Please help and show work :)
A housecat accelerates (from rest) at 0.21 m/s^2 for 7 seconds. How fast is the cat going after 7 seconds? Hint: look at the kinematic equations.

Answer: Febreze was initially not successful because it was marketed as an air freshener. At the time, people did not think their homes smelled bad, so they did not see the need for an air freshener.

Explanation:

Field investigations and classroom experiments can differ in several ways due to the unique characteristics and limitations of each setting.

Here are some reasons why field investigations can differ from classroom experiments:Real-world context: Field investigations take place in the natural environment, allowing students to observe and study phenomena in their natural setting. This context provides a more authentic experience and helps students understand the complexities and interactions of the real world. In contrast, classroom experiments often involve controlled conditions that may not accurately reflect real-world scenarios.Complexity and unpredictability: Field investigations often deal with complex and unpredictable variables, such as weather, terrain, and natural processes. This complexity can make it challenging to control and manipulate variables compared to classroom experiments, where conditions can be tightly controlled.Scale and scope: Field investigations can involve larger scales and broader scopes than classroom experiments. For example, studying the ecosystem of a forest or the geological features of a landscape requires observing and collecting data over a large area, which may not be feasible within a classroom setting.Resources and equipment: Classroom experiments often have access to a controlled and well-equipped laboratory, whereas field investigations may require specialized equipment, transportation, and logistical planning to conduct research in the field. This can add logistical challenges and resource constraints to field investigations. Ethical considerations: Field investigations may involve interactions with living organisms and ecosystems, raising ethical considerations related to environmental impact and the well-being of organisms. Classroom experiments, on the other hand, can be designed with ethical considerations in mind, ensuring the well-being and safety of participants.Overall, field investigations provide students with valuable opportunities to engage with the natural world, understand its complexity, and develop skills in observation, data collection, and critical thinking. Classroom experiments, on the other hand, offer controlled environments for testing specific hypotheses and concepts. Both approaches have unique benefits and play important roles in science education, providing complementary learning experiences for students.

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The answer is C

Please dont get mad at me if this is not right im pretty sure it is
hope this helps!

Explanation:

See drawing

Using V= I * resistance

v = (.025 A) * (r+400 Ω)      <====given

and

v = (.060A) * ( r+100 Ω)

equate the two equations

.025(r + 400) = .060(r+100)    <==== solve for r = 114 .3 Ω

 then use this value in either of the equations to calculate v = 12.9 v

Q1: A: The force applied would be the same, but the distance would be shorter.

Q2: D: mechanical advantage

Q3: A: Both levers place the fulcrum in between the applied force and the object being lifted.

Q4: C: The man is farther from the fulcrum than the rock is.

Q5: 96.1%

Q6: the friction of the individual pulleys

Q7: 10 kg moving at 5 m/s

Q8: The first rider has more mass.

Q9: mass and speed

Q10: The kinetic energy of the 15 balls will be greater than the kinetic energy of one ball.

Q11: from the electric fields of charged subatomic particles

When using a dial test indicator (DTI), it is essential to ensure that the gauge is positioned at a right angle to the contact surface for accurate readings.

However, in certain situations, it may be challenging to achieve a perfect right angle alignment. In such cases, adjustments can be made to compensate for the misalignment and obtain accurate measurements.To adjust the DTI reading when the gauge is not positioned at a right angle to the contact surface, the following steps can be taken:Determine the misalignment angle: Measure the angle at which the DTI is misaligned from the right angle position. This can be done using a protractor or by estimating the deviation visually.Calculate the correction factor: Based on the misalignment angle, calculate the correction factor using trigonometric functions such as sine or cosine. The correction factor accounts for the difference between the actual displacement and the displacement measured by the DTI.Apply the correction factor: Multiply the correction factor by the DTI reading to adjust the measurement. This compensates for the misalignment and provides a more accurate reading.It's important to note that adjusting the DTI reading can introduce some degree of error, especially if the misalignment is significant. Therefore, it is always preferable to position the gauge at a right angle to the contact surface whenever possible to obtain the most precise measurements.

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Someone might choose to use film rather than digital methods, for photography because some film methods allow you to print a picture from anywhere without electricity. Option (A) is right.

Film Photography is the practice of capturing images using photographic film, which is a light-sensitive material coated with an emulsion containing silver halide crystals.

