A child in danger of drowning in a river is being carried downstream by a current that flows uniformly with a speed of 2.0 m/s. The child is 200 m away from the show and 1500 m upstream of the boat dock from which the rescue team sets out. If their boat speed is 8.0 m/s with respect to the water, at what angle should the pilot leave the shore to go directly to the child?

Explanation:

To calculate the work done by the elevator, we need to use the formula:

Work = Force x Distance x cos(theta)

Where:

Force = the weight of the person, which is 600 N

Distance = the vertical distance the elevator lifts the person, which is 6 meters

theta = the angle between the force and the direction of movement, which is zero since the force is acting in the same direction as the movement of the elevator (thus, cos(theta) = 1)

Plugging in the values, we get:

Work = 600 N x 6 m x cos(0) = 3600 J

Therefore, the elevator did 3600 Joules of work lifting the person.

A math question that is written as one or more sentences and asks students to use their mathematical understanding to solve an issue from "real world" is known as a word problem.

One of the first applications of math that we encounter are word problems. Many grade school students find word problems to be the most stressful type of math problem. A great selection of word problems for each of the four fundamental math operations can be found on this page.

There are word problems for addition, subtraction, multiplication, and division. Each type of problem starts out simple and straightforward and progresses to require more complex reasoning—a skill that is required on many standardized tests. Along the way, students will encounter a variety of operations that call for them to determine the kind of story problem they must resolve.

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The final speed of the sphere's center of mass is approximately 2.72 m/s.

Using conservation of energy to find the final speed v of the sphere's center of mass.

At the beginning, the sphere has only potential energy, and at the end, the sphere has only kinetic energy.

The work done by friction (rolling without slipping) is equal to the change in kinetic energy.

The potential energy at the beginning is:

[tex]U = mgh[/tex]

where m is the mass of the sphere,

g is the acceleration due to gravity, and

h is the vertical distance the sphere drops, which is 0.697 m.

The kinetic energy at the end is:

[tex]K = (1/2)mv^2[/tex]

where v is the final speed of the sphere's center of mass.

The work done by friction is:

[tex]W = K - U[/tex]

The work done by friction is negative, since it acts against the motion of the sphere. For a sphere rolling without slipping, the work done by friction is:

[tex]W = -(7/10)mgd[/tex]

where d is the distance the center of mass has dropped, which is equal to h/2 in this case.

Setting these two expressions for W equal to each other, we have:

[tex]K - U = -(7/10)mgd[/tex]

[tex](1/2)mv^2 - mgh = -(7/10)mg(h/2)[/tex]

Simplifying and solving for v, we get:

[tex]v = \sqrt{(10gh/7)}[/tex]

Plugging in the given values, we have:

[tex]v = \sqrt{(10 * 9.81 * 0.697 / 7)}[/tex]

[tex]v \approx 2.72 m/s[/tex]

Therefore, the final speed of the sphere's center of mass is approximately 2.72 m/s.

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The power developed by a man who climbs a rope at a constant speed can be calculated using the following formula:

power = force x velocity

The force required to lift the man’s weight can be calculated using the formula:

force = mass x gravity

where mass is the man’s mass and gravity is the acceleration due to gravity (9.8 m/s²).

force = 110 kg x 9.8 m/s² = 1078 N

The power developed by the man can then be calculated as:

power = force x velocity = 1078 N x 2.5 m/s = 2695 W

Therefore, the power developed by the man is 2695 W.

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The distance required to brake to a panic stop is given by the equation: d = v^2 / 2a

Where d is the distance, v is the initial velocity, and a is the acceleration. For the first situation, we have: v = 9.0 m/s, a = -9.0 m/s^2 (assuming a constant acceleration), d = v^2 / 2a = 9.0^2 / 2(-9.0) = 4.5 m. For the second situation, we have: v = 29 m/s, a = -9.0 m/s^2, d = v^2 / 2a = 29^2 / 2(-9.0) = 45.5 m. Therefore, the distance required to brake to a panic stop from a speed of 29 m/s is 45.5 meters.

