The Earth’s radius is 6378.1 kilometers. If you were standing at the equator, you are essentially undergoing uniform circular motion with the radius of your circular motion being equal to the radius of the Earth. You are an evil mad scientist and have come up with the simultaneously awesome and terrifying plan to increase the speed of the Earth’s rotation until people at the Earth’s equator experience a centripetal (radial) acceleration with a magnitude equal to g, (9.81 m/s2 ), effectively making them experience weightlessness. If you succeed in your dastardly plan, what would be the new period of the Earth’s rotation?

a. 2.7 minutes b. 84 minutes c. 48 minutes d. 76 minutes

Answer:

n₁ = 3

Explanation:

The energy of the states in the hydrogen atom is explained by the Bohr model, the transitions heal when an electron passes from a state of higher energy to another of lower energy,

       ΔE = [tex]E_{nf}[/tex] - E₀ = - k²e² / 2m (1 / [tex]n_{f}[/tex]²2 - 1 / n₀²)

The energy of this transition is given by the Planck equation

       E = h f = h c / λ

       h c / λ = -k²e² / 2m (1 / no ²- 1 / no²)

       1 / λ = Ry (1/ [tex]n_{f}[/tex]² - 1 / n₀²)

Let's apply these equations to our case

     λ = 821 nm = 821 10⁻⁹ m

     E = h c / λ

     E = 6.63 10⁻³⁴ 3 10⁸/821 10⁻⁹

     E = 2.423 10⁻¹⁹ J

Now we can use the Bohr equation

Let's reduce to eV

       E = 2,423 10⁻¹⁹ J (1eV / 1.6 10⁻¹⁹) = 1,514 eV

       [tex]E_{nf}[/tex] - E₀ = -13.606 (1 /  [tex]n_{f}[/tex]² - 1 / n₀²)   [eV]

Let's look for the energy of some levels

n         [tex]E_{n}[/tex] (eV)          [tex]E_{nf}[/tex] - E[tex]E_{ni}[/tex] (eV)

1         -13,606           E₂-E₁ = 10.20

2        -3.4015           E₃-E₂ = 1.89

3        -1.512              E₄- E₃ = 0.662

4        -0.850375

We see the lines of greatest energy for each possible series, the closest to our transition is n₁ = 3 in which a transition from infinity (n = inf) to this level has an energy of 1,512 eV that is very close to the given value

Answer:

[tex]1.13333\times 10^{-8}\ s[/tex] and [tex]0.7\times 10^{-8}\ s[/tex]

Explanation:

Light which travels from the crackers reaches the detector at [tex]c=3\times 10^{8}\ m/s[/tex]

[tex]\Delta x_1[/tex] = Distance at which event 1 leaves a char mark = 3.4 m

[tex]\Delta x_2[/tex] = Distance at which event 2 leaves a char mark = 2.1 m

The speed of light in a medium is a universal constant

[tex]c=\dfrac{\Delta x_1}{\Delta t_1}\\\Rightarrow \Delta t_1=\dfrac{\Delta x_1}{c}\\\Rightarrow \Delta t_1=\dfrac{3.4}{3\times 10^8}\\\Rightarrow \Delta t_1=1.13333\times 10^{-8}\ s[/tex]

[tex]c=\dfrac{\Delta x_2}{\Delta t_2}\\\Rightarrow \Delta t_2=\dfrac{\Delta x_2}{c}\\\Rightarrow \Delta t_2=\dfrac{2.1}{3\times 10^8}\\\Rightarrow \Delta t_2=0.7\times 10^{-8}\ s[/tex]

The pulse will be detected at [tex]1.13333\times 10^{-8}\ s[/tex] and [tex]0.7\times 10^{-8}\ s[/tex]

In the reference frame of the detector, both light pulses will be detected approximately 1.13 x 10^-8 seconds after they occur.

In the reference frame of the detector, event 1 leaves a mark at a distance of 3.40 m and event 2 leaves a mark at a distance of 2.10 m. Since the events are simultaneous in the reference frame of the detector, the time it takes for each light pulse to be detected will be the same for both events. To calculate the time, we can use the speed of light as the distance traveled divided by the speed of light. Therefore, the time it takes for each light pulse to be detected is t = distance/speed of light.

