The wavelength of the 3-K radiation from outer space is 4.85 x 10^-3 m or 4.85 mm.
The energy of a photon with a wavelength of 1.25 cm can be calculated using the formula:
E = hc/λ
where h is Planck's constant (6.626 x 10^-34 J s), c is the speed of light (2.998 x 10^8 m/s), and λ is the wavelength in meters. We need to convert the wavelength to meters:
1.25 cm = 0.0125 m
Substituting these values into the formula, we get:
E = (6.626 x 10^-34 J s) x (2.998 x 10^8 m/s) / (0.0125 m)
E = 1.59 x 10^-20 J
To convert this energy to electron volts (eV), we can use the conversion factor:
1 eV = 1.602 x 10^-19 J
Dividing the energy by this conversion factor, we get:
E = (1.59 x 10^-20 J) / (1.602 x 10^-19 J/eV)
E = 0.099 eV
Therefore, the energy of the microwave photon with a wavelength of 1.25 cm is 0.099 eV.
The energy of the 3-K radiation can be calculated using the formula:
E = k_B T
where k_B is the Boltzmann constant (1.381 x 10^-23 J/K) and T is the temperature in Kelvin. We can substitute the temperature T = 3 K into the formula:
E = (1.381 x 10^-23 J/K) x (3 K)
E = 4.143 x 10^-23 J
To find the wavelength of this radiation, we can use the formula:
E = hc/λ
Rearranging the formula to solve for λ, we get:
λ = hc/E
Substituting the values for h, c, and E, we get:
λ = (6.626 x 10^-34 J s) x (2.998 x 10^8 m/s) / (4.143 x 10^-23 J)
λ = 4.85 x 10^-3 m
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A vehicle that weights 400 n on the surface of the earth is traveling in outer space at a speed of 400 m/s. It can be stopped by applying a constant force of 20 n for:_____.
A vehicle that weights 400 n on the surface of the earth is traveling in outer space at a speed of 400 m/s. It can be stopped by applying a constant force of 20 n for 800 secs.
Newtons 3 law whilst one object exerts pressure on any other object, the second object exerts a force on the first item this is identical in significance, but opposite in course; for each movement, there is an identical, but opposite response; referred to as the law of action-reaction.
You can use the equal formulation d = rt because of this distance equals the price instance's time. To solve for speed or rate use the components for pace, s = d/t this means that pace equals distance divided by way of time. To clear up for time use the components for time, t = d/s this means that time equals distance divided by way of velocity.
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Find the minor measurement of the vernier scale by taking 49, 1mm divisions of the main scale and dividing it into 50 vernier divisions.
length of V-50 = 49mm
length of V-1 = 49/50mm
= 0.98mm
so,
minor measurement = (M-1) - (V-1)
= 1mm -0.98mm
= 0.02mm
☆ Therefore,
The minor measurement of the vernier scale is 0.02mm.
☆...hope this helps...☆
_♡_mashi_♡_
What happens within a protostar to create a star?
Clouds of gas in space
A protostar resembles a star, but its core is still too cool for fusion to occur. The protostar's heating as it contracts is the only source of brightness. The light that protostars release is typically blocked by dust, making them challenging to study in the visible spectrum.
By the time a protostar is produced, the cloud has flattened and a protostellar disk is spinning around it. The cloud starts spinning as it collapses. These disks occasionally form planetary systems and are thought to slow the protostar's rotation. The protostar produces a powerful magnetic field as it revolves.
Additionally, a strong protostellar wind—a movement of particles into space—is produced by the magnetic field. A lot of protostars also release gas into space in the form of fast-moving streams or jets.
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A sky-diver jumps from a stationary balloon. His initial downwards acceleration is 10m/s².
Fig. 1.1 shows the directions of the air resistance and the weight of the sky-diver.
The mass of the sky-diver is 60 kg and his weight is 600 N.
(a) Explain, using ideas about the forces, why his initial downwards acceleration is 10m/s².
