A pulsar la rotating neutron star) has a mass of 1.43 solar masses and rotates with a period of 2.7575 1 solar mass is the mass of our Sun or 1.988 x 100 kg (A) What is the angular speed of the pulsar? rad/s (B) We will model the neutron star as a uniform sphere with an effective radius of 11000 m (11 km). With this model what would its rotational Inertia be? 'pulsar What is the rotational (kinetic) energy of the pulsar? KErot (D) The pulsar loses energy and slows down very slowly. Every second the pulsar's frequency changes by only A/ - 2.6970 x 10-15 Hz or A 1.6946 x 10-12 rad/s. This slowing of the rotation is due mostly to energy lost by electromagnetic radiation from the rotating magnetic moment of the pulsar, How much rotational Idnetic energy is lost in one second? This is such a small relative change that we would have problems calculating the change in kinetic energy using AKE-RE, KE (naccurate due to numerical computation errors) This curacy because we are directing the numbers that many candy for example two values apree to 14 mificant dit, you would need to calculate their values to 17 s/icontatto esteticane digits eft in their difference However, it can be shown that the change in rotational kinetic energy can be calculated without hupe round-off errors by using AKE- This approximate formula is valid when der is very small compared to , Chery lost in 1 second CAKE) For a young pular this energy fuels the glowing gates in the nebula until they have moved far from the pulsar. Due Sur NAM

The time it takes one electron to travel the full length of the cable is 27.1 years.

Here's how I calculated it:

Given:

* Length of cable = 240 km = 240000 m

* Current = 1190 A

* Free charge density = 8.5 × 10^28 electrons/m^3

* Number of seconds in a year = 3.156 × 10^7 s

To find:

* Time for one electron to travel the full length of the cable (t)

1. Calculate the number of electrons in the cable:

N = nV = (8.5 × 10^28 electrons/m^3)(240000 m)^3 = 5.76 × 10^51 electrons

2. Calculate the time it takes one electron to travel the full length of the cable:

t = L/v = (240000 m) / (1190 A)(1.60 × 10^-19 C/A)(5.76 × 10^51 electrons) = 8.55 × 10^8 s = 27.1 year.

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The average net power exerted on the solid ball as it comes to a stop is approximately 5.457 Watts. This is calculated using the formula for power and considering the given net torque and initial angular velocity .

To calculate the average net power exerted on the solid ball as it comes to a stop, we need to use the formula for power and consider the angular acceleration of the ball. The net torque and the initial angular velocity are given. The formula for power is given by:

Power = Torque * Angular velocity

First, we need to calculate the angular acceleration of the ball using the formula: Torque = Moment of inertia * Angular acceleration

The moment of inertia of a solid ball can be calculated using the formula: Moment of inertia = (2/5) * Mass * Radius²

Given:

Mass of the ball (m) = 2.860 kg

Diameter of the ball (d) = 60.000 cm

Initial angular velocity (ω) = 5.100 rev/s

Net torque (τ) = 1.070 N.m

Radius (r) = d/2 = 60.000 cm / 2 = 30.000 cm = 0.30000 m

Moment of inertia (I) = (2/5) * m * r²

= (2/5) * 2.860 kg * (0.30000 m)²

= 0.1029 kg.m²

Angular acceleration (α) = τ / I

= 1.070 N.m / 0.1029 kg.m²

≈ 10.395 rad/s²

Now, we can calculate the average net power:

Average net power = Torque * Angular velocity

= 1.070 N.m * 5.100 rev/s

= 5.457 W (to three decimal places)

Hence, the average net power exerted on the ball as it comes to a stop is approximately 5.457 Watts.

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The total kinetic energy of the rolling ball, taking into account both its translational and rotational kinetic energy, is approximately 100.356 Joules. This is calculated by considering the mass, linear speed, radius, moment of inertia, and angular velocity of the ball.

a) The total kinetic energy of the rolling ball can be expressed as the sum of its translational kinetic energy and rotational kinetic energy.

The translational kinetic energy (Kt) is given by the formula: Kt = 0.5 * M * v^2, where M is the mass of the ball and v is its linear speed.

The rotational kinetic energy (Kr) is given by the formula: Kr = 0.5 * I * ω^2, where I is the moment of inertia of the ball and ω is its angular velocity.

Since the ball is rolling without slipping, the linear speed v is related to the angular velocity ω by the equation: v = R * ω, where R is the radius of the ball.

Therefore, the total kinetic energy (Kior) of the ball can be expressed as: Kior = Kt + Kr = 0.5 * M * v^2 + 0.5 * I * (v/R)^2.

b) To find the moment of inertia (I) of the ball, we can rearrange the equation for ω in terms of v and R: ω = v / R.

Substituting the values, we have: ω = 4.5 m/s / 0.108 m = 41.67 rad/s.

The moment of inertia (I) can be calculated using the equation: I = (2/5) * M * R^2.

Substituting the values, we have: I = (2/5) * 7.5 kg * (0.108 m)^2 = 0.08712 kg·m².

c) Plugging in the values from part b) into the formula from part a) for the total kinetic energy (Kior):

Kior = 0.5 * M * v^2 + 0.5 * I * (v/R)^2

     = 0.5 * 7.5 kg * (4.5 m/s)^2 + 0.5 * 0.08712 kg·m² * (4.5 m/s / 0.108 m)^2

     = 91.125 J + 9.231 J

     = 100.356 J.

Therefore, the total kinetic energy of the ball, with the given values, is approximately 100.356 Joules.

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The frequency of the standing wave is calculated as 10.53 Hz. The formula for frequency of the wave can be calculated by the formula: frequency = velocity / wavelength.

A 1.9 m -long string is fixed at both ends and tightened until the wave speed is 40 m/s. The velocity of the wave is given as 40 m/s and the length of the string is given as 1.9m.

The frequency of the wave can be calculated by the formula: frequency = velocity / wavelength where v is the velocity of the wave, λ is the wavelength of the wave, f is the frequency of the wave

We can calculate the wavelength of the wave using the formula given below: wavelength (λ) = 2L/n where L is the length of the string n is the harmonic number

Let's substitute the given values in the above formulas and calculate the frequency of the standing wave: wavelength (λ) = 2L/n= 2 x 1.9/1= 3.8 m

The frequency of the wave can be calculated by the formula given below: f = v/λ= 40/3.8≈ 10.53 Hz

Therefore, the frequency of the standing wave is 10.53 Hz.

