A package is dropped from the plane which is flying with a constant horizontal velocity of va = 150 ft/s. Determine the normal and tangential components of acceleration and the radius of curvature of the path of motion (a) at the moment the package is released at a, where it has a horizontal velocity va = 150 ft/s, and (b) just before it strikes the ground at

b

Answer:

I think true but im not sure sorry if this didnt help/

Explanation:

Answer:

true 50%

Explanation:

The altitude above the Earth's surface where the acceleration due to gravity is equal to g/7 is approximately 4.9019353 × 10^7 meters, or 49,019,353 meters, or 49,019.353 kilometers.

The acceleration due to gravity, denoted as "g," is approximately 9.8 meters per second squared (m/s²) near the Earth's surface. To determine the altitude at which the acceleration due to gravity is equal to g/7, we can use the formula for the acceleration due to gravity as a function of distance from the center of the Earth.

The formula for the acceleration due to gravity (g') at a certain distance (h) from the Earth's center is given by:

g' = (G * M) / (R + h)²

where:

- G is the gravitational constant (approximately 6.67430 × 10^(-11) m³/(kg·s²)),

- M is the mass of the Earth (approximately 5.972 × 10^24 kg),

- R is the mean radius of the Earth (approximately 6,371,000 meters),

- h is the distance above the Earth's surface.

Given that g' = g/7, we can set up the equation:

g/7 = (G * M) / (R + h)²

Rearranging the equation, we can solve for h:

h = sqrt((G * M) / (g/7)) - R

Substituting the known values, we get:

h = sqrt((6.67430 × 10^(-11) * 5.972 × 10^24) / (9.8/7)) - 6,371,000

Evaluating this equation will give us the altitude above the Earth's surface where the acceleration due to gravity is equal to g/7.

the altitude above the Earth's surface where the acceleration due to gravity is equal to g/7 is approximately 4.9019353 × 10^7 meters, or 49,019,353 meters, or 49,019.353 kilometers.

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Answer:

5.84×10^-10 N

Explanation:

F=G×ms×mr/r^2

ms=50 kg

mr= 70 kg

r=20 m

F=6.67×10^-11 N×m^2/kg^2×50 kg×70 kg/(20 m)^2

F=5.84×10^-10 N

The gravitational force is [tex]5.84*10^{-10} N[/tex].

The gravitational force is a force that attracts any two objects with mass.

[tex]F=G{\frac{m_1m_2}{r^2}}[/tex]

Where,

F = force

G = gravitational constant

[tex]m_{1}[/tex] = mass of object 1

[tex]m_{2}[/tex] = mass of object 2

r = distance between centers of the masses

[tex]m_{1}[/tex] =      70kg

[tex]m_{2}[/tex] = 50kg

r       =      20 m

G     =       [tex]6.67*10^{-11} Nm^2/kg^2[/tex]

[tex]F= \frac{6.67*10^{-11}*70*50}{20^2}[/tex]

[tex]F = 5.84*10^{-10} N[/tex]

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On which planet would a 15.0 kg object weigh the most?The weight of an object is affected by its mass and the strength of gravity. On planets with a greater mass, the force of gravity is stronger than on those with a lower mass. Hence, the weight of an object is greater on planets with more mass, according to Newton's second law of motion.

The weight of a 15 kg object is the largest on the planet Jupiter because Jupiter has the highest mass among all the planets of our solar system. The answer is written below:Jupiter is the planet where a 15.0 kg object would weigh the most.Round the answer to the nearest whole numberOn Jupiter, the acceleration due to gravity is 24.79 m/s². The weight of an object on Jupiter is calculated using the formula:

F = m × gwhereF is the force of gravity (in newtons),m is the mass of the object (in kilograms), andg is the acceleration due to gravity (in meters per second squared).On Jupiter, the weight of a 15.0 kg object is:F = 15.0 kg × 24.79 m/s²= 371.85 N = 372 N (to the nearest whole number)Therefore, on Jupiter, a 15.0 kg object would weigh about 372 N.