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The equation "wavelength multiplied by frequency equals" is known as the wave equation and is often written as:

λν = c, where λ is the wavelength of a wave, ν is the frequency of the wave, and c is the speed of light. Therefore, the correct answer is (c) Speed. The wave equation relates the fundamental properties of a wave, which are its wavelength, frequency, and speed. The speed of a wave depends on the properties of the medium through which it is traveling, and in the case of electromagnetic waves like light, it travels at a constant speed in a vacuum, denoted by the symbol "c." The wave equation is a fundamental relationship in physics and is used to describe a wide range of phenomena, from the behavior of light and sound waves to the properties of quantum particles.

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The voltage output (in V) of a transformer used for rechargeable flashlight batteries, if its primary has 510 turns, its secondary 8 turns, and the input voltage is 125 V is 7.8 V.

A transformer is a device that is used to transfer electrical energy from one electrical circuit to another using electromagnetic induction. It's made up of two or more windings wound over a magnetic core. It has two coils, primary and secondary, that are wound around a magnetic core to create mutual induction.

According to Faraday's law of electromagnetic induction, the voltage induced in the secondary coil of the transformer is proportional to the number of turns in the secondary coil. Therefore, the voltage induced in the secondary coil can be calculated as follows:

Voltage induced in the secondary coil = (number of turns in the secondary coil/number of turns in the primary coil) × input voltage

Voltage induced in the secondary coil = (8/510) × 125 V = 1.96 V × 4 = 7.8 V

Therefore, the voltage output (in V) of a transformer used for rechargeable flashlight batteries, if its primary has 510 turns, its secondary 8 turns, and the input voltage is 125 V is 7.8 V.

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The power dissipated is 225 W

The power can also be calculated using the formula:

P = V^2 / R

where V is the voltage in volts (V).

To calculate the power consumed by a specific resistor, you will need to know either the current flowing through the resistor or the voltage across it, as well as the resistance value.

Once you have this information, simply plug in the values into one of the formulas above to calculate the power consumed by the resistor.

We have the total resistance as;

1/RT = 1/15 + 1/20 + 1/25 + 1/10

1/RT = 0.257

RT = 4 ohm

Power = V2/R

Power = (30)^2/4

= 225 W

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We can examine the motion of the enormous swing at the county fair and figure out the tension in the rope using the concepts of circular motion and energy conservation.

The giant swing at a county fair consists of a vertical central shaft with a number of horizontal arms attached at its upper end. Each arm supports a seat suspended from a cable 5.00 m long, the upper end of the cable being fastened to the arm at a point 3.00 m from the central shaft.

To analyze the motion of the swing, we can use the principles of circular motion. When the swing is in motion, the seat and the person on it move in a circular path around the central shaft.

The tension in the cable is what keeps the seat suspended and provides the necessary centripetal force to keep it moving in a circular path. The magnitude of the tension can be found using the equation:

Tension = (mass x velocity^2) / radius

In this case, the mass of the person on the swing does not affect the tension in the cable, so we can ignore it. The radius of the swing is the length of the cable, which is 5.00 m. The velocity of the swing can be found using the principles of conservation of energy:

Initial potential energy = final kinetic energy

mgh = (1/2)mv^2

where m is the mass of the person, g is the acceleration due to gravity, h is the initial height of the person, and v is the final velocity of the person.

Solving for v, we get:

v = sqrt(2gh)

where h is the initial height of the person above the lowest point of the swing.

Once we know the velocity, we can use the equation for tension to find the tension in the cable.

Therefore, by using the principles of circular motion and conservation of energy, we can analyze the motion of the giant swing at the county fair and determine the tension in the cable.

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The Giant Swing at a county fair consists of a vertical central shaft with a number of horizontal arms attached at its upper end. Each arm supports a seat suspended from a cable 5.00 m long, the upper end of the cable being fastened to the arm at a point 3.00 m from the central shaft.