For event 1, the distance is 3.40 m and for event 2, the distance is 2.10 m. The speed of light is approximately 3.00 x 10^8 m/s. Plugging in these values, we get:

t = (3.40 m) / (3.00 x 10^8 m/s) = 1.13 x 10^-8 s

Therefore, each light pulse will be detected approximately 1.13 x 10^-8 seconds after the events occur.

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Using Momentum Principle and the Energy Principle to determine the final velocities of the projectile and target are respectively;

v1 = (m1•u1 - m2•u1)/(m1 + m2)

v1 = (m1•u1 - m2•u1)/(m1 + m2)v2 = (m1•u1 + m1•u2)/(m1 + m2)

Let us denote the following;

Mass of projectile; m1

Mass of target; m2

Initial velocity of projectile; u1

Final velocity of projectile; v1

Initial velocity of target; u2 = 0 m/s

Final velocity of target; v2

From conservation of momentum, we have;

(m1•u1) + (m2•u2) = (m1•v1) + (m2•v2)

Plugging in the relevant values;

(m1•u1) + (m2•0) = (m1•v1) + (m2•v2)

(m1•u1) = (m1•v1) + (m2•v2) - - - (eq 1)

From energy principle, we have;

u1 + v1 = u2 + v2

Thus;

Since u2 is zero, then we have;

u1 + v1 = v2 - - - (eq 2)

Put (u1 + v1) for v2 in eq 1 to get;

(m1•u1) = (m1•v1) + (m2(u1 + v1))

(m1•u1) = (m1•v1) + (m2•u1 + m2•v1) - - - (eq 3)

Let us make v1 the subject;

m1•u1 - m2•u1 = m1•v1 + m2•v1

v1(m1 + m2) = m1•u1 - m2•u1

v1 = (m1•u1 - m2•u1)/(m1 + m2)

Let us make v1 the subject in eq 2;

v1 = v2 - u2

Put v2 - u2 for v1 in eq 1 to get;

(m1•u1) = (m1(v2 - u2) + (m2•v2)

(m1•u1) = m1•v2 - m1•u2 + m2•v2

m1•u1 + m1•u2 = v2(m1 + m2)

v2 = (m1•u1 + m1•u2)/(m1 + m2)

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The final velocities of the projectile and the target after an elastic collision can be arrived at by solving the equations for Conservation of Momentum and Conservation of Kinetic Energy simultaneously. Here, the target's initial velocity is zero (it's at rest). By substituting this into the equations and solving them, we can find the final velocities.

An elastic collision is a type of collision where both momentum and kinetic energy are conserved. For this problem, we can solve the final velocities using the conservation laws.

Conservation of Momentum gives us:

m1*v1 = m1*v1' + m2*v2'     -------- (1)

V1' and V2' are the final velocities of the projectile and target, respectively.

Conservation of Kinetic Energy gives us:

1/2*m1*v1^2 = 1/2*m1*(v1')^2 + 1/2*m2*(v2')^2     -------- (2)

While solving these two equations, one should remember that m2 was initially at rest, so the momentum in the Y-direction is zero. The above-mentioned equations can be solved by substitution method or any other algebraic method to get the final velocities v1' and v2'.

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Answer:

a. get warmer.

Explanation:

When the water vaporous reach the upper layer of the atmosphere they get a cooler air to which they loose their temperature and condense to form clouds as a the temperature of the air increases.

It may be noted that the water looses its high amount of latent heat of vaporization to condense into water this significantly increases the temperature of the air in contact.

Answer:

a. T₂ = 901.43 ° K

b. T₂ = 843.85 ° K

Explanation:

Given

η = 0.165 mol , Q = 690 J , V = 280 cm³ , P = 3.40 x 10 ⁶ Pa , T₁ = 700 ° K

Using the equation that describe the experiment of heat

Q = η * Cv * ΔT

a.

Nitrogen   Cv =  20.76 J / mol ° K  

ΔT = T₂ - T₁

T₂ = [ Q / ( η * Cv) ] + T₁

T₂ = [ 690 J / ( 0.165 mol * 20.76 J / mol ° K ) ] + 700 ° K

T₂ = 901.43 ° K

b.

