What is the potential energy of a 2 kg plant that is on a windowsill 1.3 m high

a) The inductance of the solenoid is approximately 0.0068 H.

b) The energy stored in the inductor is approximately 0.012 J.

a) The inductance (L) of an air-core solenoid can be calculated using the formula L = (μ₀n²A) / ℓ, where μ₀ is the permeability of free space (4π × 10⁻⁷ T·m/A), n is the number of turns, A is the cross-sectional area of the solenoid, and ℓ is the length of the solenoid.

To calculate the cross-sectional area, we need the diameter (d) of the solenoid. The formula for the cross-sectional area of a circle is A = (π/4)d². Given the diameter, we can calculate the cross-sectional area.

Using the given values of the number of turns, length, diameter, and the constants μ₀ and π, we can calculate the inductance of the solenoid.

b) The energy stored in an inductor (W) can be calculated using the formula W = (1/2)LI², where L is the inductance of the solenoid and I is the current flowing through the wire.

Using the calculated value of the inductance from part a and the given current, we can calculate the energy stored in the inductor.

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Answer:

F = 4.48N

Explanation:

In order to calculate the net gravitational force on the rocket, you take into account the formula for the gravitational force between two objects, which is given by:

\(F=G\frac{m_1m_2}{r^2}\)         (1)

G: Cavendish's constant = 6.674*10^-11 m^3kg^-1s^-2

r: distance between the objects

You have a rocket at the middle of the distance between Earth and Moon, then, you have opposite forces on the rocket.

If you assume the origin of a system of coordinates at the rocket position, with the Moon to the left and the Earth to the right, you have:

\(F=G\frac{M_em}{r_1^2}-G\frac{M_mm}{r_2^2}\)       (2)

Me: mass of the Earth = 5.98*10^24 kg

Mm: mass of the Moon = 7.35*10^22 kg

m: mass of the rocket = 1200kg

r1: distance from the rocket to the Earth = 3.0*10^8m

r: distance between rocket and Moon = 3.84*10^8m - 3.0*10^8m = 8.4*10^7m

You replace the values of the parameters in the equation (2):

\(F=Gm[\frac{M_e}{r_1^2}-\frac{M_m}{r_2^2}]\\\\F=(6.674*10^{-11}m^3kg^{-1}s^{-2})(1200kg)[\frac{5.98*10^{24}kg}{(3.0*10^8m)^2}-\frac{7.35*10^{22}kg}{(8.4*10^7m)^2}]\\\\F=4.48N\)

The net force exerted over the rocket is 4.48N

The net gravitational force on the rocket from the earth and moon is 4.48N.

The gravitational force between two objects is:

F = Gm₁m₂/r²

Where m₁, m₂, are masses, r is the distance between them and G is gravitational constant = 6.67 * 10⁻¹¹ Nm²/kg²

Let us assume that the rocket is at the origin with the Moon to the left and the Earth to the right. Hence:

\(F=G[\frac{m_em}{r_1^2} -\frac{m_mm}{r_2^2} ]\)

Where me is the mass of earth =  5.98 × 10^24 kg,

r₁ is the distance from earth to rocket = 3.0 × 10^8 m,

m is the mass of rocket = 1200 kg,

mm is the mass of moon = 7.35 × 10^22 kg,

r₂ is the distance from moon to rocket = 3.84 × 10^8 m - 3.0 × 10^8 m = 0.84 × 10^8 m

Hence:

\(F=6.67*10^{-11}[\frac{1200*5.98*10^{24}}{3*10^8}-\frac{1200*3*10^{22}}{0.84*10^8} ]\\\\F=4.48\ N\)

The net gravitational force on the rocket from the earth and moon is 4.48N.

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Answer:

The magnitude of third force is 74.4 N and direction of third force  is 72.8 degrees South.

Explanation:

Let F1, F2 and F3 are three forces exerted on an object.