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The resistors are connected in parallel between the terminals of a battery.

Note: If those resistors are connected in parallel, then their equivalent resistance is already 5 ohms, even if they're still in the drawer or in a box on the shelf. They don't have to be connected to a source of voltage for that to happen.

As you drive from the equator toward the north pole, the altitude of the celestial north pole star would increase.

This is due to the Earth's rotation on its axis, which causes the celestial pole to appear to move around the sky in a circle once every 24 hours. The altitude of the celestial north pole star is equal to the observer's latitude, so as you move towards the north pole, your latitude increases and so does the altitude of the celestial north pole star. At the equator, the celestial pole is located on the horizon, so the altitude of the celestial north pole star would be zero.

However, as you move towards the north pole, the altitude of the celestial north pole star would increase until it reaches 90 degrees at the north pole. In conclusion, the altitude of the celestial north pole star would increase as you drive from the equator toward the north pole, this is due to the Earth's rotation on its axis, which causes the celestial pole to appear to move around the sky in a circle once every 24 hours. The altitude of the celestial north pole star is equal to the observer's latitude, so as you move towards the north pole, your latitude increases and so does the altitude of the celestial north pole star.

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The ratio of the electron's speed to the speed of light is approximately 0.859.

According to special relativity, the total relativistic energy of a particle is given by the equation:

E = γmc^2

where E is the total energy of the particle, m is its rest mass, c is the speed of light in a vacuum, and γ is the Lorentz factor, which is defined as:

γ = 1 / sqrt(1 - (v^2 / c^2))

where v is the velocity of the particle.

Given that the final kinetic energy of the electron is 2.35 MeV, we can equate this energy to the relativistic energy equation:

E = γmc^2

Rearranging the equation to solve for γ:

γ = E / (mc^2)

The rest mass of an electron, m, is approximately 9.10938356 × 10^-31 kg, and the speed of light, c, is approximately 2.998 × 10^8 m/s.

Converting the final kinetic energy of the electron from MeV to joules:

2.35 MeV = 2.35 × 10^6 × 1.6 × 10^-19 J

= 3.76 × 10^-13 J

Substituting the values into the equation for γ:

γ = (3.76 × 10^-13 J) / ((9.10938356 × 10^-31 kg) × (2.998 × 10^8 m/s)^2)

Simplifying the equation:

γ ≈ 4.18 × 10^8

To find the ratio of the electron's speed to the speed of light, we can use the equation:

v / c = sqrt(1 - (1 / γ^2))

Substituting the value of γ:

v / c = sqrt(1 - (1 / (4.18 × 10^8)^2))

Simplifying the equation:

v / c ≈ 0.859

Therefore, the ratio of the electron's speed to the speed of light is approximately 0.859.

Using special relativity, the ratio of the electron's speed to the speed of light is approximately 0.859.

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Answer:

3kWx(3/60)h=0.15kwh

1. Some inventions in history which are related to magnetism include the compass, the electromagnet, and the electric motor. The discovery of magnetism dates back to around 600 B.C. in China when they found a naturally occurring magnetic rock, which is now called magnetite. They discovered that the magnetic rock had an effect on iron.

2. Faraday disc requires a magnetic field to operate. It works based on electromagnetic induction. The principles of the Faraday disc can be explained using FIGURE 2. When a magnetic field is applied perpendicular to a disc, it creates a voltage difference between the center and the outer edge of the disc. This voltage can be used to power an electrical device.3. The Faraday disc produces a magnetic force on electrons that are moving along the conducting path. The magnetic force acts perpendicular to the direction of the electron's velocity and the magnetic field.

.4. The magnitude of the magnetic force is greater on the electrons located at the rim than on the electrons located at the center of the disc. The magnetic force is directly proportional to the distance from the center of the disc. Therefore, the magnetic force is stronger at the rim than at the center.5. The work done by the magnetic force in moving a charge along the radial line between the center and the rim is given by the formula W = (1/2)mv². The relation between work done and generated emf is given by the formula W = qEMF.

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The direction of the force experienced by the negative charge when it passes the wire is South.

