Answer:
the angular velocity θ' = 27.125 rad/s
the angular acceleration is θ'' = 238.044 rad/s²
Explanation:
The plate OAB forms an equilateral triangle which rotates counterclockwise with increasing speed about point O.
If the normal and tangential components of acceleration of the centroid C at a certain instant are 68 m/s2 and 22 m/s2
From the distance (radius r) of the centroid C in the centre of the equilateral triangle to the point of rotation O; the position OC can be calculated as:
r = OC
[tex]r = \dfrac{2}{3} \sqrt {160^2 - 80^2}[/tex]
r = 0.667 × 138.564
r = 92.42 mm
r = 0.09242 m
However; the angular velocity can be determined by using the expression:
[tex]a_n = \theta'^2 r[/tex]
where;
[tex]a_n[/tex] = normal acceleration = 68 m/s²
r = 0.09242 m
[tex]\theta'^2 =[/tex] angular velocity = ???
[tex]68 = \theta'^2 * 0.09242[/tex]
[tex]\dfrac{68}{0.09242} = \theta'^2[/tex]
θ'² = 735.771478
θ' = [tex]\sqrt{735.771478}[/tex]
θ' = 27.125 rad/s
Thus; the angular velocity θ' = 27.125 rad/s
Similarly ; the angular acceleration can be determined as by the following relation:
[tex]a_t = r \theta"[/tex]
where;
[tex]a_t[/tex] = tangential components of acceleration = 22 m/s²
r = 0.09242 m
[tex]\theta"[/tex] = angular acceleration
[tex]22= 0.09242* \theta"[/tex]
[tex]\dfrac{22}{0.09242}= \theta"[/tex]
[tex]\theta"= \dfrac{22}{0.09242}[/tex]
θ'' = 238.044 rad/s²
Thus; the angular acceleration is θ'' = 238.044 rad/s²
Assign a positive velocity to the red box and negative velocity to the blue box. Are they moving in opposite direction?
Answer:
Yes, they are moving in opposite direction one to the other.
Explanation:
Velocity is a vector quantity, which means that it has both magnitude and direction. The magnitude shows the size of the velocity, and the direction shows which way it is moving in reference to a chosen reference direction. If the red box is assigned a positive velocity, and the blue box is assigned a negative velocity, as indicated in the question, then it means that the red box, and the blue box, both move in opposite direction to the other.
How do you use these muscles in your everyday life? What daily activities do you complete that mimic the movements of these exercises
Answer:
If ur talking abkut hamstrings then it would be running that mimics them xplanation:
This was on a gym class quizz and I got it wrong but turned out this was the right answer
Answer:
In this activity, I exercised my hips, thighs, knees, calves, ankles, and legs. In some exercises, I specifically worked on only one type of muscle or on a combination of muscles. For example, the lunges mainly exercised the muscles of the inner thighs while the dead lift worked the muscles of the leg as well as the back and shoulders. I haven't consciously exercised my leg muscles before, but I have often noticed their tightening during my daily body movements, like when I climb the stairs or run to catch the school bus.
Explanation:
Hope this helped:)
What is the frequency, in hertz, of a sound wave (v = 340 m/s) with a wavelength of 68 m?
Answer:
frequency = 5 Hz
Explanation:
F = v/wavelength
F = 340/68 =5Hz
The frequency would be 5 hertz, of a sound wave (v = 340 m/s) with a wavelength of 68 meters.
What is the frequency?It can be defined as the number of cycles completed per second. It is represented in hertz and inversely proportional to the wavelength.
As given in the problem we have to find out the frequency, in hertz, of a sound wave (v = 340 m/s) with a wavelength of 68 meters,
velocity of the sound = 340 meters/second
the wavelength of the sound wave = 68 meters
the velocity of the sound wave = frequency × wavelength of the sound wave
frequency of the sound wave = 340/68
= 5 hertz
Thus, the frequency of the sound wave would be 5 hertz.
To learn more about frequency from here, refer to the link;
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how to find an AC voltage whose peak value is 100V across an 80 ohm resistor. What is the current peak in tjr resistor
Answer:
1.25 A
Explanation:
V = IR
100 V = I (80 Ω)
I = 1.25 A
What is the length of a contention slot in CSMA/CD for (a) a 2-km twin-lead cable (signal propagation speed is 82% of the signal propagation speed in vacuum)
Answer:
1.99*10-4sec
Explanation:
Signal propagation speed=0.82∗2.46∗108m/s
d=2000 m
Tp=20000/0.82∗2.46∗108 sec
ContentionPeriod=2Tp=2∗20000/0.82∗2.46∗10^8
= 1.99* 10^-4seconds
Two carts connected by a 0.05 m spring hit a wall, compressing the spring to 0.02 m. The spring constant k is
N
100
m
What is the elastic potential energy stored from the spring's compression?
Choose 1 answer:
-3.0J
-0.045 J
0.090 J
0.045 J
Answer:
0.045 J
Explanation:
From the question,
The elastic potential energy stored in a spring is given as,
E = 1/2ke²...................... Equation 1
Where E = elastic potential energy, k = spring constant, e = compression.