\(\theta_1=90^{\circ}\)

\(\theta_2=60^{\circ}\)

\(|F_1|=33 N\)

\(|F_2|=44 N\)

We have to find the direction and magnitude of third force i.e F3.

\(F_{1x}=33cos(90^{\circ})=0 N\)

\(F_{1y}=33sin(90^{\circ})=33 N\)

\(F_{2x}=44cos(60^{\circ})=22 N\)

\(F_{2y}=44 sin(60^{\circ})=22\sqrt{3}=38.11 N\)

Now,

x-component of  resultant

\(R_x=F_{1x}+F_{2x}=0+22=22 N\)

y-component of resultant

\(R_y=F_{1y}+F_{2y}=33+38.11=71.11 N\)

\(|R|=\sqrt{R^2_x+R^2_y}\)

\(|R|=\sqrt{(22)^2+(71.11)^2}=74.4 N\)

\(\theta=tan^{-1}(\frac{R_y}{R_x})\)

\(\theta=tan^{-1}(\frac{71.11}{22})=72.8^{\circ}\) South

Hence, the magnitude of third force is 74.4 N and direction of third force  is 72.8 degrees South.

The magnitude and direction of the third force is;

F3 = 74.44 N

θ3 = 72.81° in the south direction

We are given;

F1 = 33 N

F2 = 44 N

θ1 = 90°

θ2 = 60°

Let the third force be F3 which will serve as the resultant

Let's first find the x and y component of the forces.

F1x = F1 cos θ1

F1x = 33 × cos 90

F1x = 0 N

F1y = F1 sin θ1

F1y = 33 × sin 90

F1y = 33 N

F2x = F2 cos θ2

F2x = 44 × cos 60

F2x = 22 N

F2y = F2 sin θ2

F2y = 44 × sin 60

F2y = 38.11 N

Thus, the resultant of the x component is;

F3x = F1x + F2x

F3x = 0 + 22

F3x = 22 N

The resultant of the y component is;

F3y = F1y + F2y

F3y = 33 + 38.11

F3y = 71.11 N

Thus, magnitude of resultant of the F3 force is;

F3 = √((F3x)² + (F3y)²)

F3 = √(22² + 71.11²)

F3 = 74.44 N

The direction of the resultant of F3 is;

θ3 = tan^(-1) F3y/F3x

θ3 = tan^(-1) 71.11/22

θ3 = 72.81° in the south direction

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The distribution factor (DF) is used to distribute the load from a beam to its supports or the adjacent beams of the grid. Distribution factor of CC1 = 0.38, Distribution factor of CC2 = 0.38.

The DF of a member is the portion of the load that is transferred to the adjacent member or support. Distribution factor of member CB = 0.25, Distribution factor of member CD = 0.21We are required to calculate the distribution factor for members CC1 and CC2, given they are equal. Distribution of the load on CC1 and CC2. The load acting on member CD = 20 kN. The distribution factor of CD = 0.21. Therefore, the load transferred to member CC1 = 0.21 × 20 kN = 4.2 kNAnd, the load transferred to member CC2 = 0.21 × 20 kN = 4.2 kN. Now, let's distribute the load acting on CB.The load acting on member CB = 30 kN. The distribution factor of CB = 0.25Therefore, the load transferred to member CC1 = 0.25 × 30 kN = 7.5 kN And, the load transferred to member CC2 = 0.25 × 30 kN = 7.5 kN. We know that the loads on CC1 and CC2 are equal. Hence, the distribution factor for members CC1 and CC2 is as follows: Distribution factor of CC1 = (4.2 + 7.5) kN / 30 kN = 0.3833 ≈ 0.38Distribution factor of CC2 = (4.2 + 7.5) kN / 30 kN = 0.3833 ≈ 0.38.

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The weight of the automobile is calculated to be as 12.46 kN.The weight of the automobile can be found by adding the weights supported by the front and rear wheels.