At the very instant when the negative charge passes the wire, it experiences force in the south direction. This is because of the interaction between the magnetic field due to the current-carrying wire and the electric field of the charge. Since the current-carrying wire produces a magnetic field around it, it exerts a magnetic force on the negative charge, perpendicular to both the magnetic field and the direction of the charge. This force causes the negative charge to experience a southward force. Hence, option (g) South is the correct answer.

When a long straight wire carries a current, it produces a magnetic field around it. When a charge moves through this magnetic field, it experiences a magnetic force due to the interaction of the magnetic field and the electric field of the charge. The magnitude of this force is given by F = qvBsinθ, where F is the force, q is the charge, v is the velocity of the charge, B is the magnetic field, and θ is the angle between the velocity of the charge and the direction of the magnetic field. In this case, the negative charge is moving vertically down and perpendicular to the direction of the wire. Therefore, the angle between the velocity of the charge and the direction of the magnetic field is 90 degrees, and the sine of 90 degrees is 1. So, the magnitude of the force experienced by the charge is F = qvB. Since the charge is negative, the force is in the opposite direction of the velocity, which is towards the south.

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In a large country, the distribution of the heights of married men is normally distributed with a mean of 68.9 inches and a standard deviation of 2.9 inches.

The normal distribution is a commonly observed pattern in various natural phenomena, including human characteristics such as height. In this case, the heights of married men in the country follow a normal distribution.

The mean height of married men is given as 68.9 inches, which represents the center or average height of the distribution. The standard deviation of 2.9 inches indicates the spread or variability of heights around the mean. A smaller standard deviation implies a narrower and more concentrated distribution, while a larger standard deviation indicates a wider and more dispersed distribution.

Using the normal distribution properties, we can make various calculations. For example, we can find the probability of a married man's height falling within a certain range by calculating the area under the curve. The z-score formula (z = (x - μ) / σ) is often used to standardize values and determine their relative position within the distribution.

In a large country, the heights of married men follow a normal distribution with a mean of 68.9 inches and a standard deviation of 2.9 inches. This information allows us to analyze and make predictions about the heights of married men using the properties of the normal distribution.

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When the V filter is used, the color that is detected is green. The V filter selectively allows green light to pass through while blocking other wavelengths.

The V filter selectively allows green light to pass through while blocking other wavelengths. Therefore, when the V filter is applied, only green light is transmitted and detected.

Additionally, if multiple V filters are used, the green color will appear darker. This occurs because each V filter further restricts the passage of light, allowing only a narrower range of green wavelengths to pass through. As a result, the intensity of the transmitted green light decreases, creating a darker shade of green.

It's important to note that the perception of color is subjective and can vary depending on individual differences in visual perception. The statement regarding the darkening of green with multiple V filters assumes ideal conditions and may vary in practical scenarios.

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Answer:

B is the answer

trust me

naturo na kasi samin yan eh

To store four times as much energy in a combination of two capacitors by adding a single capacitor to this one. Thus, the capacitance of the added capacitor should be: 3(240 pF) = 720 pF.

To begin with, the energy stored by a capacitor in an electrical circuit is given by the equation:

E = 1/2CV^2...

where C is the capacitance of the capacitor and V is the voltage across it. The energy stored in the circuit is equal to the energy stored in the capacitor,

so the formula can be rewritten as follows:

E = 1/2C(∆V)^2...

where ∆V is the voltage across the capacitor.

The energy in the capacitor can be increased by increasing the capacitance or the voltage across it, or by increasing both.

The problem specifies that it is desired to store four times as much energy in a combination of two capacitors by adding a single capacitor to this one.

So, the energy stored in two capacitors can be expressed as:

E1 + E2 = 4E...

where E is the energy stored in the single capacitor and E1 and E2 are the energies stored in the two capacitors after adding the single capacitor.

Let's say the capacitance of the single capacitor is C and the capacitance of the added capacitor is C'.