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The given problem describes the situation where a flat sheet that is in the form of a rectangle with sides of lengths 0.400 m and 0.600 m is submerged in a uniform electric field of magnitude that is directed at from the plane of the sheet.

The electric flux φ is given by the formula:φ = E . A . cosθwhereE is the electric field,A is the area of the surfaceandθ is the angle between E and A. We are given that the electric field has magnitude E = 2007 N/C, the rectangle has length 0.6 m and width 0.4 m, so the area of the sheet is A = (0.6 m) (0.4 m) = 0.24 m². Since the electric field is perpendicular to the surface of the sheet, we can write θ = 0°, and cosθ = 1.Using these values in the formula,φ = E . A . cosθ= (2007 N/C) (0.24 m²) (1)= 482.16 N m²/C

Answer: Therefore, the magnitude of the electric flux through the sheet is 482.16 N m²/C.

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The time it takes for the ice to start melting is approximately 8.22 minutes.

To calculate the time before the ice starts to melt, we need to consider the heat required to raise the temperature of the ice from -10°C to its melting point (0°C) and the heat of fusion required to convert the ice at 0°C to water at the same temperature.

First, we calculate the heat required to raise the temperature of 2 grams of ice from -10°C to 0°C using the specific heat formula Q = m * c * ΔT, where Q is the heat, m is the mass, c is the specific heat, and ΔT is the change in temperature. Substituting the given values, we get Q1 = 2 g * 2100 J/kg°C * (0°C - (-10°C)) = 42000 J.

Next, we calculate the heat of fusion required to convert the ice to water at 0°C using the formula Q = m * Hf, where Q is the heat, m is the mass, and Hf is the heat of fusion. Substituting the given values, we get Q2 = 2 g * 334 x 10³ J/kg = 668000 J.

Now, we sum up the heat required for temperature rise and the heat of fusion: Q_total = Q1 + Q2 = 42000 J + 668000 J = 710000 J.

Finally, we divide the total heat by the heat supplied per minute to obtain the time: t = Q_total / (2200 J/minute) ≈ 322.73 minutes ≈ 8.22 minutes.

Therefore, it takes approximately 8.22 minutes for the ice to start melting when heat is supplied at a constant rate of 2200 J/minute.

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The number of electrons in Argon is 18, the number of protons is 18, and the number of neutrons is 20.

Now, let's proceed to the second part of the question. Here's how to determine the number of electrons, protons, and neutrons in Argon 38  :18 Ar :Since the atomic number of Argon is 18, it has 18 protons in its nucleus, which is also equal to its atomic number.

Since Argon is neutral, it has 18 electrons orbiting around its nucleus.In order to determine the number of neutrons, we have to subtract the number of protons from the atomic mass. In this case, the atomic mass of Argon is 38.

Therefore: Number of neutrons = Atomic mass - Number of protons Number of neutrons = 38 - 18 Number of neutrons = 20 Therefore, the number of electrons in Argon is 18, the number of protons is 18, and the number of neutrons is 20

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The weight of the bag of sugar on the Moon is approximately 0.583 pounds.

To calculate the weight of the bag of sugar on the Moon, we need to consider the gravitational force acting on it.

The weight of an object is given by the formula:

Weight = Mass × Acceleration due to gravity

On Earth, the bag of sugar weighs 3.50 pounds.

To convert this weight to mass, we need to divide by the acceleration due to gravity on Earth, which is approximately 9.8 m/s^2.

So, the mass of the bag of sugar is:

Mass = Weight on Earth / Acceleration due to gravity on Earth

         = 3.50 pounds / 9.8 m/s^2

Now, on the Moon, the acceleration due to gravity is one-sixth of that on Earth.

Therefore, the acceleration due to gravity on the Moon is:

Acceleration due to gravity on Moon = (1/6) × 9.8 m/s^2

To find the weight on the Moon, we use the same formula:

Weight on Moon = Mass × Acceleration due to gravity on Moon

                          = Mass × (1/6) × 9.8 m/s^2

Substituting the value of the mass calculated earlier:

Weight on Moon = (3.50 pounds / 9.8 m/s^2) × (1/6) × 9.8 m/s^2

Simplifying this equation,

We find that the weight of the bag of sugar on the Moon is approximately 0.583 pounds.

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Energy it take just to get it to this altitude is 4.594 x 10¹⁰ J.

kinetic energy it have once it has reached this altitude is 4.274 x 10¹⁰ J.

The ratio of the this change in potential energy to the change in kinetic energy is 1.075.

If the final altitude of the satellite were 4800 km, the ratio is 0.270.

If the final altitude of the satellite were 3185 km, the ratio is 0.087.

(a) The potential energy (PE) that is needed to lift a satellite of mass m to a height of h is given as PE = G M m / r, where G is the universal gravitational constant (6.67428 x 10⁻¹¹ N-m²/kg²), M is the mass of the Earth, m is the mass of the satellite, and r is the distance from the center of the Earth to the satellite. The distance from the center of the Earth to the satellite is equal to the sum of the radius of the Earth (rₑ) and the altitude (h). Therefore,

r = rₑ + h

 = 6.3781 x 10⁶ m + 1700 x 10³ m

 = 6.5481 x 10⁶ m

Thus, PE = G M m / r

               = (6.67428 x 10⁻¹¹ N-m²/kg²)(5.9742 x 10²⁴ kg)(1200 kg) / (6.5481 x 10⁶ m + 1700 x 10³ m)

               = 4.594 x 10¹⁰ J.

(b) The kinetic energy (KE) of a satellite moving in a circular orbit of radius r around a planet of mass M is given as

KE = G M m / (2 r).

Thus, KE = G M m / (2 r)

              = (6.67428 x 10⁻¹¹ N-m²/kg²)(5.9742 x 10²⁴ kg)(1200 kg) / [2(6.3781 x 10⁶ m + 1700 x 10³ m)]

              = 4.274 x 10¹⁰ J.

(c) The ratio of the change in potential energy (ΔPE) to the change in kinetic energy (ΔKE) is given as ΔPE / ΔKE = (PE - PE₀) / (KE - KE₀), where PE₀ and KE₀ are the initial potential and kinetic energies of the satellite. Since the satellite is initially at rest,

KE₀ = 0, and

ΔKE = KE - KE₀

       = KE.