Since the scale balances at 7.23 kN when only the front wheels are on it, this means that the weight supported by the front wheels is 7.23 kN. Similarly, the weight supported by the rear wheels is 5.23 kN when only the rear wheels are on the scale. Therefore, the weight of the automobile is:

Weight = Weight supported by front wheels + Weight supported by rear wheels
Weight = 7.23 kN + 5.23 kN
Weight = 12.46 kN

So, the weight of the automobile is 12.46 kN.

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Answer:

The power of water transferred each hour through a 500m high dam if 500m² are cleared every hour is approximately 4.41 GW

Explanation:

To answer this question, we need to know the density of water, the gravitational acceleration, and the efficiency of the dam. Let's assume that the density of water is 1000 kg/m³, the gravitational acceleration is 9.81 m/s², and the efficiency of the dam is 100%.

The power of water transferred each hour through the dam is given by the formula:

Power = Flow rate x Density x Gravity x Height x Efficiency

where Flow rate is the volume of water that passes through the dam each second, Density is the density of water, Gravity is the gravitational acceleration, Height is the height of the dam, and Efficiency is the efficiency of the dam.

First, let's calculate the flow rate:

Flow rate = Area x Velocity

where Area is the cleared area of 500m² and Velocity is the speed of water passing through the dam.

Assuming that the water is moving at a constant speed, we can use the formula:

Velocity = Height / Time

where Time is the time it takes for the water to pass through the dam.

Since the height of the dam is 500m and we want to know the power transferred each hour, we can convert the time to seconds as follows:

Time = 1 hour / 3600 seconds per hour = 0.000277778 hours

So, the velocity of the water is:

Velocity = 500m / 0.000277778 hours = 1,800,000 m/s

Now we can calculate the flow rate:

Flow rate = 500m² x 1,800,000 m/s = 900,000 m³/s

Finally, we can calculate the power of water transferred each hour through the dam:

Power = Flow rate x Density x Gravity x Height x Efficiency

Power = 900,000 m³/s x 1000 kg/m³ x 9.81 m/s² x 500m x 1

Power = 4,405,500,000 watts or approximately 4.41 GW

Therefore, the power of water transferred each hour through a 500m high dam if 500m² are cleared every hour is approximately 4.41 GW.

A group of two or more atoms.

Mass.

Because mass doesn't depend on weight but weight depends on mass.

1) push down on the end of the lever, and 2) 3/4 of the way from the fulcrum

When a stone is hurled horizontally from the top of a 31.2 m tall cliff, its horizontal initial velocity is 39.25 m/s.

By measuring the ball's diameter d and dividing it by the time t it takes for it to cross the photogate, one can also calculate the ball's starting horizontal velocity. Therefore, Vo = d/t. The kinematics equations of motion can be used to calculate the horizontal velocity of a projectile motion made by a person or an item.

Height of cliff is 31.2 m

distance is 21.4

t = √2h/g

t = √2×31.4/9.8

t = √62.8/9.8

t = 0.80

V = 31.4/0.08

V = 39.25 m/s.

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Answer:

5

step by step explanation ;

(8) times (7)

equals 56

The solubility of lead nitrate is 52 grams per 100 milliliters of water at 0°C. Since the question does not specify the temperature, we will assume it is 0°C.To calculate the number of millimeters of water required to dissolve 5 g of lead nitrate,

we must first calculate the number of milliliters of water required to dissolve 5 g of lead nitrate.52 grams of lead nitrate can dissolve in 100 milliliters of water. So,5 g of lead nitrate can dissolve in x milliliters of water.= (5 g / 52 g) × 100 mL= 9.6 mLTherefore, 5 g of lead nitrate can dissolve in 9.6 mL of water. However, the question is asking for the number of millimeters of water, not milliliters. Since 1 milliliter of water weighs 1 gram,9.6 milliliters of water will weigh 9.6 grams.Thus, 5 g of lead nitrate can dissolve in 9.6 millimeters of water.

main answer:5 g of lead nitrate can dissolve in 9.6 millimeters of water. :52 grams of lead nitrate can dissolve in 100 milliliters of water. So, 5 g of lead nitrate can dissolve in x milliliters of water.= (5 g / 52 g) × 100 mL= 9.6 mLTherefore, 5 g of lead nitrate can dissolve in 9.6 mL of water. However, the question is asking for the number of millimeters of water, not milliliters. Since 1 milliliter of water weighs 1 gram,9.6 milliliters of water will weigh 9.6 grams.Thus, 5 g of lead nitrate can dissolve in 9.6 millimeters of water.