Then, the total capacitance of the two capacitors can be expressed as:

C total = C + C'...

and the energy stored in each capacitor can be expressed as:

E = 1/2C(∆V)^2 and

E' = 1/2C'(∆V')^2...

where ∆V is the voltage across the single capacitor, and ∆V' is the voltage across the added capacitor.

The voltage across each capacitor is the same,

so ∆V = ∆V'.

Substituting these equations into the first equation,

we get:

1/2C(∆V)^2 + 1/2C'(∆V)^2 = 4[1/2C(∆V)^2].

which can be simplified to: C' = 3C...

Therefore, the capacitance of the added capacitor should be 3 times the capacitance of the single capacitor.

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The electric energy dissipated in the process is 1.167 J.

The electric energy dissipated in the process can be calculated using the formula:

Electric energy dissipated = (average power) * (time)

To find the average power, we need to calculate the average induced emf and the current in the loop.

The average induced emf can be calculated using Faraday's law of electromagnetic induction:

Average induced emf = (change in magnetic flux) / (change in time)

The change in magnetic flux can be calculated by subtracting the final magnetic flux (0) from the initial magnetic flux. Since the loop is perpendicular to the magnetic field, the magnetic flux through the loop is given by:

Initial flux = B * A

= 0.365 T * (0.265 m)²

= 0.3425 T·m²

The change in time is given as 34.0 ms, which is equal to 0.034 s.

Therefore, the average induced emf is:

Average induced emf = (0 - 0.3425 T·m²) / 0.034 s = -10 V/s

Using Ohm's law, we can calculate the current in the loop:

Current = emf / resistance = -10 V/s / 4.00 Ω

= -2.5 A

The average power is given by:

Average power = (current²) * resistance

= (-2.5 A)² * 4.00 Ω

= 25 W

Finally, the electric energy dissipated is:

Electric energy dissipated = (average power) * (time)

= 25 W * 0.034 s

= 1.167 J

Therefore, the electric energy dissipated in the process is 1.167 J.

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The position of the body at time t is given by the equation s(t) = (1/2)(9.8)(t²) + C if the velocity is v = ds/dt.

To find the body's position at time t, we need to integrate the velocity function with respect to time.

Given that v = 9.8t, we can integrate this function to find the position function, s(t).

∫v dt = ∫(9.8t) dt

Applying the power rule of integration, we have

s(t) = (1/2)(9.8)(t²) + C

Here, C is the constant of integration, which represents the initial position of the body. Since the initial position is not given in the question, we'll leave it as C for now.

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The orbital radius of the farther asteroid will be less than twice the orbital radius of the closer one.

The orbital radius of an object in orbit around a central body depends on its orbital period. Kepler's third law states that the square of the orbital period of a planet or asteroid is directly proportional to the cube of its orbital radius. Mathematically, this relationship can be expressed as [tex]T^2[/tex] ∝ [tex]R^3[/tex], where T is the orbital period and R is the orbital radius.

In this scenario, if one asteroid has an orbital period of 2 years and another has an orbital period of 4 years, we can compare their orbital radii. Let's assume the orbital radius of the closer asteroid is R1. According to Kepler's third law, [tex](2)^2[/tex]∝ [tex]R1^3[/tex]. Similarly, let's assume the orbital radius of the farther asteroid is R2. Therefore, [tex](4)^2[/tex] ∝ [tex]R2^3[/tex].

By comparing these two equations, we can see that [tex](4)^2/(2)^2 = R2^3/R1^3[/tex], which simplifies to [tex]4 = R2^3/R1^3[/tex]. Taking the cube root of both sides gives us [tex]R2/R1 = 3\sqrt4 = 1.587[/tex]. This means that the orbital radius of the farther asteroid will be less than twice the orbital radius of the closer one, indicating that it will be closer to the central body.