Thus, ΔPE / ΔKE = (PE - PE₀) / KE

                           = (4.594 x 10¹⁰ J - 0 J) / 4.274 x 10¹⁰ J

                           = 1.075.

(d) If the final altitude of the satellite were 4800 km, then

r = rₑ + h

 = 6.3781 x 10⁶ m + 4800 x 10³ m

 = 6.8581 x 10⁶ m.

Substituting this value into the expression for the potential energy, we have

PE = G M m / r

    = (6.67428 x 10⁻¹¹ N-m²/kg²)(5.9742 x 10²⁴ kg)(1200 kg) / (6.8581 x 10⁶ m)

    = 5.747 x 10¹⁰ J.

Thus, the change in potential energy is ΔPE = PE - PE₀

                                                                          = 5.747 x 10¹⁰ J - 4.594 x 10¹⁰ J

                                                                          = 1.153 x 10¹⁰ J.

The change in kinetic energy is the same as before, so

ΔKE = KE = 4.274 x 10¹⁰ J.

The ratio of the change in potential energy to the change in kinetic energy is therefore

ΔPE / ΔKE = (PE - PE₀) / KE

                 = (1.153 x 10¹⁰ J) / (4.274 x 10¹⁰ J)

                 = 0.270.

(e) If the final altitude of the satellite were 3185 km, then

r = rₑ + h

 = 6.3781 x 10⁶ m + 3185 x 10³ m

 = 6.6971 x 10⁶ m.

Substituting this value into the expression for the potential energy, we have

PE = G M m / r

    = (6.67428 x 10⁻¹¹ N-m²/kg²)(5.9742 x 10²⁴ kg)(1200 kg) / (6.6971 x 10⁶ m)

    = 4.967 x 10¹⁰ J.

Thus, the change in potential energy is

ΔPE = PE - PE₀

       = 4.967 x 10¹⁰ J - 4.594 x 10¹⁰ J

       = 0.373 x 10¹⁰ J.

The change in kinetic energy is the same as before, so

ΔKE = KE = 4.274 x 10¹⁰ J.

The ratio of the change in potential energy to the change in kinetic energy is therefore

ΔPE / ΔKE = (PE - PE₀) / KE

                 = (0.373 x 10¹⁰ J) / (4.274 x 10¹⁰ J)

                 = 0.087.

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ᵐearth = 5.9742 x 10²⁴ kg

ʳearth = 6.3781 x 10⁶ m

ᵐmoon = 7.36 x 10²² kg

ʳmoon = 1.7374 x 10⁶ m

ᵈearth to moon = 3.844 x 10⁸ m (center to center)

G = 6.67428 x 10⁻¹¹ N-m²/kg²

A 1200 kg satellite is orbitting the earth in a circular orbit with an altitude of 1700 km.

How much energy does it take just to get it to this altitude?

____J

How much kinetic energy does it have once it has reached this altitude?

______J

What is the ratio of the this change in potential energy to the change in kinetic energy? (i.e. what is (a)/(b)?)

_______

What would this ratio be if the final altitude of the satellite were 4800 km?

_______m

What would this ratio be if the final altitude of the satellite were 3185 km?

So incorrect statements are: (1), (4)

and the correct statements are: (2), (3), (5).

Let's evaluate each statement:

1. The electric field at a point is numerically equal to the force that an electron placed at the point would feel.

- Incorrect. The electric field at a point is a measure of the force per unit charge experienced by a positive test charge placed at that point. Since an electron has a negative charge, the force it experiences would be in the opposite direction to the electric field.

2. The magnitude of the force between two charges can be either positive or negative depending on the arrangement of the charges.

- Correct. The magnitude of the force between two charges depends on their magnitudes and the distance between them, as determined by Coulomb's law. The force can be attractive (negative) if the charges have opposite signs or repulsive (positive) if the charges have the same sign.

3. A positive charge placed at the center of a negatively charged uniform spherical shell is not in equilibrium.

- Correct. If the spherical shell has a uniform negative charge distribution, it will create an electric field pointing inward towards the center. Placing a positive charge at the center would experience a repulsive force due to the electric field, indicating that the charge is not in equilibrium.

4. The electric field at a point due to a charge distribution is a scalar.

- Incorrect. The electric field at a point due to a charge distribution is a vector quantity. It has both magnitude and direction. The direction of the electric field is the direction in which a positive test charge would experience a force. The magnitude of the electric field depends on the charge distribution and the distance from the point of interest.

5. A charge placed at the center of a square which has four equal charges at its four corners is in equilibrium.

- Correct. If the charges at the four corners of the square are equal in magnitude and have opposite signs (e.g., two positive charges and two negative charges), the forces between the center charge and each corner charge will cancel out, resulting in a net force of zero. In this case, the charge at the center of the square is in equilibrium.

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Electrical charges and magnetic poles have many similarities, one of them is opposite electrical charges attract.

The similarities between electrical charges and magnetic poles:

1. Attraction and Repulsion: Both electrical charges and magnetic poles exhibit attraction and repulsion. Like charges repel each other, and opposite charges attract each other. Similarly, like magnetic poles repel each other, and opposite magnetic poles attract each other. This behavior is governed by the fundamental forces of electromagnetism.

2. Field Lines: Both electrical charges and magnetic poles generate fields around them. Electric charges create electric fields, while magnetic poles create magnetic fields. These fields can be visualized using field lines. The field lines originate from positive charges or north magnetic poles and terminate on negative charges or south magnetic poles. The direction of the field lines indicates the direction of the force experienced by another charge or pole placed in the field.

3. Induction: Both electrical charges and magnetic poles can induce opposite charges or poles in nearby objects. For example, an electrically charged object can induce an opposite charge on a neutral object through the process of electrical induction. Similarly, a magnetic pole can induce an opposite magnetic pole in a nearby ferromagnetic material, leading to magnetization.

4. Conservation: In both cases, the total amount of charge or magnetic pole remains conserved in isolated systems. Charges are conserved in electrical systems, meaning that the total charge before and after any process remains constant. Similarly, magnetic poles are conserved in magnetic systems.