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Answer:the universe will expand forever. there are small ripples in the microwave background, the seeds of galaxies.

Explanation:

The Acceleration of the system is 6.41 m/s².

Given,

α= 15°, m₁ = 7kg

β= 65°, m₂ = 11 kg

Let, a be the acceleration and T is the tensions at the end it's the cord.

Let, the mass m₂ be coming down along the inclined plane along the inclined surface towards downward m₂g sin β and the tension in the upward direction,

Resultant force, m₂a=m₂g sin β -T

11a=((11) ×g sin 65°) -T  ...(i)

Now, considering the motion of m₁ which moves downwards, the forces are m₁g sinα, and T both are acting downwards.

Resultant force m₁a = m₁g sin α+T

7a =7g sin 15°+T  ...(ii)

Solving both the equations by adding them,

18a=11gsin 65°+7g sin 15°-T+T

18a=11gsin 65°+7g sin 15°=115.45

a=115.45/18=6.41 m/s²

Hence, the Acceleration of the system is 6.41 m/s².

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A bullet and a base ball

Objects with a small mass can produce a large acceleration if they experience a large force. This is described by Newton's second law of motion, which states that the acceleration of an object is directly proportional to the net force acting on the object and inversely proportional to its mass.

It's important to note that while these objects can produce a large acceleration, they can also cause significant damage or harm if not handled properly. It's important to always take appropriate safety precautions when working with objects that have the potential to produce large forces and accelerations.

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When two heavy blocks were connected by a uniform rope that has a mass of 4.00 kg, acceleration of the system was found to be 2.7 m/s²; the tension at the top of heavy rope was 136.5 N;  the tension exerted at the midpoint of the rope was112.5N.

Tension is defined as the act of stretching or straining or the condition of  a substance being stretched or strained.

a) Acceleration of the system can be calculated as

a=Fn/m

={200-(16x9.8)}/16

=2.7m/s²

so therefore the acceleration of the system was calculated and found to be=2.7m/s²

b) the tension of the heavy rope at the top can be calculated as

200-50-T1=5(2.7)

T1=136.5N

the tension at the top of heavy rope was 136.5 N

c)the tension at the middle point of the rope was

T2-9g = 9(a)

         =9(g+a)

         =9(9.8+2.7)

         =112.5 N

so here the tension exerted at the midpoint of the rope was 112.5N.

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Ari's final vertical jump height is 5.2 m.

Ari's final vertical jump height is calculated from the law of conservation of energy as shown below.

P.E ( of Ari ) = P.E ( of Bari )

m1gh1 = m2gh2

where;

h1 = ( m2h2 ) / ( m1 )

h1 = ( 70 x 4 ) / ( 50 )

h1 = 5.6 m

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If the Sun were replaced by a star that is 3 times brighter, the habitable zone in our solar system would undergo significant changes. The habitable zone, also known as the Goldilocks zone, refers to the region around a star where conditions are suitable for the presence of liquid water and potentially life as we know it.

A star that is 3 times brighter than the Sun would emit more radiation, including visible light and other forms of electromagnetic radiation. This increase in brightness would have several effects on the habitable zone:

Expansion of the habitable zone: The increased brightness of the star would lead to the habitable zone expanding farther away from the star. This means that planets previously located outside the habitable zone might now fall within it, potentially increasing the number of planets with the potential for liquid water and habitability.