Nitrogen   Cp = 29.07 J / mol ° K

T₂ = [ Q / ( η * Cv) ] + T₁

T₂ = [ 690 J / ( 0.165 mol * 29.07 J / mol ° K ) ] + 700 ° K

T₂ = 843.85 ° K

The Final temperature of the air when the volume of the cylinder is constant is 700 K.

To find the final temperature of the air when the volume of the cylinder is constant, we can use the ideal gas law equation:

P₁V₁/T₁ = P₂V₂/T₂

Since the volume is constant, V₁ = V₂.

The initial pressure (P₁) and temperature (T₁) are given, and the final pressure (P₂) is also given (3.40×106 Pa).

We can rearrange the equation to solve for T₂:

T₂ = (P₂T₁) / P₁

Substituting the given values, we have:

T₂ = (3.40×106 Pa * 700 K) / (3.40×106 Pa)

T₂ = 700 K

So, the final temperature of the air is 700 K when the volume of the cylinder is constant.

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To calculate the time it takes for the coffee pot to bring the water to a boil, we need to determine the amount of heat required and use the power of the coffee pot. By calculating the heat required for both the water and the aluminum pot, and using the power of the coffee pot, we can determine that it takes approximately 96.077 seconds for the coffee pot to bring the water to a boil.

To determine how long it takes for the coffee pot to bring the water to a boil, we need to calculate the amount of heat required. The heat required to raise the temperature of the water from 15?C to 100?C is given by the formula Q = mcΔT, where Q is the heat, m is the mass, c is the specific heat, and ΔT is the change in temperature. Using the values given, we can calculate the heat required.

The heat required for the water is Q = (0.85 kg)(4186 J/kg?C)(100?C - 15?C) = 43707.5 J. The heat required for the aluminum coffee pot is Q = (0.36 kg)(900 J/kg?C)(100?C - 15?C) = 28350 J. Adding the two heats together gives us a total heat of 72057.5 J.

The power of the coffee pot is given as 750 W. Power is defined as the rate at which work is done or the rate at which energy is transferred. So, we can use the formula P = E/t, where P is power, E is energy, and t is time. Rearranging the formula, we can solve for t:

t = E/P = 72057.5 J / 750 W = 96.077 seconds.

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Answer:

 a₂ = m₁ / m₂ a₁

Explanation:

For this exercise we note that the attraction between the two stars is an action and reaction force, therefore it has the same magnitude, but it is applied to each of the bodies

Let's apply Newton's second law on the star 1

     F₁ = m₁ a₁

Newton's second law in star 2

       F₂ = m₂ a₂

       | F₁ | = | F₂ |

      m₁  a₁ = m₂  a₂

      a₂ = m₁ / m₂ a₁

Answer:

C) antimony and bismuth

Explanation:

Answer:

its C

Explanation:

The spring constant of the spring is approximately 234.50 N/m.

To find the spring constant of the spring, we can use Hooke's Law, which states that the force exerted by a spring is proportional to the displacement of the spring from its equilibrium position. In this case, we know that the mass of the block is 6.7 kg and the displacement of the spring is 0.28 m. Therefore, we can use the equation F = kx, where F is the force, k is the spring constant, and x is the displacement. Plugging in the given values, we get:

F = kx
F = (6.7 kg)(9.8 m/s^2)
F = 65.66 N
65.66 N = k(0.28 m)
k = (65.66 N)/(0.28 m)
k ≈ 234.50 N/m

Therefore, the spring constant of the spring is approximately 234.50 N/m.

Answer:

C/V

Explanation:

Mathematically it is known that one farad is defined as the capacitance across which, when charged with one coulomb, there is a potential difference of one volt. In other words, the Farad is Coulomb/Volt or C/V

Answer:

B. Temperatures remain high as lowered pressures decrease melting temperatures.

Explanation:

Rocks , like ice , contracts on melting so , their melting point decreases on decrease in pressure . When hot solid rocks are pushed towards the surface of the earth, due to decrease in pressure , their melting point decreases , so the solid rocks starts melting without any addition of  heat . Their  present temperature becomes more than their melting point.