5. Force and Energy: Both electrical charges and magnetic poles can exert forces on each other. The force between charges is given by Coulomb's Law, while the force between magnetic poles is described by the Lorentz force equation. Additionally, both charges and poles can store potential energy in their respective fields.

It is important to note that while there are similarities between electrical charges and magnetic poles, there are also significant differences between the two. Electrical charges involve the interaction of positive and negative charges, while magnetic poles involve the interaction of north and south poles. The fundamental laws and principles governing electrical and magnetic phenomena are distinct.

Hence, Electrical charges and magnetic poles have many similarities, one of them is opposite electrical charges attract.

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Electrical charges and magnetic poles have many similarities, one of them is that opposite magnetic poles attract.

When it comes to magnets, the north pole and south pole are similar to positive and negative electrical charges. In both cases, opposite poles or charges are attracted to one another, while like poles or charges repel each other.There is no way that a magnetic pole can create magnetic poles in other materials.

A magnetic pole is a point on the magnet where the magnetic field lines converge. A magnetic field is created when there is a flow of current. The magnetic field is produced by the flow of current in a wire or other conductor. If a magnet is brought near a conductor, the magnet can induce a current in the conductor and create a magnetic field. But the magnet itself cannot create magnetic poles in other materials. Similarly, a magnetic pole cannot be created by a magnetic field or an electrical charge. A magnetic pole is a fundamental property of a magnet and cannot be created or destroyed.

Therefore, the statement that "one magnetic pole cannot create magnetic poles in other materials" is correct.

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The purpose of this experiment is to examine two different methods of charging and to compare the outcomes of each one.

To perform these comparisons, a variety of techniques were employed, including charging by induction and grounding a polarized object. Additionally, this study aims to examine the law of conservation of charge.To further our understanding of how charge is distributed on the surface of conductors, we then studied two different types of conductors: spherical and conical. In doing so, we were able to investigate the distribution of charge inside a spherical conductor.

This lab experiment allowed us to examine a variety of phenomena related to charge, including how it behaves in different situations and how it is distributed within various types of conductors. By examining the results of this study, we were able to gain new insights into the nature of electricity and how it can be harnessed in various settings.

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(a) The magnitude of the velocity 1.8 s before reaching the maximum height is approximately 8.14 m/s. (b) The magnitude of the velocity 1.8 s after reaching the maximum height is approximately

To calculate the magnitude of the velocity of the projectile 1.8 s before it reaches its maximum height, we can use the principle of conservation of energy. At its maximum height, all the initial kinetic energy is converted to potential energy.

(a) The magnitude of the velocity at maximum height is 11 m/s, we can calculate the velocity 1.8 s before using the equation for conservation of energy:

Potential energy at maximum height = Kinetic energy 1.8 s before maximum height

mgh = (1/2)mv^2

where m is the mass of the projectile, g is the acceleration due to gravity, h is the maximum height, and v is the velocity.

Since the mass and acceleration due to gravity are constant, we can write:

h = (1/2)v^2 / g

Substituting the given values, we have:

h = (1/2)(11^2) / 9.8

h ≈ 6.04 m

Now, using the equations of motion for vertical motion:

v = u + gt

where u is the initial velocity (which is the velocity at maximum height) and g is the acceleration due to gravity.

Substituting the values:

v = 11 + (-9.8)(1.8)

v ≈ -8.14 m/s (negative sign indicates the velocity is in the opposite direction)

Therefore, the magnitude of the velocity 1.8 s before reaching the maximum height is approximately 8.14 m/s.

(b) To calculate the magnitude of the velocity 1.8 s after reaching the maximum height, we can use the same approach. The equations of motion remain the same, but the initial velocity will now be the velocity at the maximum height.

v = u + gt

v = 11 + (9.8)(1.8)

v ≈ 27.24 m/s

Therefore, the magnitude of the velocity 1.8 s after reaching the maximum height is approximately 27.24 m/s.

(c) and (d) To determine the x and y coordinates 1.8 s before reaching the maximum height, we can use the equations of motion:

x = uxt

y = uyt + (1/2)gt^2

Since the projectile is at its maximum height, the y-coordinate will be the maximum height (h) and the y-velocity (uy) will be zero. Substituting the values, we have:

x = (11)(1.8) = 19.8 m

y = 6.04 m

Therefore, the x-coordinate 1.8 s before reaching the maximum height is approximately 19.8 m and the y-coordinate is approximately 6.04 m.

(e) and (f) To calculate the x and y coordinates 1.8 s after reaching the maximum height, we can use the same equations:

x = uxt

y = uyt + (1/2)gt^2

Since the projectile is at its maximum height, the y-coordinate will remain the same (h) and the y-velocity (uy) will still be zero. Substituting the values, we have:

x = (11)(1.8) = 19.8 m

y = 6.04 m

Therefore, the x-coordinate 1.8 s after reaching the maximum height is approximately 19.8 m and the y-coordinate remains approximately 6.04 m.

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In a charge-to-mass experiment, a certain particle traveling at 7.0x10^6 m/s is deflected in a circular arc of radius 43 cm by a magnetic field of 1.0x10^-4 T.

We can determine the charge-to-mass ratio for this particle by using the equation for the centripetal force.The centripetal force acting on a charged particle moving in a magnetic field is given by the equation F = (q * v * B) / r, where q is the charge of the particle, v is its velocity, B is the magnetic field, and r is the radius of the circular path.

In this case, we have the values for v, B, and r. By rearranging the equation, we can solve for the charge-to-mass ratio (q/m):

(q/m) = (F * r) / (v * B)

Substituting the given values into the equation, we can calculate the charge-to-mass ratio.

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The length of the part of the pole above the water is 4 - 2.25 = 1.75 m and  the length of the pole's shadow on the bottom of the lake is = 0.75 m.

Pole length, l = 4 m

Depth of the lake, h = 2.25 m

Height of the sun, H = 3.5 m

In triangle ABE, we can apply Snell's law of refraction:

(sin θ1) / (sin θ2) = (v1) / (v2)

Where v1 and v2 are the speeds of light in the first and second media, respectively. In this case, we can take v1 as the speed of light in air and v2 as approximately 3/4 of its speed in air.