Increased radiation and heat: With a brighter star, the habitable zone would experience higher levels of radiation and heat. This could lead to higher surface temperatures on planets within the habitable zone, potentially pushing the upper limits of what is considered habitable for life as we know it.

Changes in planetary atmospheres: The increased radiation and heat from a brighter star could impact the composition and stability of planetary atmospheres. There could be an increase in atmospheric evaporation, loss of volatile compounds, and changes in the balance of greenhouse gases, potentially affecting the overall habitability of planets.

It's important to note that the specific effects on the habitable zone would depend on various factors, including the characteristics of the star and the composition and properties of the planets within the system. Additionally, the concept of habitability is complex and depends on many factors beyond just the presence of liquid water, such as atmospheric conditions and the presence of suitable chemical building blocks for life.

In summary, if the Sun were replaced by a star 3 times brighter, the habitable zone in our solar system would likely expand and experience increased radiation and heat. This would have implications for the potential habitability of planets within the zone and could lead to changes in planetary atmospheres.

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.c ...energy

Explanation:

Amplitude does not affect wavelength. It also does not affect wave speed. Amplitude is the energy of the wave measured from the rest position to the top of the crest. A wave with more energy has a higher up crest/ higher amplitude.

As the amplitude of a mechanical wave increases, the quantity which increases is: C. energy.

A mechanical wave can be defined as a type of wave that requires a medium for its propagation and it creates a disturbance in the medium. Thus, a good example of mechanical wave is sound wave, which is typically transported in a perpendicular direction.

Amplitude refers to the maximum displacement of a wave when measured from its equilibrium position. Also, the amplitude of a mechanical wave has no effect on the following:

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The proton and electron, being opposite charges, would accelerate toward each other. Thus, 3rd option is correct.

According to Coulomb's law, the force between two charged particles is given by:

F = (k * |q₁ * q₂|) / r²

where F is the force between the particles, k is the electrostatic constant, q₁ and q₂ are the magnitudes of the charges of the particles (in this case, the charge of the proton and the charge of the electron), and r is the distance between the particles.

In the case of a proton and an electron, the proton has a positive charge (+e) and the electron has a negative charge (-e), where e is the elementary charge. Since the charges are opposite in sign, the product q₁ * q₂ is negative.

Therefore, the force between the proton and the electron is attractive, causing them to accelerate toward each other. This acceleration will continue until they collide or until external factors come into play (such as the presence of other particles or forces) that may alter their motion.

It's important to note that in a typical atomic or molecular system, electrons are usually bound to nuclei due to the attractive electrostatic forces between them. However, if an electron and a proton are initially separated and have no other influences, they will accelerate toward each other due to their opposite charges.

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Answer:

No, it is not necessary for them to have same mass.

Explanation:

Let both bodies have a density d1 and d2 respectively.

Since their volumes are equal V1 = V2

we know that, https://tex.z-dn.net/?f=%5Cfrac%7Bmass%7D%7Bvolume%7D

Hence, d1 =  and d2 =    

Taking the ratio of densities,we get

This implies that unless the bodies have same densities, the mass of the two bodies will not be same.

Answer:1.36078

Explanation:

Answer:

B- He is correct. Evaporation is a physical change, but dissolving salt in water is a chemical change. The change in mass is evidence that a chemical change occurred.

Explanation: Dissolving Salt in water is a Chemical Change, Because the Salt arrangement is different in solid state than dissolved in water. As we can see in the image below, once the Salt is dissolved, it is separated into its ions, Na+ and Cl- Now, The evaporation process is a physical change, because the water doesn´t change its configuration H20 and it only changes its form, as we can see in the image below.

Answer:

B) He is correct. Evaporation is a physical change, but dissolving salt in water is a chemical change. The change in mass is evidence that a chemical change occurred.