Advantages

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The nucleus never has a negative charge, and its charge cannot change from positive to negative.

The nucleus of an atom has a charge that can either be positively charged or neutral, as stated in the sentence. Positively charged protons and neutral neutrons make up the nucleus.

As a result, it can either be neutral or have a positive charge when there are an equal amount of protons and neutrons. The other claims made in the question are untrue. The charge of the nucleus is always positive and cannot ever become negative.

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The speed of the sound wave is 660 meters per second.

To calculate the speed of the sound wave, we need to use the formula:
Speed = Frequency x Wavelength
Here, the frequency of the sound wave is given as 880Hz, and the wavelength is given as 0.75. To get the answer, we just need to plug these values into the formula and solve for the speed:
Speed = 880 x 0.75
Speed = 660 meters per second
It's important to note that the speed of sound depends on the medium through which it is traveling. In air, the speed of sound is approximately 343 meters per second at standard temperature and pressure. However, this value can change depending on factors such as temperature, humidity, and altitude.
Understanding the speed of sound is important in various fields, such as music, engineering, and physics. For example, in music, the speed of sound determines the pitch of a note, while in engineering, it can be used to design and optimize acoustic systems. In physics, it's used to study the properties of waves and to explain phenomena such as Doppler effect and sonic booms.

Therefore, the speed of the sound wave is 660 meters per second.

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Answer:

Black's Law Dictionary defines a retroactive law as a law “that looks backward or contemplates the past, affecting acts or facts that existed before the act came into effect.” While Congress often considers legislation that would apply retroactively, the Constitution imposes some limited constraints on such laws.

1 km = 1000 m

1 hr = 3600 s

So, 18 km/hr = (18 * 1000) / (3600) m/s = 5 m/s

And 2 m/s^2 = 2 m/s^2

Now, we can use the formula for final velocity (v) when an object starts with an initial velocity (u) and accelerates at a constant rate (a) for a given time (t):

v = u + at

Plugging in the values, we get:

v = 5 + (2 * 5) m/s v = 15 m/s

To convert this back to km/hr, we use the inverse conversions: v = (15 * 3600) / (1000) km/hr v = 54 km/hr

Therefore, the car’s velocity in km/hr after 5 seconds is 54 km/hr.

The mass of the person standing at the top of Mt. Everest with 4336500 joules of potential energy is approximately 49.1 kilograms.

To find the mass of the person standing at the top of Mt. Everest, we can use the formula for potential energy:
Potential Energy (PE) = mass (m) x acceleration due to gravity (g) x height (h)
We know that the potential energy (PE) is 4336500 joules, the height (h) is 8850 meters, and the acceleration due to gravity (g) is 9.8 m/s^2. So, we can rearrange the formula to solve for the mass (m):
m = PE / (g x h)
Substituting the given values, we get:
m = 4336500 J / (9.8 m/s^2 x 8850 m)
m ≈ 49.1 kg
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B. The equivalent resistance of the two resistors is 20.0 ohms.

C. The voltmeter will measure the voltage across the 30.0 ohm resistor.

D. The 30.0 ohm resistor will stop working.

B. The total resistance of the circuit is 60.0 ohms. This is because the 30.0 ohm resistor and the 60.0 ohm resistor are in parallel, and the equivalent resistance of two resistors in parallel is equal to the product of the resistors divided by the sum of the resistors.

R_T = 1/(1/R_1 + 1/R_2 + ...)

In this case, the product of the resistors is:

30.0 ohms × 60.0 ohms = 1800 ohms,

and the sum of the resistors is:

30.0 ohms + 60.0 ohms = 90.0 ohms.

Therefore, the equivalent resistance of the two resistors is 1800 ohms / 90.0 ohms = 20.0 ohms.

C. Meter 1 is an ammeter, and it is connected in series with the 30.0 ohm resistor. This means that the ammeter will measure the current flowing through the 30.0 ohm resistor.

Meter 2 is a voltmeter, and it is connected in parallel with the 30.0 ohm resistor. This means that the voltmeter will measure the voltage across the 30.0 ohm resistor.

D. When the switch is opened, the 30.0 ohm resistor will stop working. This is because the switch is in series with the 30.0 ohm resistor, and when the switch is opened, the circuit is broken.

The 60.0 ohm resistor will continue to work, because it is in parallel with the switch, and the current will continue to flow through the 60.0 ohm resistor even when the switch is opened.