Explanation:
a. The force acting down is gravity, on Earth gravity is 10 m/s^2. When the skydiver jump, the acceleration will start out as -10 m/s^2, but it will eventually equals the air resistance , which is called terminal velocity.
A car goes around a curve at a constant speed. what is the direction of the net force on the car?
Answer:
The direction of the net force is at a tangent to the circle.
If the ocular lens magnifies a specimen 10x, and the objective lens used magnifies the specimen 35x. What is the total magnification being used to observe the specimen?
Answer:
350x^2
Explanation:
I'm assuming you're saying you're using them together, like stacking them on top of each other, so if that's the case then this is a simple multiplication problem, which can be written as 10x(35x). Solve it and you get 350x^2.
A bungee jumper jumps from a bridge and starts accelerating towards a lake below. What energy transfer is he experiencing?
Group of answer choices
a. Elastic potential to kinetic
b. Kinetic to elastic potential
c. Kinetic to gravitational potential
d. Gravitational potential to kinetic
D. The energy transfer he is experiencing is Gravitational potential to kinetic.
Energy transferred experienced by the bungee jumper
The bungee jumper possesses gravitational potential energy due to his position above the ground level (on a bridge).
As he starts accelerating towards a lake below, his gravitational potential energy will be converted into kinetic energy.
Thus, the energy transfer he is experiencing is Gravitational potential to kinetic.
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An astronaut in training is seated at the end of a horizontal arm 8 meters long. How many revolutions per second must the arm make for the astronaut to experience a horizontal acceleration of 4.0g
Answer:
a = v^2 / R where a is the centripetal acceleration and R the radius
v = 2 π R N / t where (N / t) is the revolutions / sec giving distance / time
a = 4 g horizontal acceleration
4 g = (2 π R N / t)^2 / R = 4 π^2 * R * (N / t)^2
(N / t)^2 = g / (π^2 * R)
N / t = 1 / π * (g / R)^1/2 = .318 * (9.80 m / s^2 / 8 m)^1/2 = .352 / sec
Arrange these objects according to size, largest to smallest. Group of answer choices traditional dwarf elliptical, ultra-compact dwarfs, globular clusters ultra-compact dwarfs, traditional dwarf galaxies, globular clusters globular clusters, traditional dwarf elliptical, ultra-compact dwarfs globular clusters, ultra-compact dwarfs, traditional dwarf elliptical
Among the given objects, the largest one is dwarf elliptical, then ultra-compact dwarf and then globular clusters. Thus, option a is correct.
To find the correct answer, we need to know about the galaxies.
Are galaxies bigger than globular clusters?Globular clusters are spherical groups of stars that are primarily located in the spiral galaxy' stellar halo. There are 150 or so globular clusters in our Milky Way galaxy, some of which house our galaxy's oldest stars.Globular clusters are much smaller than galaxies.a system of gas and dust, as well as millions or billions of stars, kept together by gravitational attraction is called galaxy.Ultracompact dwarfs are more extended and have higher surface brightness than typical dwarf nuclei.Dwarf elliptical galaxies are elliptical galaxies that are smaller than ordinary elliptical galaxies. They are quite common in galaxy groups and clusters and are usually companions to other galaxies.Thus, we can conclude that, option a is correct.
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Draw a vector representing the direction of the electric field. The orientation of the vector will be graded. The location and length of the vector will not be graded.
1. E is a vector pointing to the right.
2. E is a vector pointing to the left.
3. E is a vector pointing to the right.
4. E is a vector pointing to the left
What is a Vector ?A vector is an object that has both a magnitude and a direction
Geometrically, we can picture a vector as a directed line segment, whose length is the magnitude of the vector and with an arrow indicating the direction. The direction of the vector is from its tail to its head.Examples of vectors in nature are velocity, momentum, force, electromagnetic fields, and weight.Learn more about Vectors here:
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A 540 gram object is attached to a vertical spring, causing the spring’s length to change from 70 cm to 110 cm.
What is the change in the object’s gravitational potential energy?
Group of answer choices
-2.1 J
-0.84 J
2.1 J
0.84 J
The change in the object’s gravitational potential energy is 2.11 J.