A) Find the time of one revolution of the swing if the cable supporting a seat makes an angle of 30 degrees with the vertical?

B) Does the angle depend on the weight of the passenger for a given rate of revolution?

Answer:

The mass flow rate of water from the tank can be found using the Bernoulli's equation which states that the total energy of a fluid remains constant along a streamline. Applying Bernoulli's equation between the surface of the water in the tank and the orifice at the bottom, neglecting the height difference, we get:

P/ρ + gh + 1/2 * V^2 = P₂/ρ + 1/2 * V₂^2

where P is the pressure at the surface of the water in the tank, ρ is the density of water, g is the acceleration due to gravity, h is the height of the water surface above the orifice, V is the velocity of water at the surface of the water in the tank, P₂ is the pressure at the orifice, and V₂ is the velocity of water at the orifice. Since the orifice is at the bottom of the tank, h is zero, and V₂ is the velocity of water discharging from the orifice, which can be calculated using the continuity equation:

A₁V = A₂V₂

where A₁ is the cross-sectional area of the tank, and V is the velocity of water at the surface of the water in the tank.

Combining these equations and solving for the mass flow rate, we get:

m_dot = A₂ * sqrt(2 * ρ * (P - P₂))

Now, if a pipe of area A₂ and length H/2 is attached to the orifice, the velocity of water at the end of the pipe will be different than the velocity of water discharging from the orifice. We can use the Bernoulli's equation again between the orifice and the end of the pipe to calculate the velocity of water at the end of the pipe:

P₂/ρ + 1/2 * V₂^2 = P₃/ρ + gh + 1/2 * V₃^2

where P₃ is the pressure at the end of the pipe, and V₃ is the velocity of water at the end of the pipe. Again, neglecting the height difference, we get:

P₂/ρ + 1/2 * V₂^2 = P₃/ρ + 1/2 * V₃^2

Since the pipe is attached to the orifice, the pressure at the end of the pipe is atmospheric pressure Po, and the velocity of water at the end of the pipe can be calculated using the continuity equation:

A₂V₂ = A₃V₃

where A₃ is the cross-sectional area of the pipe.

Combining these equations and solving for the increase in mass flow rate, we get:

Δm_dot = A₃ * sqrt(2 * ρ * (P₂ - Po))

To find the pressure at point C for the two cases, we need to apply the Bernoulli's equation between the surface of the water in the tank and point C. Neglecting the height difference again, we get:

P/ρ + 1/2 * V^2 = Pc/ρ + 1/2 * Vc^2

where Pc is the pressure at point C, and Vc is the velocity of water at point C. For the first case, where the pipe is not attached to the orifice, we can assume that the velocity of water at point C is negligible, i.e., Vc = 0. Solving for Pc, we get:

Pc = P - 1/2 * ρ * V^2

For the second case, where the pipe is attached to the orifice, we can assume that the velocity of water at point C

The correct statement is "This wave is oscillating but not traveling."

The principle of superposition states that if two functions each separately satisfy the wave equation, then the sum (or difference) also satisfies the wave equation.

This principle follows from the fact that every term in the wave equation is linear in the amplitude of the wave.

In this case, the sum of two waves y_1 (x, t) + y_2 (x, t) can be written as y_s (x, t) = y_e (x) * y_t (t), where y_e (x) is the envelope function that depends only on position and y_t (t) is the time function that depends only on time.

To find y_e (x) and y_t (t), we need to separate the position and time dependence of the sum wave y_s (x, t). Since y_1 (x, t) and y_2 (x, t) are both trigonometric functions with unit amplitude, we can write them as:

y_1 (x, t) = A_1 * cos(k_1 * x - omega_1 * t)
y_2 (x, t) = A_2 * cos(k_2 * x - omega_2 * t)

The sum of these two waves is:

y_s (x, t) = A_1 * cos(k_1 * x - omega_1 * t) + A_2 * cos(k_2 * x - omega_2 * t)