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Answer:

Yes and no

Explanation:

Momentum in a typical mechanical system made up of macroscopic elements cannot be "hidden" from human sight. But momentum can be concealed in other systems. For instance, in an electromagnetic system, where the electromagnetic field is often inaudible to human vision, momentum can be transferred to it.

Explanation:

A disc of mass 3kg and radius 50cm rotate on a horizontal plane zcity of 40rad if it is brought to rest in 5sec calculate it's angular acceleration

Complete the following statement: A collision is elastic if...

...the total kinetic energy of the system is conserved before and after the collision, and there is no net loss of energy due to deformation or other non-conservative forces. In an elastic collision, the objects involved bounce off each other without any permanent deformation or generation of heat.

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Electromagnetic radiation of a specific wavelength or energy is called a photon.

Electromagnetic radiation of a specific wavelength or energy is called a photon. A photon is a fundamental particle of light and is the smallest unit of electromagnetic radiation. It has no mass, but carries energy and momentum. Photons travel through space at the speed of light and can exhibit both wave-like and particle-like behaviors.The energy of a photon is directly proportional to its frequency or inversely proportional to its wavelength. This relationship is known as the Planck-Einstein equation and can be expressed as E = hf, where E is the energy of the photon, h is Planck's constant, and f is the frequency of the radiation.When a photon interacts with matter, it can be absorbed, reflected, or scattered. The energy of the photon is transferred to the absorbing material, causing an excitation or ionization of the atoms or molecules. This process is the basis for many scientific techniques, such as spectroscopy, where the absorption or emission of photons by a material is used to identify its chemical composition and structure. Understanding the properties of photons and their interactions with matter is essential for many fields of science, including optics, physics, and chemistry.

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Answer:

Physics

Explanation:

The period of a pendulum is the time it takes for the pendulum to complete one full oscillation, or swing back and forth. The frequency of a pendulum, on the other hand, is the number of oscillations it completes in one second. The frequency is calculated by taking the reciprocal of the period, or 1/period.

If the period of a pendulum decreases from 8 seconds to 2 seconds, it means that the pendulum is swinging back and forth more quickly, as it is taking less time to complete one full oscillation. To find the new frequency of the pendulum, we can use the formula:

frequency = 1 / period

Initially, when the period was 8 seconds, the frequency was:

frequency = 1 / 8 = 0.125 Hz

After the period decreased to 2 seconds, the new frequency can be calculated as:

frequency = 1 / 2 = 0.5 Hz

So, the frequency of the pendulum increases from 0.125 Hz to 0.5 Hz when the period decreases from 8 seconds to 2 seconds. This means that the pendulum is oscillating at a faster rate, completing more oscillations in one second.

Answer:

(9) - 10 N, Up

(10) - 5 N at 37 degrees

Explanation:

Refer to the attached image.

Answer:

To determine the force acting at an angle of 20 degrees to the northwest, we need to break it down into its horizontal and vertical components. Since "northwest" is at an angle of 45 degrees to both north and west, we can find the horizontal and vertical components of the force by using trigonometry.

The horizontal component of the force can be found by multiplying the force by the cosine of the angle:

F_horizontal = 45 N * cos(20°) ≈ 42.9 N

The vertical component of the force can be found by multiplying the force by the sine of the angle:

F_vertical = 45 N * sin(20°) ≈ 15.4 N

Therefore, the force acting at an angle of 20 degrees to the northwest can be resolved into a horizontal component of about 42.9 N to the west, and a vertical component of about 15.4 N to the north.

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(a) To determine the acceleration of the ball while it is in flight, we can use the equation of motion:

v = u + at

where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.

In this case, the ball is thrown straight up, so its final velocity at the highest point is 0 m/s. The initial velocity is unknown, the acceleration is due to gravity and is approximately -9.8 m/s^2 (negative since it acts in the opposite direction of motion), and the time of flight is 2.21 s.

Using the equation, we can solve for the acceleration:

0 = u - 9.8 * 2.21

u = 9.8 * 2.21

u ≈ 21.658 m/s

Therefore, the acceleration of the ball, while it is in flight, is approximately 21.658 m/s^2 in the upward direction.

(b) When the ball reaches its maximum height, its velocity is 0 m/s. This occurs when the ball is momentarily at rest before falling back down. Therefore, the magnitude of the velocity when the ball reaches its maximum height is 0 m/s.