Change in the object's gravitational potential
ΔP.E = mg(hf - hi)
where;
m is mass of the objecthf is final height hi is initial heightΔP.E = 0.54 x 9.8(1.1 - 0.7)
ΔP.E = 2.11 J
Thus, the change in the object’s gravitational potential energy is 2.11 J.
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For vibrational motion, what term denotes the maximum displacement from the equilibrium position?.
For vibrational motion, the amplitude denotes the maximum displacement from the equilibrium position.
What is vibrational motion?
The motion in which there are some vibrations about the fixed position called mean position is termed vibrational motion. For example, the motion of a wave on the string has perpendicular vibration.
What is amplitude?
The amplitude is defined as the maximum displacement of the vibration from its mean position. For vibrational motion, the mean position is the equilibrium position. So amplitude is the term that denotes the maximum displacement from the equilibrium position. For example, the displacement of the string wave from its equilibrium position is the amplitude of the string wave.
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In a given neuron, the current membrane potential is 60 mv. what does this tell you about the current potential compared to the resting membrane potential of 70 mv?
The membrane is depolarized compared to the resting membrane potential.
Through conformational changes from closed, nonconducting states to an open, current-conducting state, membrane depolarization activates sodium channels. Na+ channels open slowly and change from an open state to a nonconducting, rapidly inactivated state as a result of delayed openings, which contribute to the declining fraction of INa induced by prolonged depolarization. Additionally, sodium channels can move swiftly from the closed state to the fast-inactivated state. When the membrane is depolarized, inactivated channels are prevented from opening.
The distribution of channels between the closed and slow-inactivated states, however, limits the number of excitable sodium channels as a function of the membrane potential since slow inactivation acts at greater negative potentials than fast inactivation.
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a bat emitts a sonar sound wave (343 m/s) that bounces off a mosquito 8.42 m away
The time lapse between when the bat emits the sound and when it hears the echo is 0.05 s.
From the question given above, the following data were obtained:
Velocity of sound (v) = 343 m/s
Distance (x) = 8.42 m
Time (t) =?
We can obtain obtained the time as illustrated below:
v = 2x / t
343 = 2 × 8.42 / t
343 = 16.84 / t
Cross multiply
343 × t = 16.84
Divide both side by 343
t = 16.84/343
t = 0.05 s
Thus, the time between when the bat emits the sound and when it hears the echo is 0.05 s.
How does a bat know how far away something is?A bat emits a sound wave and carefully listens to the echoes that return to it. The returning information is processed by the bat's brain in the same way that we processed our shouting sound with a stopwatch and calculator. The bat's brain determines the distance of an object by measuring how long it takes for a noise to return.
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Correction question:
A bat emits a sonar sound wave (343 m/s) that bounces off a mosquito 8.42 m away. How much time elapses between when the bat emits the sound and when it hears the echo? (Unit = s)
Consider a pipe of length l that is open at both ends. What are the wavelengths of the three lowest-pitch tones produced by this pipe?.
The correct option is D.
The three lowest-pitch tones generated by this pipe have wavelengths of = 2 L, L, 2 L/3
What is meant by wavelength?Measuring a wave's size from one peak to the next is all that is required. The wavelength is only the separation between the crests of the successive waves, if one thinks of a sound wave as being similar to a water wave.
According to the given Information:The length of a pipe with an opening at either end and a length l is given by:
[tex]l=\frac{n \lambda}{2}[/tex]
or
[tex]\lambda=\frac{2 l}{n}[/tex]
Initial pitch tone: n = 1.
[tex]\lambda_{1}=2 l[/tex]
The first pitch tone has n = 2
[tex]\begin{aligned}\lambda_{2} &=\frac{2 l}{2} \\\lambda_{2} &=l\end{aligned}[/tex]
The initial pitch tone has n = 3.
[tex]\begin{aligned}&\lambda_{3}=\frac{2 l}{3} \\&\lambda_{3}=\frac{2 l}{3}\end{aligned}[/tex]
The three lowest-pitch tones generated by this pipe have wavelengths of 2 L, L, and 2 L/3, respectively.
