hz

The relationship between the fundamental frequency and the second overtone is given by

f₂ = 5 f₀

f₂ = 5 × 308 = 1540 Hz

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Answer:

The final charge on each square plate is +10

Explanation:

The law of conservation of charge states that the charge in a system, isolated from the environment, can only be transferred to another system, as the charges can neither be destroyed nor created

Therefore, the sum of the charges in a system is always constant such that the net charge in the universe is constant

Therefore, given the amount of charges in the 1 cm side = -5

The amount of charge in the 2 cm side = +15

The net charge when they touch = The charge on the 1 cm + The charge on the 2 cm

The net charge when they touch = -5 + 15 = +10

The final charge on each square plate = +10.

Answer:

This question is incomplete, as it lacks options. The options are:

A. chemical energy is the same and additional energy is produced as heat and light.

B. chemical energy is more than the amount of heat and light

C. the amount of chemical energy is less than the amount of heat and light energy

D. the amount of chemical energy equals the amount of heat and light energy.

The answer is D

Explanation:

According to the first law of thermodynamics, which is the LAW OF CONSERVATION OF ENERGY, energy is neither created nor destroyed but can only be changed from one form to another. This law states that no energy in a system is lost, hence, the input energy must be equal to the output energy.

When a log of wood is burnt, the chemical energy stored in it is converted to heat energy and light energy. Based on the law of conservation of energy, the amount of chemical energy present in the log before the burning process must equal the amount of light and heat energy it changed to after the burning process.

That is, chemical energy = light energy + heat energy.

Answer:

the height of the cliff is approximately 42.7 meters.

Explanation:

To calculate the height of the cliff, we can use the equations of motion. When the rock is thrown straight up, it will reach its highest point before falling back down.

We can use the equation for vertical displacement:

h = u * t + (1/2) * g * t^2

Where:

h is the height of the cliff, u is the initial velocity, t is the time taken, g is the acceleration due to gravity (approximately 9.8 m/s²).

h = 8.17 m/s * 2.25 s + (1/2) * 9.8 m/s² * (2.25 s)^2

h ≈ 42.68375 m

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To solve this problem, we need to first find the speed of the block at the bottom of the inclined plane, using conservation of energy.

The potential energy of the block at the top of the inclined plane is:

PE_top = mgh = 3.1 kg * 9.81 m/s^2 * 0.46 m = 14.2 J

where m is the mass of the block, g is the acceleration due to gravity, h is the height of the block above the horizontal surface.

At the bottom of the inclined plane, all of the potential energy has been converted into kinetic energy:

KE_bottom = 1/2 * mv^2

where v is the speed of the block at the bottom.

Using conservation of energy, we can equate the potential energy at the top to the kinetic energy at the bottom:

PE_top = KE_bottom

mgh = 1/2 * mv^2

v = sqrt(2gh)

Substituting the values, we get:

v = sqrt(2 * 9.81 m/s^2 * 0.46 m) = 3.01 m/s

Now we can use the frictional force to find the distance the block slides on the horizontal surface before coming to rest. The frictional force is given by:

f_friction = friction coefficient * f_normal

where f_normal is the normal force exerted by the horizontal surface on the block.

The normal force is equal to the weight of the block, which is:

f_normal = mg

where g is the acceleration due to gravity.

Substituting the values, we get:

f_normal = 3.1 kg * 9.81 m/s^2 = 30.411 N

The frictional force is then:

f_friction = 0.31 * 30.411 N = 9.43 N

The deceleration of the block due to friction is given by:

a_friction = f_friction / m = 9.43 N / 3.1 kg = 3.04 m/s^2

We can use the equation of motion to find the distance the block slides before coming to rest:

v_f^2 = v_i^2 + 2ad

where v_f is the final velocity (zero in this case), v_i is the initial velocity (3.01 m/s), a is the deceleration due to friction (-3.04 m/s^2), and d is the distance traveled by the block.

Substituting the values and solving for d, we get:

d = (v_f^2 - v_i^2) / (2a) = (0 - (3.01 m/s)^2) / (2 * (-3.04 m/s^2)) = 0.472 m

Therefore, the block slides 0.472 meters on the horizontal surface before coming to rest.

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Two football players are running towards each other in a straight line (exact opposite directions). Player A is running at 3.3 m/s and has a mass of 105 kg. Player B is 126 kg.

The players collide and their net momentum after the collision is 0 Ns. How fast was Player B running before they collided? The law of conservation of momentum states that in a closed system, the total momentum remains constant. Therefore, the total momentum of both players before collision equals the total momentum of both players after collision. This means: mA * VA + mB * VB = (mA + mB) * V, where VA and VB are the initial velocities of A and B, respectively, and V is their final velocity after the collision (which is 0).

So, we can rearrange the above equation to solve for VB. VB = (mA * VA) / mB Here, mA = 105 kg and VA = 3.3 m/s, and mB = 126 kg. Substituting the values, we get: VB = (105 * 3.3) / 126= 2.75 m/s Therefore, Player B was running at 2.75 m/s before they collided.

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The acceleration of the sled of mass 300 kg is 0.57 m/s².

Acceleration is the rate of change of the velocity of a body.

To calculate the acceleration of the sled, we use the formula below.

Formula:

Where:

From the question,

Given:

Substitute these values into equation 1

Hence, the acceleration of the sled is 0.57 m/s².

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Rotation period for a mass of 0.550 kg at the same radius is 1.45 s.

The rotation period of a mass in circular motion is given by:

T = 2πR/v

where T is the period, R is the radius of the circular path, and v is the velocity of the mass.