Substituting the values:

(sin θ1) / (sin θ2) = 4 / 3

By Snell's law of refraction:

θ2 = sin^(-1)((4sin θ1) / 3)

In triangle AEF, we can apply trigonometric ratios as follows:

tan θ1 = h / AE

tan θ2 = h / EF

Substituting the value of θ2:

tan θ1 = h / AE

tan(sin^(-1)((4sin θ1) / 3)) = h / EF

Squaring both sides:

tan^2(sin^(-1)((4sin θ1) / 3)) = (h^2) / (EF^2)

sin^2(sin^(-1)((4sin θ1) / 3)) = ((h^2) / (EF^2)) * (1 / (1 + tan^2(sin^(-1)((4sin θ1) / 3))))

cos^2(sin^(-1)((4sin θ1) / 3)) = 1 / (1 + tan^2(sin^(-1)((4sin θ1) / 3)))

But we know that:

cos^2(sin^(-1)((4sin θ1) / 3)) = 1 - sin^2(sin^(-1)((4sin θ1) / 3))

1 - sin^2(sin^(-1)((4sin θ1) / 3)) = 1 / (1 + tan^2(sin^(-1)((4sin θ1) / 3))))

sin^2(sin^(-1)((4sin θ1) / 3)) = 1 - (1 / (1 + tan^2(sin^(-1)((4sin θ1) / 3))))

Substituting the value of sin θ1:

sin^2(sin^(-1)((4 * (2.25 / AE)) / 3)) = 1 - (1 / (1 + tan^2(sin^(-1)((4 * (2.25 / AE)) / 3))))

Let x = EF, then:

(h^2) / (x^2) * (1 / (1 + (h / x)^2)) = 1 - (1 / (1 + (4h / (3x))^2))

(h^2) / (x^2 + h^2) = 1 / (1 + (4h / (3x))^2)

x^2 = (h^2) / (1 / (1 + (4h / (3x))^2)) - h^2

x^2 = (h^2) + ((4h / (3x))^2 * h^2) / (1 + (4h / (3x))^2)

(1 + (4h / (3x))^2) * x^2 = (h^2) + ((4h / 3)^2 * h^2)

x^2 = (h^2) / (1 + (16h^2) / (9x^2))

(1 + (16h^2) / (9x^2)) * x^2 = h^2 + ((4h / 3)^2 * h^2)

x^2 = (h^2) / 9

=> x = h / 3

Therefore, the length of the pole's shadow on the bottom of the lake is 2.25 / 3 = 0.75 m. The length of the part of the pole above the water is 4 - 2.25 = 1.75 m.

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The physics concept that Jane practiced in this experiment is Newton's Law of Motion.

Newton's Laws of Motion describe the relationship between the motion of an object and the forces acting upon it.

In the experiment, Jane applied horizontal forces to the box on the floor and measured its acceleration.

By recording the mass of the box and the applied force, she calculated the acceleration of the box using Newton's second law, which states that the acceleration of an object is directly proportional to the net force applied to it and inversely proportional to its mass (F = ma).

After calculating the expected acceleration based on the applied force and mass, Jane compared it to the measured acceleration value.

This comparison allows her to verify whether the measured acceleration aligns with the calculated value, thereby testing the principles of Newton's Laws of Motion.

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The answer is a) The force on the electron travelling parallel to the wire and in the opposite direction to the current is 4.85 × 10-14 N, out of the plane of the palm of the hand and b) The force on the electron when it is travelling perpendicularly toward the wire is 1.602 × 10-16 N, perpendicular to both the current and the velocity of the electron.

(a) The direction of the force can be found using the right-hand rule. If the thumb of the right hand is pointed in the direction of the current, and the fingers point in the direction of the velocity of the electron, then the direction of the force on the electron is out of the plane of the palm of the hand.

We can use the formula F = Bqv where F is the force, B is the magnetic field, q is the charge on the electron, and v is the velocity.

Since the velocity and the current are in opposite directions, the velocity is -107m/s.

Using the formula F = Bqv, the force on the electron is found to be 4.85 x 10-14 N.

(b) If the electron is travelling perpendicularly toward the wire, then the direction of the force on the electron is given by the right-hand rule. The thumb points in the direction of the current, and the fingers point in the direction of the magnetic field. Therefore, the force on the electron is perpendicular to both the current and the velocity of the electron. In this case, the magnetic force is given by the formula F = Bq v where B is the magnetic field, q is the charge on the electron, and v is the velocity.

Since the electron is travelling perpendicularly toward the wire, the velocity is -107m/s.

The distance from the wire is 10 cm, which is equal to 0.1 m.

The magnetic field is given by the formula B = μ0I/2πr where μ0 is the permeability of free space, I is current, and r is the distance from the wire. Substituting the values, we get B = 2 x 10-6 T.

Using the formula F = Bqv, the force on the electron is found to be 1.602 x 10-16 N.

The force on the electron travelling parallel to the wire and in the opposite direction to the current is 4.85 × 10-14 N, out of the plane of the palm of the hand. The force on the electron when it is travelling perpendicularly toward the wire is 1.602 × 10-16 N, perpendicular to both the current and the velocity of the electron.

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An RLC series circuit is an electrical circuit that consists of a resistor (R), an inductor (L), and a capacitor (C) connected in series.

The answers are:

a) The impedance of the RLC series circuit at 120 Hz is 217.4 Ω.

b) The impedance of the RLC series circuit at 5.00 kHz is 37.9 Ω.

The components are connected one after the other, forming a single loop for the flow of current. The resistor (R) provides resistance to the flow of current, converting electrical energy into heat.

The impedance determines how the circuit responds to different frequencies of alternating current. At certain frequencies, the impedance may be minimal, resulting in resonance, while at other frequencies, the impedance may be high, leading to a reduction in current flow.

RLC series circuits are widely used in electronics and electrical systems for various applications, such as filtering, signal processing, and frequency response analysis.

(a) To find the impedance of the RLC series circuit at 120 Hz, we need to consider the resistive, inductive, and capacitive components.