To develop this problem we will proceed to convert all units previously given to the international system for which we have to:

[tex]140 lb = 63.5 kg \rightarrow 63.5kg (9.8m/s) =622.3 N[/tex]

[tex]120 lb = 54.4 kg \rightarrow 54.4kg (9.8m/s)= 533 N[/tex]

[tex]170 lb = 77.1 kg \rightarrow 77.1 kg (9.8m/s) =756 N[/tex]

PART A ) From the given values the minimum acceleration will be given for 120Lb and maximum acceleration when 170Lb is reached therefore:

[tex]F = 756 - 622.3[/tex]

[tex]F = 133.7N[/tex]

Through the Newtonian relationship of the Force we have to:

[tex]F= ma[/tex]

[tex]a = \frac{F}{m}[/tex]

[tex]a = \frac{133.7}{63.5}[/tex]

[tex]a = 2.1m/s^2[/tex]

PART B) For the maximum magnitude of the acceleration downward we have that:

[tex]F = 622.3 - 533[/tex]

[tex]F = 89.3N[/tex]

Through the Newtonian relationship of the Force we have to:

[tex]F= ma[/tex]

[tex]a = \frac{F}{m}[/tex]

[tex]a = \frac{89.3}{63.5}[/tex]

[tex]a = 2.1m/s^2[/tex]

[tex]a = 1.04 m/s^2[/tex]

Answer:

U. With no variation.

Explanation:

Note- since temperature remains constant when pressure becomes twice and volume becomes half, and internal energy of ideal gas is function of only temperature so it remains constant. The internal energy is independent of the variables stated in the exercise.

Answer:

Option (a) is correct.

Explanation:

For us to be able to visualize an object, there are basic processes that takes place in a split second.

First Light which strike the object reflect into our eye. This reflected light passes through the pupil as the iris gives way for it. This light moves through other layers and impinges on the retina and stimulates it. This is the process of within the eye processing. The light is then converted to a signal by the optic nerve and sent to the brain for interpretation. The image produced by the retina is inverted. The brain helps to interpret this image as upright. This is the reason why we see objects just the way they are

To solve this problem it is necessary to apply the concepts related to the Power defined from the Stefan-Boltzmann equations.

The power can be determined as:

[tex]P = \sigma T^4[/tex]

Making the relationship for two states we have to

[tex]\frac{P_1}{P_2} = \frac{T_1^4}{T_2^4}[/tex]

Since the final power is 8 times the initial power then

[tex]P_2 = 8P_1[/tex]

Substituting,

[tex]\frac{1}{8} = \frac{T_1^4}{T_2^4}[/tex]

[tex]T_2 = T_1 8*(\frac{1}{4})[/tex]

[tex]T_2 = 1,68T_1[/tex]

The temperature increase would then be subject to

[tex]\Delta T = T_2-T_1[/tex]

[tex]\Delta T = 1.68T_1 -T_1[/tex]

[tex]\Delta T = 0.68T_1[/tex]

The correct option is D, about 68%

Final answer:

The correct option is (c) 100%. To increase the power radiated from an object by a factor of 8, the temperature must be doubled, corresponding to a 100% increase. This is based on the Stefan-Boltzmann law.

Explanation:

The question relates to the physics concept of radiation and how it varies with temperature. Specifically, it involves the Stefan-Boltzmann law, which states that the power radiated from an object is proportional to the fourth power of its absolute temperature.

If the power radiated from an object is to increase by a factor of 8, we use the relationship P ≈ T^4, where P is power and T is the temperature in Kelvin. To calculate the necessary temperature increase, if the initial power is P and the final power is 8P, then (T_final)^4 = 8 × (T_initial)^4.

By taking the fourth root on both sides, we find that T_final = 2× T_initial. Therefore, the temperature must be doubled, or the temperature increase is by 100%. Hence, the correct answer is option (c) 100%

Answer:

The right option is (d) substance undergoing a change of state

Explanation:

Latent Heat: Latent heat is the heat required to change the state of a substance without change in temperature. Latent heat is also known as hidden heat because the heat is not visible. The unit is Joules (J).

Latent heat is divided into two:

⇒ Latent Heat of fusion

⇒ Latent Heat of vaporization.

Latent Heat of fusion: This is the heat  energy required to convert a substance from its solid form to its liquid form without change in temperature. E.g (Ice) When ice is heated, its temperature rise steadily until a certain temperature is reached when the solid begins to melts.