For the first mass with a mass of 0.45 kg, radius R of 0.140 m, and period T of 1.45 s, we can calculate the velocity as follows:

v = 2πR/T = 2π(0.140 m)/(1.45 s) = 0.6066 m/s

Now, we can use the velocity and radius values to find the period for the second mass with a mass of 0.550 kg:

T = 2πR/v = 2π(0.140 m)/(0.6066 m/s) = 1.45 s

Therefore, the rotation period for a mass of 0.550 kg at the same radius is 1.45 s.

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Answer:

Explanation:

An intensive property is a property of matter that does not change as the amount of matter changes. It is a bulk property, which means it is a physical property that is not dependent on the size or mass of a sample. In contrast, an extensive property is one that does depend on sample size.

Therefore, the algebra inverse Laplace transform of \(5s - 8s^2 + 16\) is:

\(L^{-1}\) \(5s - 8s^2 + 16\) = 5δ(t) - 16t + 16

So, the answer is: f(t) = 5δ(t) - 16t + 16

We can use linearity and the differentiation property of the Laplace transform to find the inverse Laplace transform of 5s - \(8s^2\) + 16. Using Theorem 7.2.1, we have:

Laplace transform of a function, we can use the Laplace transform operator \(L^{-1}\), which takes a function of t as input and produces a function of s as output, where s is a complex variable representing the frequency of the function.

\(L^{-1}\){5s} = 5\(L^{-1}\){s} = 5δ(t)

\(L^{-1}\){ \(8s^2\)} = 8\(L^{-1}\){s} = 8(1/Γ(3))

\(d^2/dt^2 t^2\) = 16t

\(L^{-1}\)1{16} = 16\(L^{-1}\){1} = 16δ(t)

Therefore, the inverse Laplace transform of  \(5s - 8s^2 + 16\) is:

\(L^{-1}\){ \(5s - 8s^2 + 16\)} = 5δ(t) - 16t + 16

So, the answer is: f(t) = 5δ(t) - 16t + 16

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The following set of quantum numbers (ordered n , ℓ , mℓ , ms ) are possible for an electron in an atom are 3, 2, -3, 1/2; 3, 2, 2, -1/2; 5, 3, 4, 1/2; 2, 2, 2, 1/2".

The set of quantum numbers (ordered n, ℓ, mℓ, ms) are possible for an electron in an atom are as follows:3, 2, -3, 1/23, 2, 2, -1/25, 3, 4, 1/22, 2, 2, 1/2

The quantum numbers are a set of numbers that can be used to identify an electron's location.

In atoms, the principal quantum number (n), the angular momentum quantum number (l), the magnetic quantum number (ml), and the electron spin quantum number (ms) are all used.

Principal Quantum Number(n) - It specifies the energy level of an electron in an atom.

Angular Momentum Quantum Number (l) - It specifies the shape of the orbital in which the electron is present.

Magnetic Quantum Number (ml) - It specifies the orientation of the orbital in which the electron is present.

Electron Spin Quantum Number (ms) - It specifies the spin of an electron in the orbital.

In the given options, 4 sets of quantum numbers are possible for an electron in an atom.

They are 3, 2, -3, 1/2; 3, 2, 2, -1/2; 5, 3, 4, 1/2; 2, 2, 2, 1/2, and hence the correct answer is "DETAIL ANS: 3, 2, -3, 1/2; 3, 2, 2, -1/2; 5, 3, 4, 1/2; 2, 2, 2, 1/2".

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Answer:

b

Explanation:

\(\qquad \qquad \huge \pink {\sf{☁Answer☁}} \\ \\ \)

\(\pink{\boxed{\sf{↪Option.c↩}}}\)

→Matter✓

\(\pink{\boxed{\sf{↪Option.d↩}}}\)

→The sum of the kinetic and potential energy of an object✓

\(\sf{\:мѕнαcкεя\: ♪...}\)

Explanation:

v = u + at

24 = 10 + a * 7

24 = 10 * 7 a

a = 24 - 10/7

a = 14/7

a = 2m/sec^2

hope it helps you

Answer:

First calculate change in velocity:

final velocity = 20 m/s.

initial velocity = 30 m/s.

change in velocity = (20 - 30) = -10 m/s.

acceleration = -10 ÷ 25.

acceleration = -0.4 m/s 2

The acceleration value is negative here because the car is slowing down or decelerating.

Explanation:

Answer:

his acceleration rate is -0.00186 m/s²

Explanation:

Given;

initial position of the car, x₀ = 100 miles = 160, 900 m ( 1 mile = 1609 m)

time of motion, t₀ = 60 minutes = 60 mins x 60 s = 3,600 s

final position of the car, x₁ = 150 miles = 241,350 m

time of motion, t₁ = 100 minutes = 100 mins x 60 s = 6,000 s

The initial velocity is calculated as;

u = 160, 900 m / 3,600 s

u = 44.694 m/s

The final velocity is calculated as;

v = 241,350 m / 6,000 s

v = 40.225 m/s

The acceleration is calculated as;

\(a = \frac{\Delta V}{\Delta t} = \frac{v- u}{t_1 - t_ 0} = \frac{40.225 - 44.694}{6000-3600} = -0.00186 \ m/s^2\\\\\)

Therefore, his acceleration rate is -0.00186 m/s²

Answer:

1) True

2) None of the above

Explanation:

Q # 1:

Weight is actually the gravitational force or force of gravity. So, it's true.

Q # 2:

None of above because force of gravity can not be negative. In order to have a negative weight, the body has to be "repelled" by gravity.

\(\rule[225]{225}{2}\)

Hope this helped!

Answer:

In the interaction between the bus and the bug, we can call the force "bus pushes bug" the action force. Then the force "bug pushes bus" is the reaction force. Since these forces are a Newton's Third Law action/reaction force pair, they are exactly the same size.

Explanation:

Newtons 3rd law