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The Ka value of the acid HA is approximately 0.0692 for the equation ha(aq) + [tex]h_2o[/tex](l)⇌[tex]a^-[/tex](aq) + [tex]h_3o[/tex] (aq)

The pH of a solution can be related to the concentration of hydronium ions ([tex]H_3O^+[/tex]) using the equation:

pH = -log[[tex]H_3O^+[/tex]]

Given that the pH of the solution is 0.30, we can calculate the concentration of [tex]H_3O^+[/tex]:

[[tex]H_3O^+[/tex]] = [tex]10^{(-pH)[/tex] = [tex]10^{(-0.30)[/tex]

Now, let's assume that the initial concentration of HA is [HA] M. At equilibrium, a fraction x of HA will ionize to form [tex]A^-[/tex] and [tex]H_3O^+[/tex] ions.

The equilibrium concentrations of HA, [tex]A^-[/tex], and [tex]H_3O^+[/tex] can be expressed as follows:

[HA] = [HA] - x

[A-] = x

[[tex]H_3O^+[/tex]] = x

The equilibrium expression for the dissociation of HA can be written as:

Ka = ([A-] × [[tex]H_3O^+[/tex]]) / [HA]

Substituting the equilibrium concentrations, we have:

Ka = (x × x) / ([HA] - x)

Given that the initial concentration of HA is 1.30 M, we can substitute this value into the equation.

Now, let's solve for x by using the approximation that x is much smaller than [HA]. This assumption is valid for weak acids.

1.30 ≈ 1.30 - x

Solving for x:

x ≈ 0.30

Now we can substitute the values into the equilibrium expression to calculate Ka:

Ka = (0.30 × 0.30) / (1.30 - 0.30)

Ka ≈ 0.0692

Therefore, the Ka value of the acid HA is approximately 0.0692.

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The question is -

If the pH of a 1.30 M solution of the acid HA is 0.30, what is the Ka of the acid?

The equation described by the Ka value is,

HA(aq) + H_2O(l) ⇌ A−(aq)+H_3O+(aq)

The radius of the bubble just before it reaches the surface can be determined using the temperature difference between the bottom and top of the bubble.

The explanation of the answer involves the application of the ideal gas law and the assumption of isothermal conditions during the ascent of the bubble.

To calculate the radius of the bubble, we can make use of the ideal gas law, which states that the pressure of a gas is directly proportional to its temperature. Assuming the bubble follows ideal gas behavior and that the conditions during its ascent are isothermal, we can equate the pressures at the bottom and top of the bubble.

Using the equation P = ρgh, where P is the pressure, ρ is the density of the liquid, g is the acceleration due to gravity, and h is the height of the bubble, we can set up an equation for the pressure difference between the bottom and top of the bubble. Since the temperature is directly related to the pressure, we can express this pressure difference in terms of the temperature difference.

Using the ideal gas law, PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature, we can express the pressure difference in terms of the radius of the bubble.

By equating the pressure difference equation with the ideal gas law equation, we can solve for the radius of the bubble just before it reaches the surface.

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The radius of the output cylinder is 0.11 m.

Radius of the input cylinder, r₁ = 0.01 m

Input force applied, F₁ = 200 N

Mass of the output cylinder, m₂ = 2500 kg

Since more collisions with the piston occur when the area is increased but the number of molecules per cubic centimetre remains constant, the force is proportional to the area.

Force applied on the output cylinder = Weight of the output cylinder

F₂ = m₂g

F₂ = 2500 x 9.8

F₂ = 245 x 10²N

We know that the force applied on an object is directly proportional to the area of the object.

F ∝ A

So, F₁/F₂ = A₁/A₂

F₁/F₂ = (r₁/r₂)²

200/24500 = (r₁/r₂)²

Therefore, the radius of the output cylinder is,

r₂ = r₁√(24500/200)

r₂ = 0.01 x√122.5

r₂ = 0.01 x 11.06

r₂ = 0.11 m

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The distance from the wire at which a current of 7.07 A produces a magnetic field of 8.21 μT is approximately 0.287 meters (or 28.7 cm).

We can use Ampere's law to calculate the magnetic field produced by a current-carrying wire at a certain distance from it.

Ampere's law states that the magnetic field around a long, straight wire is directly proportional to the current and inversely proportional to the distance from the wire.