Explanation:

hope this helps!! :)

Answer:

electrical to mechanical

Explanation:

Answer:

When the rock has fallen 12.0 m;

The kinetic energy of the rock is approximately 3,531.6 J

The potential energy of the rock is approximately 6,768.9 J

Explanation:

The question relates to the characteristic constant total mechanical energy of a body

The mass of the rock that falls, m = 30.0 kg

The height of the cliff from which the rock falls, h₁ = 35.0 m

The required information = The kinetic and potential energy when the rock has falling 12.0 m

The kinetic energy is given by the formula, K.E. = 1/2×m×v²

The potential energy is given by the formula, P.E. = m·g·h

Where;

g = The acceleration due to gravity ≈ 9.81 m/s²

The velocity of the rock after falling through a given height, h is given by the formula, v² = 2·g·Δh

The total mechanical energy of the rock, M.E. = K.E. + P.E. = Constant

At the height of the cliff before falling, Δh =0, therefore v₁ = 0, therefore, K.E. = 1/2×m×v₁² = 0 J

The potential energy at the cliff before the rock begins to fall, P.E. is goven as follows;

P.E. = 30.0 kg × 35.0 m × 9.81 m/s² = 10,300.5 J

At the top of the cliff, M.E. = K.E. + P.E. = 0 J+ 10,300.5 J = 10,300.5 J

∴ M.E. = 10,300.5 J

When the rock has fallen, 12.0 m, Δh = 12.0 m, the speed of the rock, v₂, is given as follows;

v₂² = 2 × 9.81 m/s² × 12.0 m = 235.44 m²/s²

v₂ = √(235.44 m²/s²) ≈ 15.344 m/s

∴ When the rock has fallen 12.0 m, K.E., is given as follows;

K.E. = 1/2×m×v₂²

K.E. = 1/2 × 30.0 kg × 235.44 m²/s² = 3,531.6 J

When the rock has fallen 12.0 m the kinetic energy, K.E. = 3,531.6 J

When the rock has fallen 12.0 m, M.E. = P.E. + K.E.

M.E. = Constant = 10,300.5 J

K.E. = 3,531.6 J

∴ 10,300.5 J = P.E. + 3,531.6 J

P.E. = 10,300.5 J - 3,531.6 J = 6,768.9 J

∴ When the rock has fallen 12.0 m, the potential energy, P.E. = 6,768.9 J.

Answer: The student changed more than one independent variable.

Explanation:

     What is the difference between dependent and independent variables?

           The independent variable is the variable you change. For example in this experiment, changing the temperature of the water.

           The dependent variable is the variable you don't change yourself, but changes based on the independent variable. In this experiment, it would be the time it takes for the sugar to dissolve.

     If we change two independent variables (in this student's experiment, the temperature and amount of water), then we cannot clearly see the answer to our question as we do not have the correct data.

     This is why you should only change one independent variable when conducting an experiment.

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Answer:The answer is C: any fundamental force

Explanation:

A P E X

Based on the measurements, the maximum height reached by the puck in trial 6 is determined as 5.1 m.

The maximum height reached by the puck is calculated by applying the principle of conservation of energy as shown below;

From the trials, the puck was thrown upwards with initial velocity of 10 m/s and the mass of the puck is given as 400 g.

With these values we can predict the maximum height the puck will reach by applying the principle of conservation of energy.

Potential energy of the puck at maximum height = Kinetic energy of the puck at minimum height

P.E = K.E

mgh = ¹/₂ mv²

h = ¹/₂ (v² / g )

where;

The maximum height reached by the puck is calculated as follows;

h =  ¹/₂ (10² / 9.8 )

h = 5.1 m

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Answer:

Ans : C

Explanation:

C. The force of gravity acting on the chair

Answer:

D. the force from the chair on the student

Explanation:

Newton's third law: If an object A exerts a force on object B, then object B must exert a force of equal magnitude and opposite direction back on object A.

This law represents a certain symmetry in nature: forces always occur in pairs, and one body cannot exert a force on another without experiencing a force itself. 

Hope this helps :)