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I understand that the question you are looking for is :
Consider a pipe of length L that is open at both ends. What are the wavelengths of the three lowest-pitch tones produced by this pipe?
Answer
A. 2L, L, L/2
B. 2L, L, L/2
C. 4L, 2L, L
D. 2L, L, 2L/3
E. 4L, 4L/3, 4L/5
A 150 g piece of metal has a specific heat capacity of 0.845 J/g°C. If it takes 3.30x 103 J to heat the metal to 120°C, the initial temperature of the metal was
The initial temperature of the metal was 94.24⁰C.
Initial temperature of the metal
The initial temperature of the metal is calculated as follows;
Q = mcΔθ
where;
Δθ is change in temperatureΔθ = Q/mc
Δθ = (3,300) / (0.854 x 150)
Δθ = 25.76⁰C
Δθ = T₂ - T₁
T₁ = T₂ - Δθ
T₁ = 120 - 25.76
T₁ = 94.24⁰C
Thus, the initial temperature of the metal was 94.24⁰C.
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If a 100 ω resistor is placed across a 0. 10 μf charged capacitor which is initially charged to 3 v. How long does it take it to discharge to 2v to 1v?
1. The time taken to discharge to 2 V is 2×10⁻⁹ s
2. The time taken to discharge to 1 V is 5×10⁻¹⁰ s
Energy stored in a capacitorThe energy stored in a capacitor is given by
E = ½CV²
But
E = Pt
Thus,
Pt = ½CV²
Where
E is the energy C is the capacitorV is the voltageP is the power t is the timeWith the formula (Pt = ½CV²), we can determine the time in each case. Detail below:
1. How to determine the time required to discharge to 2 VData obtained from the question include:
Power (P) = 100 wCapacitor (C) = 0.10 μF = 1×10⁻⁷ FVoltage (V) = 2 VTime (t) = ?Pt = ½CV²
100 × t = ½ × 1×10⁻⁷ × 2²
Divide both sides by 100
t = (½ × 1×10⁻⁷ × 2²) / 100
t = 2×10⁻⁹ s
Thus, the time required to discharge to 2 V is 2×10⁻⁹ s
2. How to determine the time required to discharge to 1 VData obtained from the question include:
Power (P) = 100 wCapacitor (C) = 0.10 μF = 1×10⁻⁷ FVoltage (V) = 1 VTime (t) = ?Pt = ½CV²
100 × t = ½ × 1×10⁻⁷ × 1²
Divide both sides by 100
t = (½ × 1×10⁻⁷ × 1²) / 100
t = 5×10⁻¹⁰ s
Thus, the time required to discharge to 1 V is 5×10⁻¹⁰ s
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How does the distance between adjacent orbits in a hydrogen atom vary with increasing values of the orbital number n?
The distance between adjacent orbits in a hydrogen atom vary with increasing values of the orbital number n because:
Option C: The energy difference between adjacent orbit radii decreases with increasing values of the principal quantum number.What determines the distance of the electrons from the nucleus?In an atom, an electron is known to be attracted to a given nucleus by the use of "electromagnetic force".
Note that similar to a baseball, the faster the electron is said to go, the farther away from the nucleus it is known to be seen. Therefore, the electrons in an atom are known to be in a state where they are moving a lot and very fast, so they are said to be far away from their nucleus.
Note also that energy difference between what we call consecutive levels tends to often decreases as well as increases.
Therefore, The distance between adjacent orbits in a hydrogen atom vary with increasing values of the orbital number n because:
Option C: The energy difference between adjacent orbit radii decreases with increasing values of the principal quantum number.Learn more about hydrogen atom from
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See full question below
Consider the hydrogen atom. How does the distance between adjacent orbits in a hydrogen atom vary with increasing values of the orbital number n?
The distance difference between adjacent orbit radii varies with increasing values of the principal quantum number.