Mathematically, Ampere's law can be written as:

B = (μ₀ * I) / (2π * r),

where:

B is the magnetic field (in Tesla),

μ₀ is the permeability of free space (4π × 10^(-7) T·m/A),

I is the current in the wire (in Amperes), and

r is the distance from the wire (in meters).

We are given the current I = 7.07 A and the magnetic field B = 8.21 μT. To find the distance r, we need to rearrange the equation and solve for r:

r = (μ₀ * I) / (2π * B).

Now we can substitute the given values and calculate the distance:

r = (4π × 10^(-7) T·m/A * 7.07 A) / (2π * 8.21 × 10^(-6) T)

 ≈ (2.828 × 10^(-6) T·m²/A) / (1.646 × 10^(-5) T)

r ≈ 0.1715 m.

Therefore, the distance from the wire is approximately 0.1715 meters or 17.15 cm.

The distance from the wire at which a current of 7.07 A produces a magnetic field of 8.21 μT is approximately 0.287 meters (or 28.7 cm).

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Answer:

Conductor

Explanation:

A battery holds all of the energy in itself. So without the battery, the circuit cannot work.

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Answer:

i think its its c but im not positive :>

Explanation:

(a)The hydrostatic pressure on the bottom of the tank is 35,910 Pa

(b)The hydrostatic force on the bottom of the tank is 913,104 N

(c)The hydrostatic force on one end of the tank is 117.6 N

(a) Hydrostatic pressure on the bottom of the tank. The hydrostatic pressure is given by the formula: P = ρghWhereP is pressureρ is density g is acceleration due to gravity h is height. We are given: Length of the tank, l = 6 m Width of the tank, w = 4 m. Height of the tank, h = 5 m. Density of kerosene, ρ = 820 kg/m3Depth of kerosene, d = 4.5 m Acceleration due to gravity, g = 9.8 m/s2We need to find the hydrostatic pressure at the bottom of the tank, which is:P = ρghP = 820 * 9.8 * 4.5P = 35,910 Pa. Therefore, the hydrostatic pressure on the bottom of the tank is 35,910 Pa.

(b) Hydrostatic force on the bottom of the tank .The hydrostatic force on the bottom of the tank is given by the formula: F = ρgVWhereF is forceρ is density g is acceleration due to gravity V is volume We are given: Length of the tank, l = 6 m Width of the tank, w = 4 m Height of the tank, h = 5 m Density of kerosene, ρ = 820 kg/m3Depth of kerosene, d = 4.5 m Acceleration due to gravity, g = 9.8 m/s2We need to find the hydrostatic force on the bottom of the tank, which is: F = ρgVThe volume of the tank is given by: lwh = 6 × 4 × 5 = 120 m3The volume of the kerosene is given by: ldw = 6 * 4* 4.5 = 108 m3.The volume of the kerosene is less than the volume of the tank. So the kerosene fills only a part of the tank and the hydrostatic force acts only on the part that is filled with kerosene. The volume of the kerosene is the displaced volume of the kerosene, so: V = 108 m3The hydrostatic force is: F = ρgVF = 820 * 9.8 * 108F = 913,104 N. Therefore, the hydrostatic force on the bottom of the tank is 913,104 N.

(c) Hydrostatic force on one end of the tank We need to find the hydrostatic force on one end of the tank. The end that has dimensions of 4 m x 5 m. Let us assume that the direction along the length of the tank is x, and the direction along the width of the tank is y. The force on one end of the tank will act in the x-direction only, and is given by: F = PA where P is pressure A is area We already know the hydrostatic pressure on the bottom of the tank. We can also find the hydrostatic pressure at the end of the tank, which is at the same height as the bottom of the tank. The depth of kerosene at this end of the tank is given by:4.5 - 5 = -0.5 m. The negative depth indicates that there is no kerosene at this end of the tank. So the hydrostatic pressure is due to the weight of the air above this end of the tank. The hydrostatic pressure at this end of the tank is given by: P = ρghP = 1.2 * 9.8 * 0.5P = 5.88 Pa. The area of the end of the tank is given by: A = lw A = 4 * 5A = 20 m2The hydrostatic force on one end of the tank is: F = PAF = 5.88 * 20F = 117.6 N. Therefore, the hydrostatic force on one end of the tank is 117.6 N.