Using the trigonometric identity for the sum of two cosines, we can write this as:

y_s (x, t) = (A_1 + A_2) * cos((k_1 + k_2) * x / 2 - (omega_1 + omega_2) * t / 2) * cos((k_1 - k_2) * x / 2 - (omega_1 - omega_2) * t / 2)

The first cosine term depends only on position and the second cosine term depends only on time, so we can write:

y_e (x) = (A_1 + A_2) * cos((k_1 + k_2) * x / 2)
y_t (t) = cos((omega_1 - omega_2) * t / 2)

Therefore, the envelope function y_e (x) and the time function y_t (t) are:

y_e (x) = (A_1 + A_2) * cos((k_1 + k_2) * x / 2)
y_t (t) = cos((omega_1 - omega_2) * t / 2)

Regarding the statement about the superposition wave y_s (x, t), the correct statement is "This wave is oscillating but not traveling."

This is because the envelope function y_e (x) depends only on position and the time function y_t (t) depends only on time, so the wave is not traveling in any direction.

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Answer: the acceleration needed will be 2.1 m/s²

Explanation:

A.9 kg toy car has a momentum of 5 kg*m/s. What is the car's speed?

The momentum of an object is defined as the product of its mass and velocity:

p = m * v

where p is the momentum, m is the mass, and v is the velocity.

To find the car's speed, we can rearrange this equation to solve for v:

v = p/m

Substituting the given values, we get:

v = 5 kg*m/s / 9 kg

v ≈ 0.56 m/s

Therefore, the speed of the car is approximately 0.56 m/s.

It would take 2 seconds for a total charge of 1.67C to pass through a  filament of the bulb

What is Charge?

Charge is a fundamental property of matter that determines how it interacts with electromagnetic fields. There are two types of charge, positive and negative, and opposite charges attract while like charges repel. The unit of charge is the Coulomb (C), which is defined as the amount of charge that flows in one second through a wire carrying a constant current of one Ampere (A).

Electrons carry a negative charge, while protons carry a positive charge. Neutrons, on the other hand, have no charge. Atoms typically have an equal number of electrons and protons, making them electrically neutral. When an atom gains or loses one or more electrons, it becomes charged and is called an ion.

The current in the light bulb is 0.835 A, then we can write:

Q = I * t

1.67 C = 0.835 A * t

Solving for t, we get:

t = 1.67 C / 0.835 A

t = 2 seconds

Therefore, it would take 2 seconds for a total charge of 1.67C to pass through a  filament of the bulb

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The average molecular kinetic energy of the liquid will be greater than the average translational kinetic energy of the gas due to the fact that liquid molecules are much closer together than gas molecules and thus experience a greater force of attraction between them. This force of attraction reduces the total kinetic energy of the liquid molecules, resulting in an average molecular kinetic energy that is greater than that of the gas.

In terms of the average speed, it is impossible to determine which substance has a greater average speed from the information given. We would need to know the temperature of the containers and the number of molecules of each substance in order to calculate the average speed of each. This is because the average speed is dependent on the temperature of the system, which affects the kinetic energy of the particles, and the number of molecules present, which affects the ratio of molecules moving at different speeds.

Therefore, in order to accurately compare the average speed of the two substances, we would need to know the temperature of the containers, the number of molecules of each substance present, and the mass of the molecules. With this additional information, we could then use the Kinetic Theory of Gases to calculate the average speed of each substance.

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Answer:

First, we need to calculate the amount of heat transferred when the frozen piece of alcohol melts and then heats up to the final temperature.

The heat transferred, Q, can be calculated using the formula:

Q = mcΔT

where:

m = mass of the frozen alcohol (10 g)

c = specific heat of the alcohol (0.57 cal/g°C)

ΔT = change in temperature (7°C - (-11°C) = 18°C)

Q = (10 g)(0.57 cal/g°C)(18°C) = 102.6 cal

Next, we need to calculate the amount of heat released by the liquid alcohol as it cools from 22°C to 7°C.