(c) To find the initial velocity of the ball, we can use the equation:

v = u + at

At the highest point, the final velocity is 0 m/s, the acceleration is -9.8 m/s^2 (due to gravity), and the time is 2.21 s.

0 = u - 9.8 * 2.21

u = 9.8 * 2.21

u ≈ 21.658 m/s upward

Therefore, the initial velocity of the ball is approximately 21.658 m/s upward.

(d) The maximum height reached by the ball can be determined using the equation for vertical displacement:

s = ut + (1/2)at^2

At the highest point, the final displacement is 0 m, the initial velocity is 21.658 m/s upward, and the time of flight is 2.21 s.

0 = 21.658 * 2.21 + (1/2) * (-9.8) * (2.21)^2

0 = 47.864 + (-5.5294)

5.5294 = 47.864

Therefore, there seems to be an error in the calculations as the equation does not hold true. Please check the given values and equations to ensure accuracy.

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B) A pendulum is a mass suspended on the bottom of a string.

A pendulum is a simple mechanical device that consists of a mass (known as the bob) suspended from a fixed point by a string, wire, or rod. When the bob is pulled to one side and released, it swings back and forth under the influence of gravity, forming a regular pattern of motion. The time it takes for the pendulum to complete one full swing (i.e., from one extreme position to the other and back again) is known as its period. The period of a pendulum is affected by the length of the string and the strength of gravity. The longer the string, the longer the period, and the stronger the gravity, the shorter the period.Pendulums have a wide range of practical applications, such as timekeeping, as seen in grandfather clocks. They are also used in scientific experiments to measure time intervals and gravitational acceleration.

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Answer:

Explanation:

I frequency = 278 Hz

Answer:

(1) - [tex]W=20,000 \ J[/tex]

(2) - [tex]W=8.00 \times 10^6 \ J[/tex]

(3) - [tex]W=7500\ J[/tex] for part a and [tex]P=375 \ W[/tex] for part b

Conceptual:

What is work?

- Work is simply energy transferred from one place to another. Work is a scalar quantity and is measured in Joules, J.  

[tex]\boxed{\left\begin{array}{ccc}\text{\underline{Formulas used to calculate work:}}\\\\W=F\Delta r\cos\theta \ \text{(Constant} \ \vec F)\\\\W=\int\limits^{r_2}_{r_1} {F\cos\theta} \, dr \ \text{(Varible} \ \vec F)\end{array}\right}[/tex]

** "F" is the applied force on the object, "Δr" is the displacement of the object, "θ" is the measured angle between the applied force and direction of displacement. For our purposes we are assuming θ is 0, which would make the cosine equal 1.**

What is power?

- Power is the measure of work being done per second, J/s. A joule per second is a Watt, W.

[tex]\boxed{\left\begin{array}{ccc}\text{\underline{Formula used to calculate power:}}\\\\P=\frac{W}{t} \end{array}\right}[/tex]

Step-by-step:

Question (1):

Given:

[tex]F= 1000 \ N\\\\\Delta r=20 \ m\\\\Eqn\rightarrow W=F \Delta r[/tex]

Plug the known values into the equation.

[tex]\Longrightarrow W=(1000)(20)\\\\\therefore \boxed{\boxed{W=20,000 \ J}}[/tex]

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Question (2):

Given:

[tex]F= 8000 \ N\\\\\Delta r=1000 \ m\\\\Eqn\rightarrow W=F \Delta r[/tex]

Plug the known values into the equation.

[tex]\Longrightarrow W=(8000)(1000)\\\\\therefore \boxed{\boxed{W=8.00 \times 10^6 \ J}}[/tex]

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Question (3):

Given (part a):

[tex]F= 500 \ N\\\\\Delta r=15 \ m\\\\Eqn\rightarrow W=F \Delta r[/tex]

Plug the known values into the equation.

[tex]\Longrightarrow W=(500)(15)\\\\\therefore \boxed{\boxed{W=7500\ J}}[/tex]

Given (part b):

[tex]W= 7500 \ J\\\\t=20 \ s\\\\Eqn\rightarrow P=\frac{W}{t}[/tex]

Plug the known values into the equation.

[tex]\Longrightarrow P=\frac{7500}{20} \\\\\therefore \boxed{\boxed{P=375 \ W}}[/tex]

Thus, all given questions are solved.