Latent Heat of vaporization: This is the heat required to change a liquid substance to vapor without a change in temperature. The latent heat depend on the mass of the liquid and the nature of the liquid. E.g When water is heated from a known temperature  its boiling point (100°C) When more heat is supplied to its boiling temperature, it continue to boil without a change in temperature.

From The above, Latent heat brings about  a change of state of a substance at a steady temperature.

The right option is (d) substance undergoing a change of state

The substance that best represents a temperature remaining constant despite an inward heat flow is one that is undergoing a change of state, such as during the melting of ice or boiling of water. So, the correct option is D.

The best description of a substance in which the temperature remains constant while it is experiencing an inward heat flow is a substance undergoing a change of state (option d). This occurs during a process called phase transition.

Here's an example: when ice melts into water, it absorbs heat from the surroundings but the temperature stays constant at 0 degrees Celsius until all of the ice has melted.

Similarly, when water is boiling, its temperature holds steady at 100 degrees Celsius until all the water has transitioned into steam, despite the continued input of heat.

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Answer:

both have plasma membrane, ribosomes, cytoplasm, and DNA

Explanation:

Both eukaryotic and prokaryotic cells contain ribosomes, which are responsible for protein synthesis. Eukaryotic cells are more complex with a true nucleus, found in animals and plants, while prokaryotic cells, found in bacteria, are simpler and lack a true nucleus.

Option A is the correct answer. Both eukaryotic and prokaryotic cells contain ribosomes. The ribosomes in both types of cells serve the same function: protein synthesis, or assembly of amino acids to form proteins. Eukaryotic cells are those with a true nucleus and complex structures, found in animals, plants, fungi, and protists, while prokaryotic cells are smaller, simpler cells without a true nucleus, found in bacteria and archaea.

Unlike eukaryotic cells, prokaryotic cells don't contain a membrane-bound nucleus or mitochondria. A capsule is found in some bacteria, not all prokaryotes and no eukaryotes.

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Answer:

Explanation:

Given

density of cylinder is [tex]\rho _c=713 kg/m^3[/tex]

Length of first cylinder is [tex]L_1=20 cm[/tex]

radius [tex]r_1=5 cm[/tex]

For cylinder 2 [tex]L_2=10 cm [/tex]

[tex]r_2=10 cm[/tex]

[tex]h_1[/tex] and [tex]h_2[/tex] are the height above water

E

as object is floating so its weight must be balanced with buoyant force

[tex]\rho _c\frac{\pi }{4}d_1^2L_1g=\rho _w\frac{\pi }{4}d_1^2(L_1-h_1)g----1[/tex]

For 2nd cylinder

[tex]\rho _c\frac{\pi }{4}d_2^2L_2g=\rho _w\frac{\pi }{4}d_2^2(L_2-h_2)g----2[/tex]

Dividing 1 and 2 we get

[tex]\frac{L_1}{L_2}=\frac{L_1-h_1}{L_2-h_2}[/tex]

[tex]\frac{20}{10}=\frac{20-h_1}{10-h_2}[/tex]

[tex]2h_2=h_1[/tex]

[tex]\\\Rightarrow\frac{h_2}{h_1}=\frac{1}{2}[/tex]                            

The  effective diameter d of your pupil should be considered as the 0.0031 when maximum distance should be 1.30 meters.

Since

y = Length of pixel = 281 μm

L = Distance to screen = 1.3 m

Wavelength = 550 nm

d = Pupil diameter

So,

[tex]tan = y / l\\\\= tan^-1 * 281*10^-6/1.3[/tex]

Now

sin = 1.22 wavelength / d

d = 1.22 * wavelength / sin

[tex]= 1.22*550*10^-9/ sin ( tan^-1 * 281*10^-6/1.3)[/tex]

= 0.0031 m

hence, The  effective diameter d of your pupil should be considered as the 0.0031 when maximum distance should be 1.30 meters.

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The effective diameter of the pupil of one's eye when barely able to resolve two adjacent pixels on a computer screen from a certain distance, where the point of resolution is diffraction limited, is approximately 1.7 mm.

This question involves the use of the formula for the resolution of an optical system, which is the Rayleigh’s criterion. The formula can be written as follows: θ = 1.22 * λ/D, where θ is the smallest angular separation that can be resolved, λ is the wavelength of light, and D is the diameter of the aperture (i.e., the pupil in this case).