The heat released, Q, can be calculated using the formula:

Q = mcΔT

where:

m = mass of the liquid alcohol (150 g)

c = specific heat of the alcohol (0.57 cal/g°C)

ΔT = change in temperature (22°C - 7°C = 15°C)

Q = (150 g)(0.57 cal/g°C)(15°C) = 128.25 cal

Since the heat released by the liquid alcohol (128.25 cal) is greater than the heat absorbed by the frozen alcohol (102.6 cal), we can assume that the excess heat released by the liquid alcohol is due to the latent heat of fusion of the frozen alcohol.

The latent heat of fusion (L) can be calculated using the formula:

L = Q/m

where:

Q = excess heat released by the liquid alcohol (128.25 cal - 102.6 cal = 25.65 cal)

m = mass of the frozen alcohol (10 g)

L = 25.65 cal / 10 g = 2.565 cal/g

Therefore, the latent heat of fusion for this alcohol is 2.565 cal/g.

Answer:

8.0 I mean for 10,  10 is wrong

Explanation:

Multiplying the force on the object by the time over which the force acts, multiplying the mass of the object by the velocity change of the object

Answer:

[tex]10\; {\rm kg \cdot m\cdot s^{-1}}[/tex].

(Assume that there are no other unbalanced force on this object.)

Explanation:

When an object of mass [tex]m[/tex] travels at a velocity of [tex]v[/tex], the momentum [tex]p[/tex] of that object would be [tex]p = m\, v[/tex].

In this question, the [tex]m = 6.0\; {\rm kg}[/tex] object was initially at a velocity of [tex]v = 3.0\; {\rm m\cdot s^{-1}}[/tex]. The initial momentum of this object would be:

[tex]\begin{aligned} p &= m\, v \\ &= (6.0\; {\rm kg})\, (3.0\; {\rm m\cdot s^{-1}}) = 18\; {\rm kg\cdot m\cdot s^{-1}}\end{aligned}[/tex]

The question states that the external force is "resistive". In other words, this force opposes the motion of this object. Hence, while the magnitude of this force is [tex]40\; {\rm N}[/tex], the vector value of this force would be [tex](-40)\; {\rm N}[/tex].

To find the impulse [tex]P[/tex] that a force [tex]F[/tex] has exerted, multiply the force vector by the duration [tex]\Delta t[/tex] over which this force was applied.

For example, the [tex](-40)\; {\rm N}[/tex] force in this question was applied over a period of [tex]\Delta t = 0.20\; {\rm s}[/tex]. This force would have exerted an impulse [tex]J[/tex] of:

[tex]\begin{aligned}J &= F\, \Delta t \\ &= (-40\; {\rm N})\, (0.20\; {\rm s}) = (-8.0)\; {\rm N\cdot s}\end{aligned}[/tex].

Note that [tex]1\; {\rm N}[/tex] is equivalent to [tex]1\; {\rm kg\cdot m\cdot s^{-2}}[/tex]. Therefore, the unit of impulse [tex]{\rm N\cdot s}[/tex] would be equivalent to [tex]{\rm (kg\cdot m\cdot s^{-2})\cdot s}[/tex], which simplifies to [tex]{\rm (kg\cdot m\cdot s^{-1})}[/tex].

Thus, the impulse on this object [tex](-8.0)\; {\rm N\cdot s}[/tex] would be equivalent to [tex](-8.0)\; {\rm kg\cdot m\cdot s^{-1}}[/tex].

The total impulse on an object is also equal to the change in the momentum of the object. Assuming that there are no other unbalanced force on this object, the total impulse on this object would be [tex](-8.0)\; {\rm kg\cdot m\cdot s^{-1}}[/tex]. The momentum of this object would become:

[tex]\begin{aligned}& 18.0\; {\rm kg \cdot m\cdot s^{-1} + (-8.0)\; {\rm kg\cdot m\cdot s^{-1}} \\ =\; & 10\; {\rm kg\cdot m\cdot s^{-1}} \end{aligned}[/tex].