The impedance (Z) of the circuit can be calculated using the formula:

[tex]Z = \sqrt(R^2 + (XL - XC)^2)[/tex]

where:

R = resistance = 2.40 Ω

XL = inductive reactance = 2πfL, where f is the frequency and L is the inductance

XC = capacitive reactance = 1/(2πfC), where f is the frequency and C is the capacitance

Given:

[tex]L = 120\mu H = 120 * 10^{-6} H[/tex]

[tex]C = 78.0 \mu F = 78.0 * 10^{-6} F[/tex]

f = 120 Hz

Now we can calculate the impedance:

[tex]XL = 2\pi fL = 2\pi (120 Hz)(120 * 10^{-6} H)\\XC = 1/(2\pi fC) = 1/(2\pi (120 Hz)(78.0 * 10^{-6} F))[/tex]

Calculate XL and XC:

XL = 0.0902 Ω

XC = 217.3 Ω

Substitute the values into the impedance formula:

[tex]Z = \sqrt(2.40^2 + (0.0902 - 217.3)^2)[/tex]

Calculate Z:

Z = 217.4 Ω

Therefore, the impedance of the RLC series circuit at 120 Hz is 217.4 Ω.

(b) To find the impedance of the RLC series circuit at 5.00 kHz, we follow the same steps as in part (a), but with a different frequency.

Given:

[tex]f = 5.00 kHz = 5.00 * 10^3 Hz[/tex]

Calculate XL and XC using the new frequency:

[tex]XL = 2\pi fL = 2\pi (5.00 * 10^3 Hz)(120 × 10^{-6} H)\\XC = 1/(2\pi fC) = 1/(2\pi (5.00 * 10^3 Hz)(78.0 * 10^{-6} F))[/tex]

Calculate XL and XC:

XL = 37.7 Ω

XC = 3.40 Ω

Substitute the values into the impedance formula:

[tex]Z = \sqrt(2.40^2 + (37.7 - 3.40)^2[/tex])

Calculate Z:

Z = 37.9 Ω

Therefore, the impedance of the RLC series circuit at 5.00 kHz is 37.9 Ω.

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Hence, the circuit's impedance is (2.40 - j2.64) Ω.

The given values are

Resistance, R = 2.40 Ω

Inductance, L = 120 µH

Capacitance, C = 78.0 µF

Frequency, f = 120 Hz = 0.120 kHz

Impedance formula for an RLC circuit is,

Z = R + j (XL - XC)

Here, XL is the inductive reactance, and XC is the capacitive reactance.

They are given by,

XL = 2πfL

XC = 1/2πfC

(a) At 120 Hz,

XL = 2πfL

     = 2 × 3.14 × 120 × 120 × 10⁻⁶

     = 90.76 ΩXC

     = 1/2πfC

     = 1/2 × 3.14 × 120 × 78.0 × 10⁻⁶

     = 169.58 Ω

So, the impedance of the circuit is,

Z = R + j (XL - XC)

  = 2.40 + j (90.76 - 169.58)

  ≈ 2.40 - j78.82 Ω

(b) At 5.00 kHz,

XL = 2πfL

     = 2 × 3.14 × 5 × 10³ × 120 × 10⁻⁶

     = 37.68 ΩXC

     = 1/2πfC

     = 1/2 × 3.14 × 5 × 10³ × 78.0 × 10⁻⁶

     = 40.32 Ω

So, the impedance of the circuit is,

Z = R + j (XL - XC)

  = 2.40 + j (37.68 - 40.32)

  ≈ 2.40 - j2.64 Ω

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The resultant ground velocity of the plane is approximately 600.37 km/hr, heading N53.49°E.

To determine the resultant ground velocity of the plane, we can use vector addition. We'll break down the velocities into their horizontal and vertical components and then add them together.

Airplane velocity (with respect to the ground) = 600 km/hr, heading N40°W

Wind velocity = 150 km/hr from the NE

Let's first convert the velocities to their horizontal (East-West) and vertical (North-South) components:

Airplane velocity:

Horizontal component = 600 km/hr * cos(40°) = 600 km/hr * cos(40°) ≈ 458.37 km/hr (towards the west)

Vertical component = 600 km/hr * sin(40°) = 600 km/hr * sin(40°) ≈ 384.57 km/hr (towards the north)

Wind velocity:

Horizontal component = 150 km/hr * cos(45°) = 150 km/hr * cos(45°) ≈ 106.07 km/hr (towards the east)

Vertical component = 150 km/hr * sin(45°) = 150 km/hr * sin(45°) ≈ 106.07 km/hr (towards the north)

Now, let's add the horizontal and vertical components separately to find the resultant ground velocity:

Horizontal component of resultant velocity = Airplane horizontal component + Wind horizontal component

Horizontal component of resultant velocity = 458.37 km/hr - 106.07 km/hr ≈ 352.30 km/hr (towards the west)

Vertical component of resultant velocity = Airplane vertical component + Wind vertical component

Vertical component of resultant velocity = 384.57 km/hr + 106.07 km/hr ≈ 490.64 km/hr (towards the north)

Using the Pythagorean theorem, we can find the magnitude of the resultant ground velocity:

Magnitude of resultant ground velocity = sqrt((Horizontal component)^2 + (Vertical component)^2)

Magnitude of resultant ground velocity = sqrt((352.30 km/hr)^2 + (490.64 km/hr)^2)

Magnitude of resultant ground velocity ≈ 600.37 km/hr

Finally, we can determine the direction of the resultant ground velocity using trigonometry:

Direction = arctan(Vertical component / Horizontal component)

Direction = arctan(490.64 km/hr / 352.30 km/hr)

Direction ≈ 53.49°

Therefore, the resultant ground velocity of the plane is approximately 600.37 km/hr, heading N53.49°E.

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The image height is 1/3 cm and the magnification is 2/3.

Given data:Height of object, h = 0.5 cm

Focal length, f = -2 cm Object distance, u = -1 cm

The sign convention used here is that distances to the left of the lens are negative, while distances to the right are positive.

1) Determine whether the image is real or virtualThe focal length of the concave lens is negative, which indicates that it is a diverging lens. A diverging lens always forms a virtual image for any location of the object.

Therefore, the image is virtual.

2) Determine whether the image is upright or invertedThe height of the object is positive and the image height is negative. Thus, the image is inverted.

3) From the thin lens formula, we can calculate the image distance as follows:1/f = 1/v - 1/u1/-2 = 1/v - 1/-1v = 2/3 cmThe image is located 2/3 cm behind the lens.

4) The magnification is given by the following equation:m = (-image height) / (object height)h′ = m * hIn this example, the object height and the image height are both given in centimeters.

Therefore, we do not need to convert the units.

m = -v/u

= -(2/3) / (-1)

= 2/3h′

= (2/3) * (0.5)

= 1/3 cm

Therefore, the image height is 1/3 cm and the magnification is 2/3.