In this question, the pixel’s size and the distance from the screen allow us to calculate θ. The tangent of θ is approximately equal to the ratio of the pixel size to the viewing distance. Hence, θ = arctan((281 * 10^-6 m) / 1.3 m).

By substituting θ, and λ (550 * 10^-9 m) into the formula and solve for D, the effective diameter of the pupil, we get: D = 1.22 * λ / θ. The result is around 0.0017 meters or 1.7 mm.

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The fraction of the volume of an iceberg that is visible when it floats in saltwater or freshwater can be calculated using Archimedes' principle. For the given densities, the iceberg would sink in both saltwater and freshwater.

When an iceberg floats in water, a fraction of its volume is submerged, which is determined by the density of water and the density of the iceberg. To calculate the fraction of the iceberg that is submerged, you can use Archimedes' principle. Archimedes' principle states that the buoyant force acting on an object submerged in a fluid is equal to the weight of the fluid displaced by the object.

In this case, the density of the iceberg is given as 917 kg/m³. Let's calculate the fraction of the volume of the iceberg that would be visible when floating in (a) the ocean (salt water, density 1024 kg/m³) and (b) a river (fresh water, density 1000 kg/m³).

Fraction submerged = (density of iceberg - density of seawater) / density of iceberg

Using the given densities:

(917 kg/m³ - 1024 kg/m³) / 917 kg/m³ = -0.1166

The negative value indicates that the iceberg would not float in saltwater. In other words, it would sink.

Fraction submerged = (density of iceberg - density of fresh water) / density of iceberg

Using the given densities:

(917 kg/m³ - 1000 kg/m³) / 917 kg/m³ = -0.0902

Again, the negative value indicates that the iceberg would sink in freshwater as well.

Answer:

Explanation:

1) TRUE; potential difference can be calculated using path integral. Since the electric field is a conservative, the potential difference can be calculated using any path.

2) TRUE; since potential due to a charge is inversely dependent on distance, at infinity the potential will be almost zero.

3) TRUE, W = q.VBA.

4) FALSE; eV is a unit for work (or) energy.

5) TRUE; since the electric force is conservative force. There will be no loss in energy, the decreased potential energy will be coverted to kinetic energy.

6) FALSE; in the direction of electric field the potential decreases.

7) FALSE; equipotential surface is perpendicular to the electric field lines.

8) FALSE; electrostatic potential is scalar quantity. It depends only on the charge and distance from it.

9) FALSE; Inside a conductor the electric field is zero but the electric potential is constant at the value that is at the surface of the conductor.

10) TRUE; as long as the field is being measured outiside the body the bodies act as point charges. So electric fields due to all types of bodies charged identically will be equal.

Answer:

1350N

Explanation:

[tex]2.75 ms = 2.75*10^{-3}s[/tex]

The force exerted on the hand would be the momentum divided by the duration of contact.

As the hand is coming to rest, final velocity would be 0

[tex]F = \frac{\Delta P}{\Delta t} = \frac{m(0 - v)}{\Delta t} = \frac{1.65*(2.25 - 0)}{2.75 * 10^{-3}} = -1350 N[/tex]

The magnitude of the force would be 1350N

Answer:

The oscillation frequency of the spring is 1.66 Hz.

Explanation:

Let m is the mass of the object that is suspended vertically from a support. The potential energy stored in the spring is given by :

[tex]E_s=\dfrac{1}{2}kx^2[/tex]

k is the spring constant

x is the distance to the lowest point form the initial position.

When the object reaches the highest point, the stored potential energy stored in the spring gets converted to the potential energy.

[tex]E_P=mgx[/tex]

Equating these two energies,

[tex]\dfrac{1}{2}kx^2=mgx[/tex]

[tex]\dfrac{k}{m}=\dfrac{2g}{x}[/tex].............(1)

The expression for the oscillation frequency is given by :

[tex]f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}[/tex]

[tex]f=\dfrac{1}{2\pi}\sqrt{\dfrac{2g}{x}}[/tex] (from equation (1))

[tex]f=\dfrac{1}{2\pi}\sqrt{\dfrac{2\times 9.8}{0.18}}[/tex]

f = 1.66 Hz

So, the oscillation frequency of the spring is 1.66 Hz. Hence, this is the required solution.