The distance difference between adjacent orbit radii increases with increasing values of the principal quantum number.
The distance difference between adjacent orbit radii remains constant with increasing values of the principal quantum number.
The distance difference between adjacent orbit radii decreases with increasing values of the principal quantum number.
d. e. Study the given diagram and calculate the following: i. work done by load ii. work done by effort iii. M.A iv. V.R v. efficiency [Friction is neglected]
i. The work done by the load is load x distance moved by load.
ii. The work done by effort is effort applied x distance moved by effort.
iii. The mechanical advantage of the simple machine is Load/effort.
iv. The velocity ratio of the simple machine is 2.
v. The efficiency of the machine is M.A/V.R x 100%.
Work done by the loadThe work done by the load is the product of the load and the distance through which the load is moved. The magnitude is calculated as follows;
Work done by the load = load x distance moved by load
Work done by effortThe work done by the effort is the product of the effort and the distance through which the effort is applied. The magnitude is calculated as follows;
Work done by effort = effort applied x distance moved by effort
Mechanical advantage of the simple machineM.A = Load/Effort
Velocity ratio of the simple machineV.R = distance moved by effort / distance moved by load
V.R = 30 cm/15 cm
V.R = 2
Efficiency of the machineE = (M.A/V.R) x 100%
Thus, the work done by the load is load x distance moved by load.
The work done by effort is effort applied x distance moved by effort.
The mechanical advantage of the simple machine is Load/effort.
The velocity ratio of the simple machine is 2.
The efficiency of the machine is M.A/V.R x 100%.
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Which of these can happen to energy in a system? Select all that apply.
Group of answer choices
a. It can be converted.
b. It can remain constant.
c. It can be destroyed.
d. It can be created.
A. What can happen to energy in a system is that it can be converted.
What is the principle of conservation of energy?The principle of conservation of energy states that energy can neither be created nor destroyed but can be converted from one form to another.
The energy of a system can be converted from one form to another.
Thus, what can happen to energy in a system is that it can be converted.
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If a car increases its velocity from +6 m/s to +30 m/s in 6 seconds, its acceleration in m/s2 is__________.
Answer:
4m/s^2Explanation:
v = 30m/s, u = 6m/s, t = 6s
Change in velocity = v(final velocity) - u (initial velocity)
v-u = 30-6 = 24m/s
acceleration = (v-u)/t
(24m/s)/6s = 4m/s^2
an interplantetary speedcarft moving at 20000m/s.how far will it travell in one day?(give your answer in km)
Explanation:
Step I: 1 day means 24 hours * 60 minutes * 60 seconds.
So 1 day have 86400 seconds.
Step I:
Now,
To calculate the travel distance 20,000m/s* 84600 is 1728000000m
Step Ill:
Now convert the meter in kilometer
Because 1 km = 1000 m
So, = 1728000000/1000 = 172800OKm
Why is it so much easier to determine the length of the day on mars than on venus?
The thick cloud cover of Venus makes it difficulty in visibility because of impenetrable in visible light which has made it tough for astronomers to measure the length of the planet's day . Whereas Mars have clear atmosphere which do not hinder visibility
The most important reason of why is it difficult to measure the length of the day on Venus is tougher than in mars is that Mars' atmosphere is generally clear because of which their is no visibility issue on mars but due to the thick cloud covering of Venus makes it impenetrable in visible light , due which visibility is not that clear and it become tough to observe the length of the day on Venus.
Mars is much larger than Venus in size which makes easier to observe
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A 5 cm radius conducting sphere has a charge density of 2.0x10-6 C/m2 on its surface. Find the electric potential of the sphere.
noob anyways I can't get u vape because first reason I lost ur money and second I felt bad for losing ur and I will not on yourself as I am thinking of a number my number is a multiple of 6.What other numbers must my number is a multiple of 6.What other numbers
The measure of arc ed is 68°. what is the measure of angle efd? 34° 68° 112° 132°
The measure of the angle EFD = 34°. That is option A.
Calculation of an angle of an intercepted arcThe radii of the given circle are CE and CD.