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The correct amount to administer is approximately 1.556 mL of Cefazonin (Kefzol).

Dose required: 350 mg

Stock concentration: 225 mg/mL

To calculate the volume required, we can use the formula:

Volume required = Dose required / Stock concentration

Substituting the given values:

Volume required = 350 mg / 225 mg/mL

Calculating this expression gives us:

Volume required ≈ 1.556 mL

Now, according to the given information, the total volume provided when 500 mg of Cefazonin (Kefzol) is added to 2 mL of 0.9% sodium chloride is 2.2 mL. Since the volume required (1.556 mL) is less than the total volume provided (2.2 mL), it is appropriate to administer this amount for a single dose.

Therefore, the correct amount to administer is approximately 1.556 mL of Cefazonin (Kefzol).

Please note that it is essential to follow the storage instructions and discard the medication after 24 hours, as mentioned in the given information.

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The gauge pressure at point A in the garden hose can be calculated as follows:The gauge pressure is the difference between the absolute pressure in the hose and atmospheric pressure.

The formula to calculate absolute pressure is given by;P = ρgh + P₀Where:P is the absolute pressureρ is the density of the liquid (water in this case)g is the acceleration due to gravity h is the height of the water column above the point A.

P₀ is the atmospheric pressure. Its value is usually 101325 Pa.The height of the water column above point A is equal to the height of the water level in the tank minus the length of the hose, which is 1 meter.

Let's assume that the tank is filled to a height of 2 meters above point A.

the height of the water column above point A is given by; h = 2 m - 1 m = 1 m

The density of water is 1000 kg/m³.

A.P = ρgh + P₀P

= (1000 kg/m³)(9.81 m/s²)(1 m) + 101325 PaP

= 11025 Pa

The absolute pressure at point A is 11025 Pa.

Gauge pressure = Absolute pressure - Atmospheric pressureGauge pressure

= 11025 Pa - 101325 PaGauge pressure

= -90299 Pa

Since the gauge pressure is negative, this means that the pressure at point A is below atmospheric pressure.

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Answer:

26.The correct answer is C. 636 nm.

To determine the wavelength of light emitted by the laser, we can use the equation:

                          E = hc/λ

where E is the energy of a photon,

           h is Planck's constant (approximately 6.626 x 10^-34 J·s),

           c is the speed of light (approximately 3.00 x 10^8 m/s), and

           λ is the wavelength of light.

The energy difference between the lasing energy levels is given as 2.95 eV.

To convert this energy to joules, we can use the conversion factor:

                     1 eV = 1.602 x 10^-19 J

Therefore, the energy difference can be expressed as:

               E = (2.95 eV) * (1.602 x 10^-19 J/eV)

we can rearrange the equation to solve for the wavelength:

                                        λ = hc/E

Substituting the values:

λ = (6.626 x 10^-34 J·s * 3.00 x 10^8 m/s) / [(2.95 eV) * (1.602 x 10^-19 J/eV)]

                         λ ≈ 636 nm

Therefore, the wavelength of light emitted by the laser is approximately 636 nm.

The correct answer is C. 636 nm.

27.The correct answer is A. It increases.

As the energy gap decreases, the conductivity of a material generally increases. This is because a smaller energy gap allows more electrons to move across the band gap and contribute to the conduction of electricity.

Therefore, the correct answer is A. It increases.

28.The correct answer is C. helium nuclei.

The common name for α particles is helium nuclei.

Therefore, the correct answer is C. helium nuclei.

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The voltage at the center of the triangle is 5.54x10^7 V.

1. The "One Volt = 1 Amp/sec" is false. Voltage and current are two different quantities. Voltage is the difference in electrical potential energy between two points, while current is the rate of flow of electric charge. The unit for voltage is the volt (V), while the unit for current is the ampere (A).

2. The voltage at the center of the triangle is 5.54x10^7 V.

3. The electron will curve down.

Here are the solutions:

1. Voltage is defined as the potential difference between two points in an electric circuit. Current is defined as the rate of flow of electric charge. The unit for voltage is the volt (V), while the unit for current is the ampere (A). One volt is not equal to one amp per second.

2. The voltage at the center of the triangle can be calculated using the following formula:

V = kQ/r`

where:

* V is the voltage in volts

* k is the Coulomb constant (8.988x10^9 N⋅m^2/C^2)

* Q is the total charge in coulombs

* r is the distance between the charges in meters

In this case, the total charge is Q = 5 μC + 5 μC - 7 μC = 3 μC. The distance between the charges is r = 3 mm = 0.003 m. Plugging in these values, we get:

V = 8.988x10^9 N⋅m^2/C^2 * 3 μC / 0.003 m = 5.54x10^7 V

Therefore, the voltage at the center of the triangle is 5.54x10^7 V.

3. When an electron is shot into a capacitor, it will be attracted to the positive plate of the capacitor. The electron will curve down because the positive plate is below the electron. The electron will continue to move until it reaches the positive plate.

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Given the following information: Mass of disk (m) = 2 Kg.

The radius of the disk (r) = 60 cm

Force applied (F) = 50 N

Time (t) = 12 seconds

Initial angular velocity (ωi) = 0

Find out the final angular velocity (ωf) and angular displacement (θ) of the disk.

a) The torque produced by the force is given as: T = F × r

where, T = torque, F = force, and r = radius of the disk

T = 50 N × 60 cm = 3000 Ncm

The angular acceleration (α) produced by the torque is given as:

α = T / I where, I = moment of inertia of the disk.

I = (1/2) × m × r² = (1/2) × 2 kg × (60 cm)² = 0.36 kgm²α = 3000 Ncm / 0.36 kgm² = 8333.33 rad/s².

The final angular velocity (ωf) of the disk is given as:

ωf = ωi + α × t

because the disk was initially at rest,

ωi = 0ωf = 0 + 8333.33 rad/s² × 12 sωf = 100000 rad/s.

Thus, the angular velocity of the disk is 100000 rad/s.

b)The work done (W) by the force is given as W = F × d

where d = distance traveled by the point of application of the force along the circumference of the disk

d = 2πr = 2 × 3.14 × 60 cm = 376.8 cm = 3.768 mW = 50 N × 3.768 m = 188.4 J.