Answer:

0.5 N

Explanation:

[tex]r_{1}[/tex] = radius of piston 1

[tex]r_{2}[/tex] = radius of piston 2

Given that :

[tex]r_{2} = 200 r_{1}[/tex]

[tex]F_{1}[/tex] = Force applied on piston 1

[tex]F_{2}[/tex] = Force applied on piston 2 = Weight being lifted = 20000 N

Using pascal's law

[tex]\frac{F_{1}}{\pi r_{1}^{2} } = \frac{F_{2}}{\pi r_{2}^{2} } \\\frac{F_{1}}{r_{1}^{2} } = \frac{20000}{(200)^{2}r_{1}^{2} } \\\\F_{1}= \frac{20000}{(200)^{2}} } \\F_{1}= 0.5 N[/tex]

Answer:(69,69,69)

Explanation:

The ratio of the intercortex of the inner circle has an area of 98:89. This is due to the fact the vector of the circle is no less than the ratio times the pie squared. Then pie squared the circumference will end up approximately at 68.83957 where you then round up to all 69's.

The speed of the metal ball in circular motion is calculated by using the equation for centripetal acceleration (v = √(rg)) and substituting the given values for radius and gravity. After calculation, the speed of the ball is determined to be approximately 5.1 m/s.

In this physics problem, we can calculate the speed of the metal ball by using the provided values and the concept of centripetal acceleration. Centripetal acceleration results from the force that makes the metal ball move in a circular pattern, which in this case is provided by the tension in the thread and gravity. In a scenario where the ball is in uniform circular motion, with the tension at an angle, forming a cone around the vertical axis, the component of tension providing the centripetal force equals the gravitational force or weight of the object.

A key equation that can be used here is: ac = v² /r, where 'ac' is the centripetal acceleration, 'v' is the velocity (or speed), and 'r' is the radius. Given that ac = g (acceleration due to gravity = 9.8 m/s²), we can modify the equation to: v = √(rg), where 'r' is the radius or the length of the thread (2.7 m).

Implementing these values into our equation we find that: v = √[(2.7 m)(9.8 m/s²)] = √26.46 = 5.1 m/s. Therefore, the speed of the metal ball when it is in circular motion is approximately 5.1 m/s.

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Answer

given,

mass of the car = 1600 Kg

velocity of the car = 21 m/s

time = 1.8 s

a) momentum = mass x velocity

P = 1600 x 21

P = 33600 kg.m/s

[tex]Force = \dfrac{change\ in \ momentum}{t}[/tex]

[tex]Force = \dfrac{33600-0}{1.8}[/tex]

F = 18666.67 N

b) [tex]Force = \dfrac{change\ in \ momentum}{t}[/tex]

[tex]Force = \dfrac{33600-0}{0.14}[/tex]

F = 240000 N

Answer:

[tex]h=3.4817\ m[/tex]

Explanation:

Given:

We know that:

[tex]P=\rho.g.h[/tex]

putting respective values

[tex]65000-101325=1905 \times 9.8\times h[/tex]

[tex]h=3.4817\ m[/tex] is the reading in terms of the fluid column height on the manometer.

Considering the rise in the level of fluid this fluid is not suitable for such huge pressures for a practically convenient manometer.

Answer:

1- Period is the time for which one full rotation is completed. Regardless of their positions on the platform, periods of all three are the same. T2 = T.

2- Similarly, T3 = T.

3- The platform is making rotational motion. So, the relation between the angular velocity and the linear velocity is

[tex]v = \omega R[/tex]

For all the people, angular velocity is the same. Their linear velocities are different.

[tex]\omega = \frac{v_1}{R} = \frac{v_2}{3R/5}\\v_2 = \frac{3v}{5}[/tex]

4- Similarly,

[tex]\omega = \frac{v_1}{R} = \frac{v_2}{R/2}\\v_2 = \frac{v}{2}[/tex]

5- Radial acceleration in constant circular motion is

[tex]a_{rad} = \frac{v^2}{R}[/tex]

For the second person:

[tex]a_2 = \frac{v_2^2}{3R/5} = \frac{(3v/5)^2}{3R/5} = \frac{9v^2/25}{3R/5} = \frac{3v^2}{5R} = 3a/5[/tex]

6- Similarly,

[tex]a_3 = \frac{v_3^2}{R/2} = \frac{(v/2)^2}{R/2} = \frac{v^2/4}{R/2} = \frac{v^2}{2R} = a/2[/tex]

Explanation:

As a result, the period is same for every object on the rotating platform, as they all complete their revolutions at the same time. Their speed and radial acceleration is different according to their distance to the center.