The measure of the given angle of the arc is = 68°
But the central angle of an intercepted arc measure the double of the inscribed angle by the same arc.
Therefore, the measure of angle of an intercepted arc is 68/2 = 34°
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Answer:
34
Explanation:
edge 2023
If you will be rinsing your regulator after removing it from the cylinder, you must make sure that the ______ ______ is firmly in place. Select one: Mouthpiece plug Alternate-air-source retainer Dust cap None of the above
If you will be rinsing your regulator after removing it from the cylinder, you must make sure that the dust cap is firmly in place.
What is Dust cap?A dust cap is a gently curved dome mounted either in concave or convex orientation over the central hole of most loudspeaker diaphragms.
Thus, if you will be rinsing your regulator after removing it from the cylinder, you must make sure that the dust cap is firmly in place.
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The galaxy M33 is the third largest galaxy in the Local Group after M31 and the Milky Way. The galaxy M33 has a moderate-sized central bulge and two spiral arms that emerge directly out of the bulge and wrap around in rather poorly defined arcs with many cross-connections between the arms. Thus, M33 is classified as a _____ normal spiral.
Galaxy M33 is classified as a Triangulum normal spiral.
The galaxy M33 is the third largest galaxy in the Local Group after M31 and the Milky Way. The galaxy M33 has a moderate-sized central bulge and two spiral arms that emerge directly out of the bulge and wrap around in rather poorly defined arcs with many cross-connections between the arms.
This spiral galaxy is located in the triangle shaped constellation earning a pet name as triangular galaxy. This galaxy star formation rate is ten times higher than average found in Andromeda galaxy. It has relatively bright apparent. This galaxy was given by Charles Messier and he classified it as Triangulum normal spiral.
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A construction worker pushes a 25 kg load in a wheelbarrow for a distance of 5.0 m, using a horizontal force of 50.0 N. How much work is done by the worker on the wheelbarrow?
Group of answer choices
a. 55 J
b. 250 J
c. 1250 J
d. 10 J
B. The amount of work done by the worker on the wheelbarrow is 250 J.
Work done by the workerThe amount of work done by the worker on the wheelbarrow is calculated as follows;
W = Fd
where;
F is applied forced is displacementW = 50 x 5
W = 250 J
Thus, the amount of work done by the worker on the wheelbarrow is 250 J.
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6 latter word and it has a s and a I and it has mass (9.______________liquids, and gases all have mass.)
Answer:
Solids
Explanation:
Solids, liquids, and gases all have mass.
A military surveillance satellite is in circular orbit around the Earth at an altitude of 1,000 km above the surface. If the Earth's mass is 5.97 x 1024 kg and its radius is 6,370 km, what is the satellite's orbital speed in m/s
The orbital speed of the satellite is 7.35*10^3 m/s.
What is orbital speed?
The speed of the satellite in its orbit is termed the orbital speed.
The orbital speed is given by the formula,
[tex]v=\sqrt{\frac{GM}{r}}[/tex]
where G is the universal gravitational constant, M is the mass of the planet and r is the distance of the satellite from the center of the planet.
Here the distance of the satellite from the center of the planet is the sum of the planet's radius and the height attained by the satellite above the ground. So
r=6370 + 1000
r=7370 km
Given the mass of the planet is 5.97*10^24 kg and the value of the gravitational constant is 6.67*10^(-11) N m^2 kg^(-2), substitute these values in the formula of the orbital speed.
Note: 1 km=1000 m
[tex]v=\sqrt{\frac{6.67\times10^{-11}\text{ N m}^2\text{kg}^{-2}\times5.97\times10^{24} \text{ kg}}{7370 \text{ km}}} \\ v=\sqrt{\frac{6.67\times10^{-11}\text{ N m}^2\text{kg}^{-2}\times5.97\times10^{24} \text{ kg}}{7370\times 1000 \text{ m}}} \\ v= 7.35\times 10^3 \text{ m/s}[/tex]
Hence the orbital velocity of the satellite is 7.35*10^3 m/s.
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