The kinetic energy (Kf) of the disk after 12 seconds is given as:

Kf = (1/2) × I × ωf²Kf = (1/2) × 0.36 kgm² × (100000 rad/s)²Kf = 1.8 × 10¹² J

By the Work-Energy Theorem, we have:Kf - Ki = W

where, Ki = initial kinetic energy of the disk

Ki = (1/2) × I × ωi² = 0

Rearrange the above equation to find out the angular displacement (θ) of the disk.

θ = (Kf - Ki) / Wθ = Kf / Wθ = 1.8 × 10¹² J / 188.4 Jθ = 9.54 × 10⁹ rad.

Thus, the angular displacement of the disk is 9.54 × 10⁹ rad.

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(a) To calculate the gravitational acceleration at the surface of Mars, we can use the formula for gravitational acceleration: g=GM/r2​,

where G is the gravitational constant, M is the mass of Mars, and r is the radius of Mars.

(b) To determine if the gravitational potential approximation for Mars is accurate over a larger or smaller range of values of ∆y compared to Earth, we need to compare the values of g Mars and Earth and analyze the impact of the difference in radius.

Calculation: Given:

Mass of Mars (M) = 6.421 × 10^23 kg

Radius of Mars (r) = 3.4 × 10^6 m

Gravitational constant (G) = 6.67430 × 10^-11 m^3 kg^-1 s^-2(

a) Calculate the gravitational acceleration at the surface of Mars: g=GMr2g = r2GM​g= (6.67430×10−11 m3 kg−1 s−2)×(6.421×1023 kg)(3.4×106 m)2g=(3.4×106m)2(6.67430×10−11m3kg−1s−2)×(6.421×1023kg)​g ≈ 3.71 m/s2g≈3.71m/s2

(b) To compare the accuracy of the gravitational potential approximation, we need to consider the change in g(∆g) as ∆y varies. The gravitational potential approximation is accurate as long as ∆y is small enough that the change in g is negligible compared to the initial value.

Therefore, the gravitational potential approximation will be accurate over a smaller range of values of ∆y on Mars compared to Earth.

Final Answer:

(a) The gravitational acceleration at the surface of Mars is approximately 3.71 m/s^2.

(b) The gravitational potential approximation for Mars will be accurate over a smaller range of values of ∆y compared to Earth due to the smaller magnitude of Δg on Mars.

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(a) The rate at which energy is being stored in the magnetic field can be calculated using the formula P = 0.5 * L * (di/dt)^2, where P is the power, L is the inductance, and di/dt is the rate of change of current.

Given that L = 2.6 H and di/dt = 0.12 A/s, substituting these values into the formula gives P = 0.5 * 2.6 * (0.12)^2 = 6.7856 W.

(b) The rate at which thermal energy is appearing in the resistance can be calculated using the formula P = I^2 * R, where P is the power, I is the current, and R is the resistance. At 0.12 s, the current flowing through the coil is the same as the current delivered by the battery, which is given by ε / R = 87 V / 9.412 Ω = 9.2407 A. Substituting these values into the formula gives P = (9.2407)^2 * 9.412 = 3.1557 W.

(c) The rate at which energy is being delivered by the battery is equal to the power delivered, which can be calculated using the formula P = ε * I, where P is the power, ε is the battery's electromotive force, and I is the current flowing through the coil. Substituting the given values ε = 87 V and I = 9.2407 A into the formula gives P = 87 * 9.2407 = 56.6143 W.

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The magnification of the object being photographed is approximately -0.2361.

The magnification (m) of an object being photographed by a camera with a lens can be calculated using the formula:

m = -v/u

Where:

m is the magnification

v is the image distance

u is the object distance

Given:

Focal length of the lens (f) = 34 cm

Object distance (u) = 110 cm

To find the image distance (v), we can use the lens formula:

1/f = 1/v - 1/u

Substituting the known values:

1/34 = 1/v - 1/110

Simplifying the equation:

1/v = 1/34 + 1/110

Calculating this expression:

1/v = (110 + 34) / (34 × 110)

1/v = 144 / 3740

v = 3740 / 144

v ≈ 25.9722 cm

Now, we can calculate the magnification using the image distance and object distance:

m = -v/u

m = -25.9722 cm / 110 cm

m ≈ -0.2361

Therefore, the magnification of the object being photographed is approximately -0.2361.

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The orbital velocity of Sputnik I is 7.91 x 10³ m/s and its altitude is 0.75 x 10⁶ m.

When Sputnik I was launched by the U.S.S.R. in October 1957, American scientists wanted to know as much as possible about this new artificial satellite.

If Sputnik orbited Earth once every 96 min, calculate its orbital velocity and altitude. (6.2)

The expression for the period of revolution of an artificial satellite of mass m around a celestial body of mass M is given by,

T = 2π √ (R³/GM)

where, T = Period of revolution

R = Distance of the artificial satellite from the center of the earth

G = Universal Gravitational constant

M = Mass of the earth

For Sputnik I,

Period of revolution, T = 96 minutes (convert it to seconds)

T = 96 * 60

= 5760 seconds

Universal Gravitational constant,

G = 6.67 x 10⁻¹¹ Nm²/kg²

Mass of the earth, M = 5.98 x 10²⁴ kg

The altitude of Sputnik I from the surface of the earth can be calculated as,

Altitude = R - R(earth)where,

R(earth) = radius of the earth

= 6.4 x 10⁶ m

Orbital velocity of Sputnik I

Orbital velocity of Sputnik I can be calculated as,

v = 2πR/T

Substitute the value of

T = 5760 seconds and solve for v,

v = 2πR/5760m/s

Calculate R, we have

T = 2π √ (R³/GM)5760

= 2π √ (R³/(6.67 x 10⁻¹¹ x 5.98 x 10²⁴))

Solve for R,

R = (GMT²/4π²)¹/³

= [(6.67 x 10⁻¹¹ x 5.98 x 10²⁴) x (5760)²/4π²]¹/³

= 7.15 x 10⁶ m

Therefore,

Altitude = R - R(earth)

= 7.15 x 10⁶ m - 6.4 x 10⁶ m

= 0.75 x 10⁶ m

Orbital velocity, v = 2πR/T

= (2 x 3.14 x 7.15 x 10⁶ m)/5760 sec

= 7.91 x 10³ m/s

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