The rotational kinematics relations allow to find the results for the questions about the movement of the three people on the turntable are:

      1 and 2) All periods are equal, T₂ = T and T₃ = T.

       3) The linear velocity of the 2nd person is: v₂ = [tex]\frac{3}{5} \ v[/tex]  

       4) The linear velocity of the 3rd person is: v₃ = ½ v

       5) The linear acceleration of the 2nd person is: a₂ = [tex]\frac{3}{5} \ a[/tex]  

      6) The linear acceleration of the 3rd person: a₃ = ½ a

Rotational kinematics studies the rotational motion of bodies looking for relationships between angular position, angular velocity, and angular acceleration.

In the case where the angular accleration is zero, the expression for the velocity is:

        [tex]w = \frac{\Delta \theta }{\Delta t}[/tex]  

Where w is the angular velocity and Δw and Δt are the variation in angle t over time.

1 and 2)

Indicates that people are on a turntable, the period is when we have a complete rotation θ = 2π rad in time, therefore the period and the angular velocity are related.

          [tex]w= \frac{2\pi }{T} \\T = \frac{2\pi }{w}[/tex]  

In the apparatus of parks the angular velocity is constant and we see that it does not depend on the radius, therefore the period for all the people is the same.

         T = T₁ = T₂

3) They indicate that the speed of the 1 person who is in the position r=R on the plate is v, let's calculate the speed for the 2 person who is in the position  r = [tex]\frac{3}{5} \ R[/tex]  

Linear and angular variables are related.

        v = w r

Let's substitute for the 1st person.

        v = w R

For the 2nd person.

        v₂= w ( [tex]\frac{3}{5} R[/tex])

We solve these two equations.

         [tex]v_2 = \frac{3}{5} \ v[/tex]  

4) We carry out the same calculation for the 3rd person.

         v₃ = w ½ R

We solve the two equations.

          v₃ = ½ v

5) Ask for radial acceleration.  

The relationship between radial and angular acceleration is.

           a = α R

We substitute for the 1st person.

          a = α R

For the second person.

          a₂ = α ( [tex]\frac{3}{5} R[/tex])

We solve the two equations

         a₂ = [tex]\frac{3}{5} \ a[/tex]

6) Ask the radial acceleration of the 3rd person.

We substitute.

         a₃ = α (½ R)

We solve.

        a₃ = ½ a

In conclusion, using the rotational kinematics relations we can find the results for the questions about the movement of the three people on the turntable are

      1 and 2) All periods are equal, T₂ = T and T₃ = T.

       3) The linear velocity of the 2nd person is: v₂ = [tex]\frac{3}{5} \ v[/tex]  

       4) The linear velocity of the 3rd person is: v₃ = ½ v

       5) The linear acceleration of the 2nd person is: a₂ = [tex]\frac{3}{5} \ a[/tex]  

      6) The linear acceleration of the 3rd person: a₃ = ½ a

Learn more about rotational kinematics here:  brainly.com/question/14524058

Answer

given,

mass of block = m = 2.3 Kg

spring constant = k = 150 N/m

speed = 2.3 m/s

a) we know,

   [tex]\omega = \sqrt{\dfrac{k}{m}}[/tex]

   [tex]\omega = \sqrt{\dfrac{150}{2.3}}[/tex]

             ω = 8.08 rad/s

      v = A ω

      2.3 = A x 8.08

         A = 0.285 m

b) maximum acceleration

        a = A ω²

        a = 0.285 x 8.08²

        a = 18.58 m/s²

c) acceleration is minimum at mean position the minimum value of acceleration is zero.

d) time period

   [tex]T = \dfrac{2\pi}{\omega}[/tex]

   [tex]T = \dfrac{2\pi}{8.08}[/tex]

            T = 0.778 s

        t = 4.1 x 0.7778

        t = 3.188 s