Answer:

F = M g       gravitational force on M

M g = G M m / R^2

g = G m / R^2        where R is radius of earth and m the mass

g = 6.67E-11 N-m^2/ kg^2 * 5.98E24 kg / (6.37E6 m)^2

g = 6.67 * 5.98 / (6.37)^2 * 10 = 9.83 m/s^2

Answer:

The temperature at which the Kelvin scale reading is double the Fahrenheit reading is 176.69 F = 353.38 K

Explanation:

K = 2F

→ 2F = 273 + (5/9)(F - 32)

→ 18F = 2457 + 5F - 160

→ 18F - 5F = 2457 - 160

→ 13F = 2297

→ F = 2297/13

→ F = 176.69° F

K = 2F

→K = 2 * 176.69

→ K = 353.38K

The graphic shows that whereas southern winds would be redirected to the right, northern wind will be redirected to the left.

The Earth wouldn't appear much like it currently does if it ceased rotating on its axis.The absence of the Coriolis Effect would prevent the spiraling of winds, the enlargement of the equatorial area, and the oceanification of the continents.

The Coriolis effect is what appears to be a deflection.Fluids that move over wide areas, like air currents, are comparable to the trajectory of a ball.The Northern Hemisphere gives them the appearance of bending to the right.In the Southern Horizon, where currents seem to bend to a left, the Coriolis force behaves in the opposite direction.

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There are approximately [tex]1.1×10^31[/tex] electrons in 1 kg of mass and the charge is [tex]-1.76×10^-7[/tex] coulombs.

The mass of an electron is given as [tex]9.1×10^-31[/tex]kg. Therefore, the number of electrons in 1 kg can be calculated as follows:

1 kg = ([tex]1 / 9.1×10^-31[/tex]) electrons

Number of electrons = 1 kg / [tex]9.1×10^-31[/tex] kg/electron

= 1 /[tex](9.1×1 0^-31)[/tex] electrons

= [tex]1.1×10^31[/tex]electrons

The charge of an electron is [tex]-1.602×10^-19[/tex] coulombs. This is the fundamental unit of negative charge in the universe. Therefore, the total charge of [tex]1.1×10^31[/tex]electrons can be calculated as follows:

Total charge = ([tex]1.1×10^31[/tex] electrons) x ([tex]-1.602×10^-19[/tex]coulombs/electron)

= [tex]-1.76×10^-7[/tex] coulombs

Therefore, the total charge of[tex]1.1×10^31[/tex]electrons is -[tex]1.76×10^-7[/tex]coulombs.

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a). Fe*cos32 = 799 N.

Fe = 799 / cos32 = 948 N. = Force exerted.

b). W = Fe * d = 948 * 22 = 20,856 J.

c). P = W / t = 20,656 / 8 s. = 2607 J/s

= 2607 Watts = 2.607 KW.

A horizontal force is one that moves in a path perpendicular to the sky. The magnitude and direction of the horizontal forces are both equivalent. The horizontal net force is zero because they are symmetrical. The absence of horizontal motion is indicated by this. Equal in magnitude and moving in the opposing direction are the vertical forces.

The normal force on a horizontal surface is an illustration of any item maintained on a horizontal surface, such as a flat table, a stand, or just the earth. Keeping literature on a bookshelf or computers on a desk at work are two examples. Gravitational force is measured in g.

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Joseph's velocity was 4.5 meters per minute

Velocity is the rate of change of displacement of an object with respect to time, and it is measured in units of distance per unit time. Velocity is a physical quantity that describes the rate at which an object changes its position. It is a vector quantity, which means it has both magnitude

We can start by using the formula:

velocity = distance / time

Joseph's velocity can be calculated by dividing the distance he rose by the time it took him:

Substitute the values in the equation

velocity = 5.4 meters / 1.2 minutes

Simplifying this expression, we get:

Velocity = 4.5 meters per minute

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Option 2 is the correct answer. The Roche limit is reached when ring particles closest to a planet fall into its atmosphere.