Answer:

Explanation:velocity is aterm used under the branchof the kinematics of motionwhich defines the correlationbetween the position of the body and direction of motion with respect to time interval. thus the velocity is ameasure of both speed and direction of motion

Answer:

F1 = K M1 M2 / R1^2

F2 = K M1 M2 / R2^2

F2 / F1 = (R1 / R2)^2 = 1/4

The new force is 1,000,000 N / 4 = 250,000 N

The new gravitational force between the planets, when they are two million miles apart, is approximately 2.66972 × [tex]10^(-10)[/tex] N.

To solve this problem, we can use the inverse square law of gravity, which states that the gravitational force between two objects is inversely proportional to the square of the distance between them.

Let's denote the initial distance between the planets as "r1" (1 million miles) and the initial gravitational force as "F1" (1,000,000 N). The final distance between the planets will be "r2" (2 million miles), and we need to find the new gravitational force, which we'll denote as "F2."

The inverse square law formula for gravitational force is:

F = G * (m1 * m2) / [tex]r^{2}[/tex]

where:

F is the gravitational force,

G is the gravitational constant ≈ 6.67430 × [tex]10^(-11)[/tex] N m²/kg²,

m1 and m2 are the masses of the two objects (in this case, the masses of the planets), and

r is the distance between the centers of the two objects.

Since the two planets are identical, their masses are the same. Let's denote the mass of each planet as "m."

Now, we can set up two equations using the given information:

For the initial setup:

F1 = G * (m * m) / [tex]r1^2[/tex]

For the final setup:

F2 = G * (m * m) / [tex]r2^2[/tex]

Since both planets have the same mass (m), we can set these two equations equal to each other:

G * (m * m) / [tex]r1^2[/tex] = G * (m * m) /  [tex]r2^2[/tex]

Now, we can solve for F2:

F2 = (G * (m * m) / [tex]r1^2[/tex]) *  [tex]r2^2[/tex]

Substitute the given values:

F2 = (6.67430  × [tex]10^(-11)[/tex] N m²/kg² * (m * m) / (1 million mile)^2 * (2 million mile[tex]s)^2[/tex]

Note: We need to convert the distances from miles to meters before plugging into the equation. 1 mile is approximately 1609.34 meters.

r1 = 1 million miles * 1609.34 m/mile ≈ 1.60934 × [tex]10^9[/tex] meters

r2 = 2 million miles * 1609.34 m/mile ≈ 3.21868  × [tex]10^9[/tex] meters

F2 = (6.67430  × [tex]10^(-11)[/tex]  N m²/kg² * (m * m) / (1.60934  × [tex]10^9[/tex] meter[tex]s)^2[/tex]) * (3.21868 × [tex]10^9[/tex] meters[tex])^2[/tex]

Now, we see that the mass (m) cancels out, leaving us with

F2 = 6.67430 × [tex]10^(-11)[/tex] N m²/kg² * 3.21868  × [tex]10^9[/tex]  meters)^2 / (1.60934  × [tex]10^9[/tex]  meters[tex])^2[/tex]

F2 = 6.67430  × [tex]10^(-11)[/tex]  N m²/kg² * (3.21868 / 1.6093[tex]4)^2[/tex]

F2 = 6.67430  × [tex]10^(-11)[/tex]  N m²/kg² * 4

F2 ≈ 2.66972  × [tex]10^(-10)[/tex]  N

Hence, the new gravitational force between the planets, when they are two million miles apart, is approximately 2.66972 × [tex]10^(-10)[/tex]  N.

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Yes, energy was lost in this scenario. The original amount of energy applied to the strings by Jose was 1,000 joules. However, the measured energy of the vibrating strings is only 800 joules.

Yes, energy was lost in this scenario. The original amount of energy applied to the strings by Jose was 1,000 joules. However, the measured energy of the vibrating strings is only 800 joules. This discrepancy indicates that 200 joules of energy were lost. The energy loss can be attributed to various factors. Firstly, friction between Jose's fingers and the strings converts some of the applied energy into heat energy. This heat energy dissipates into the surrounding environment, resulting in a loss of energy from the system. Additionally, there may be other forms of energy loss involved, such as air resistance or sound energy radiated away from the vibrating strings. These energy losses contribute to the discrepancy between the original applied energy and the measured energy of the vibrating strings. Therefore, in this case, the difference between the initial and measured energy values indicates that some energy was lost in the form of heat, sound, or other forms of energy dissipation.