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By applying this rule, you can determine the force between objects at any given distance by comparing it to the force measured at a reference distance.

F1 / F2 = [tex](r2 / r1)^2[/tex]

The rule based on the measured force between objects that are 10 meters apart to find the force between objects at any distance apart is as follows:

"The force between two objects is inversely proportional to the square of the distance between them."

Mathematically, this can be expressed as:

F1 / F2 = [tex](r2 / r1)^2[/tex]

Where:

F1 is the measured force between objects at a distance r1.

F2 is the force between objects at a distance r2.

r1 is the initial distance between the objects where the force was measured.

r2 is the new distance between the objects at which the force is to be calculated.

By applying this rule, you can determine the force between objects at any given distance by comparing it to the force measured at a reference distance.

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There are three types of electricity – baseload, dispatchable, and variable

The diver would appear to be approximately 1.24 mm tall when viewed from the end of the diving board.

To determine how tall the diver appears to be, we need to consider the effects of refraction caused by the water. Refraction occurs when light travels from one medium (air) to another medium (water) with a different refractive index.

The apparent height of an object submerged in water can be calculated using the formula

Apparent height = Actual height / Refractive index

The refractive index of water is approximately 1.33.

Given:

Actual height of the diver = 1.65 mm

Refractive index of water = 1.33

Applying the formula:

Apparent height = 1.65 mm / 1.33

Calculating the apparent height:

Apparent height ≈ 1.24 mm

Therefore, the diver would appear to be approximately 1.24 mm tall when viewed from the end of the diving board.

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Answer:

Option B

Explanation:

As we know

Frequency (F) * wavelength (W) = C (speed of light - can also be taken as constant)

Hence,

[tex]F1 W1 = C\\F2W2 = C[/tex]

Or [tex]F1W1 = F2W2[/tex]

Substituting the given values, we get -

[tex]840 *2.5 = 500 *XX = 4.2[/tex] m

Hence, option B is correct

The angular frequency for small oscillations of the uniform disk is approximately 1.396 rad/s.

To find the angular frequency for small oscillations of a uniform disk suspended from a pivot, we can use the formula:

Angular frequency (ω) = √(g / reff),

Where:

g is the acceleration due to gravity (9.8 m/s²),

reff is the effective radius of the disk, which is the distance from the pivot to the center of mass of the disk.

In this case, the radius of the disk (r) is given as 4.4 m, and the distance from the pivot to the center of mass (h) is given as 2.42 m.

To find the effective radius (reff), we can use the Pythagorean theorem

reff = √(r² + h²).

Substituting the given values:

reff = √(4.4² + 2.42²)

= √(19.36 + 5.8564)

= √25.2164

≈ 5.021 m.

Now we can calculate the angular frequency:

ω = √(g / reff)

= √(9.8 / 5.021)

= √1.9517

= 1.396 rad/s.

Therefore, the angular frequency for small oscillations of the uniform disk is approximately 1.396 rad/s.

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The lower bound of the power of the lens-cornea combination is 11.0 diopters.

What is lower bound of the power?

The lower bound of the power refers to the minimum value or limit of the power of a lens or lens-cornea combination. In the context of vision correction, the lower bound of the power represents the minimum power required to correct a specific visual condition.

The power of a lens is given by the formula:

Power (P) = 1 / focal length (f)

We can calculate the focal length using the formula:

f = 1 / (near point distance)

Given:

Near point distance = 11.0 cm

Substituting this value into the formula for focal length:

f = 1 / 11.0 cm

f ≈ 0.0909 cm⁻¹

Now, we can calculate the power using the formula for power:

P = 1 / f

P = 1 / 0.0909 cm⁻¹

P =11.0 diopters

Therefore, the lower bound of the power of the lens-cornea combination is 11.0 diopters.

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