The Roche limit is the minimum distance at which a celestial body, such as a moon or a ring system, can approach a planet before being disrupted by tidal forces. When a celestial body gets too close to a planet, the planet's tidal forces can overcome its gravitational pull, and the body may be torn apart by these forces.

Option 2 is the correct answer. The Roche limit is reached when ring particles closest to a planet fall into its atmosphere. This happens because the tidal forces from the planet become stronger than the gravitational forces holding the particles together, causing them to break apart and fall towards the planet. Once the particles are within the Roche limit, they will continue to break apart and eventually form a disk or ring around the planet.

The Roche limit is an important concept in planetary science, as it helps us understand the structure and behavior of planets, moons, and other celestial bodies. It can also help us explain the formation of planetary rings, such as those around Saturn and other gas giants in our solar system.

So, the right response is Choice 2. When ring particles closest to a planet enter its atmosphere, the Roche limit is achieved.

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Answer:

Using the thin lens equation, we have:

1/f = 1/d_o + 1/d_i

where f is the focal length of the lens, d_o is the distance of the object from the lens, and d_i is the distance of the image from the lens.

Plugging in the given values, we get:

1/f = 1/7 + 1/48

Simplifying this expression, we get:

1/f = 0.2024

Multiplying both sides by f, we get:

f = 4.94 cm

Therefore, the focal length of the convex lens is 4.94 cm.

Answer:

6.1 cm

Explanation:

The focal length of the lens is 6.1 or 6.11 cm

When the puck has descended to a height of R/2, it no longer makes touch with the sphere.

To solve this problem, we can use the conservation of mechanical energy. Initially, the mass is at rest and has only potential energy, which is given by mgh, where m is the mass of the puck, g is the acceleration due to gravity, and h is the height of the puck above the ground. When the puck slides down the sphere, it loses potential energy and gains kinetic energy, which is given by [tex](\frac{1}{2}mv^2)[/tex], where v is the velocity of the puck. However, as the puck slides down the sphere, it also experiences a normal force from the sphere, which reduces the force of gravity and affects the speed of the puck.

Let's start by finding the normal force on the puck. At the top of the sphere, the normal force is equal to the weight of the puck, which is mg. As the puck slides down the sphere, the normal force decreases until it reaches zero when the puck loses contact with the sphere. At this point, the gravitational force is equal to the centrifugal force, which is mv^2/R, where R is the radius of the sphere.

Setting these forces equal, we get:

[tex]mg = \frac{mv^2}{R}[/tex]

Simplifying, we get:

[tex]v^2 = gR[/tex]

Now we can use conservation of mechanical energy to find the height at which the puck loses contact with the sphere. At this point, all the potential energy has been converted into kinetic energy, so we have:

[tex]mgh = (\frac{1}{2})mv^2[/tex]

Substituting [tex]v^2 = gR[/tex], we get:

[tex]h = R(1 - \frac{1}{2}) = \frac{R}{2}[/tex]

Therefore, the puck loses contact with the sphere when it has fallen to a height of R/2.

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Answer:

All the colours in the white light are absorbed but red and blue colours are reflected because the object's colours are red and blue

Answer: One

Explanation:

Time period is the time taken by a wave to travel between two consecutive crests or troughs. So, in one time period, a wave travels a distance of one wavelength.

Assuming that the rope has uniform density and thickness, the center of mass of the rope will be at its geometric center. When half of the rope has lifted off the table, it means that the lifted portion is 1 meter long, and the remaining portion still lies on the table.

Thus, the center of mass of the lifted portion will be at its midpoint, which is at a distance of 0.5 meters from the end you are holding. Similarly, the center of mass of the remaining portion still lying on the table will also be at its midpoint, which is also at a distance of 0.5 meters from the end you are holding. Therefore, the center of mass of the entire rope will be at the midpoint of the lifted portion and the remaining portion, which is at a distance of 1 meter from the end you are holding. Thus, when half of the rope has lifted off the table, the center of mass of the rope will be 1 meter above the table.

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