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Answer:

The velocity of the object 7 seconds from the start of its journey is 70 ms⁻¹.

Explanation:

To find the velocity of the object after 7 seconds after the start of its journey, we first needs to consider the acceleration of the object in the each part of the motion: the initial 2 seconds and the subsequent 4 seconds.

As we have been given the displacement (s), the initial velocity (u), the final velocity (v) and the time (t), we can use the following SUVAT equations to find the acceleration (a) and the final velocity (v) of the object during the first leg of its journey (initial 2 seconds):

[tex]\textsf{Given:} \quad s=20\;\textsf{m}, \quad u=0\;\textsf{ms}^{-1}, \quad t=2\;\textsf{s}[/tex]

[tex]\begin{aligned}\textsf{Using:} \quad s&=ut+\dfrac{1}{2}at^2\\\\\implies 20&=0(2)+\dfrac{1}{2}a(2^2)\\20&=2a\\a&=10\; \sf ms^{-2}\end{aligned}[/tex]

[tex]\begin{aligned}\textsf{Using:} \quad v&=u+at\\\\\implies v&=0+10(t)\\v&=20\; \sf ms^{-1}\end{aligned}[/tex]

Therefore, the acceleration of the object for the initial 2 seconds of its journey is 10 ms⁻² and its final velocity is 20 ms⁻¹.

The initial velocity of the second leg of the journey is equal to the final velocity of the previous leg of the journey. Therefore, as we have been given the displacement (s) and the time (t), and have calculated the initial velocity (u), we can use the following SUVAT equation to find the acceleration (a) of the object during the second leg of its journey (next 4 seconds):

[tex]\textsf{Given:} \quad s=160\;\textsf{m}, \quad u=20\;\textsf{ms}^{-1}, \quad t=4\;\textsf{s}[/tex]

[tex]\begin{aligned}\textsf{Using:} \quad s&=ut+\dfrac{1}{2}at^2\\\\\implies 160&=20(4)+\dfrac{1}{2}a(4^2)\\160&=80+8a\\8a&=80\\a&=10\; \sf ms^{-2}\end{aligned}[/tex]

Therefore, the acceleration of the object for the next 4 seconds of its journey is 10 ms⁻².

We can observe that the acceleration for both legs of the journey is the same, and so the object is moving with constant acceleration of 10 ms⁻².

Assuming the acceleration remains constant through the entire journey of the object, to calculate the velocity of the object after 7 seconds from the start, use the following SUVAT equation:

[tex]u=0\; \textsf{ms}^{-1}, \quad a=10\;\textsf{ms}^{-2}, \quad t=7\;\textsf{s}[/tex]

[tex]\begin{aligned}\textsf{Using:} \quad v&=u+at\\\\\implies v&=0+10(7)\\v&=70\; \sf ms^{-1}\end{aligned}[/tex]

Therefore, the velocity of the object 7 seconds from the start of its journey is 70 ms⁻¹.

The forces acting on each satellite are the force of gravity from the Earth and the normal force from the Earth's surface. The ratio of these forces is equal to the ratio of the masses of the satellites. Since the satellites have the same mass, the ratio of the forces is 1:1.

The speed of each satellite is the same. This is because the satellites are in circular orbits, and the speed of a satellite in a circular orbit is constant.

The speed of a satellite does not depend on its mass. This is because the force of gravity from the Earth is proportional to the mass of the satellite, but the centripetal force required to keep the satellite in a circular orbit is also proportional to the mass of the satellite. Therefore, the ratio of the forces is equal to 1, and the speed of the satellite is constant.

The satellite with the larger radius will complete one orbit around the Earth in a long time. This is because the centripetal force required to keep a satellite in a circular orbit is proportional to the square of the radius of the orbit. Therefore, the satellite with the larger radius will experience a larger centripetal force, and it will take longer to complete one orbit.

Here are the formulas used to calculate the speed and period of a satellite in circular orbit:

Speed:

v = sqrt(GM/r)

where:

v is the speed of the satellite

G is the gravitational constant

M is the mass of the Earth

r is the radius of the orbit

Period:

T = 2pi sqrt(r/GM)

U

where:

T is the period of the satellite

pi is approximately 3.14159

r is the radius of the orbit

G is the gravitational constant

M is the mass of the Earth

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