Given: k = 100 N/m, e = 0.05-0.02 = 0.03 m
Substitute these values into equation 1
E = 1/2(100)(0.03²)
E = 50(9×10⁻⁴)
E = 0.045 J
Hence the right option is 0.045 J
Two carts connected by a 0.05 m spring hit a wall, compressing the spring to 0.02 m.The spring constant k is 100 N/m.
What is the elastic potential energy stored from the spring’s compression?
Answer: 0.045 J
Two charged particles are projected into a region where a magnetic field is directed perpendicular to their velocities. If the charges are deflected in opposite directions, what are the possible relative charges and directions? (Select all that apply.)
Answer:
*If the particles are deflected in opposite directions, it implies that their charges must be opposite
*the force is perpendicular to the speed, therefore it describes a circular movement, one in the clockwise direction and the other in the counterclockwise direction.
Explanation:
When a charged particle enters a magnetic field, it is subjected to a force given by
F = q v x B
where bold letters indicate vectors
this expression can be written in the form of a module
F = qv B sin θ
and the direction of the force is given by the right-hand rule.
In our case the magnetic field is perpendicular to the speed, therefore the angle is 90º and the sin 90 = 1
If the particles are deflected in opposite directions, it implies that their charges must be opposite, one positive and the other negative.
Furthermore, the force is perpendicular to the speed, therefore it describes a circular movement, one in the clockwise direction and the other in the counterclockwise direction.
A block is attached to a horizontal spring and it slides back and forth in simple harmonic motion on a frictionless horizontal surface. At one extreme end of the oscillation cycle, where the block comes to a momentary halt before reversing the direction of its motion, another block is placed on top of the first block without changing its zero velocity. The simple harmonic motion then continues. What happens to the amplitude and the angular frequency of the ensuing motion of the two-block system
Answer:
A = A₀ , w = w₀/√2
Explanation:
This is a problem that we must solve with Newton's second law. They indicate that at the end of the initial movement where the speed is zero, add a mass to the block, we assume that it has the same mass, therefore the total mass is m_total = 2 m. Let's write Newton's second law at this point
[tex]F_{e}[/tex] = m_total a
the elastic force is
F_{e} = - k x
acceleration is
a = d²x / dt²
we substitute
- k x = m_total d²x / dt²
d²x / dt² + (k / m_total) x = 0
we substitute
d²x / dt² + (k /2m) x = 0
the solution to this differential equation is
x = A cos (wt + Ф)
where
w = √ (k / 2m)
to find the constant Ф we use the velocity
v = dx / dt = - Aw sin (wt + Ф)
At the most extreme point and when the new movement begins (t = 0) they indicate that v = 0
0 = - A w sin Ф
for this expression to be zero the sine must be zero therefore Ф = 0
when replacing
x = A cos (wt)
w = 1 /√2 √ (k / m)
if we want to relate to the initial movement (before placing the block)
w₀ = √ (k / m)
w = w₀ /√ 2
The amplitude of the movement is the distance from the equilibrium point to where the movement begins, in this case it is the same as in the initial movement
A = A₀
the subscript is used to refer to the oscillations before placing the second block
we substitute to have the final equation
x = A₀ cos (w₀ t /√2)
A = A₀
w = w₀/√2
The pressure exerted by a phonograph needle on a record is surprisingly large. If the equivalent of 0.600 g is supported by a needle, the tip of which is a circle 0.240 mm in radius, what pressure is exerted on the record in N/m2?
Answer:
[tex]P=3.25x10^{4}\frac{N}{m^2}[/tex]
Explanation:
Hello,
In this case, since pressure is defined as the force applied over a surface:
[tex]P=\frac{F}{A}[/tex]
We can associate the force with the weight of the needle computed by using the acceleration of the gravity:
[tex]F=0.600g*\frac{1kg}{1000g}*9.8\frac{m}{s^2} =5.88x10^{-3}N[/tex]
And the area of the the tip (circle) in meters:
[tex]A=\pi r^2=\pi (0.240mm)^2=\pi (0.240mm*\frac{1m}{1000mm} )^2\\\\A=1.81x10^{-7}m^2[/tex]
Thus, the pressure exerted on the record turns out:
[tex]P=\frac{5.88x10^{-3}N}{1.81x10^{-7}m^2} \\\\P=3.25x10^{4}\frac{N}{m^2}[/tex]
Which is truly a large value due to the tiny area on which the pressure is exerted.
Best regards.
The acceleration due to gravity at the surface of Planet X is 10 m/s2. What is the acceleration due to gravity at an altitude of 3000 km above the surface of this planet?
Options:
a) 10 m/s²
b) 8 m/s²
c) 4.4 m/s²
d) More information is needed
Answer:
d) More information is needed
Explanation:
The acceleration due to gravity at an altitude h above the surface of a planet is given by the equation:
[tex]g = g_o \frac{h^2}{(h + R)^2}[/tex].................(1)
where the altitude above the surface of the planet X, h = 3000 km
Acceleration due to gravity at the surface of the planet, g₀ = 10 m/s²
To be able to get the acceleration due to gravity, g, at an altitude h above the surface of the planet, all parameters must be substituted into equation (1).
Unfortunately, the radius, R, of the planet, X, is not given. This means that there are no sufficient parameters(information) to find the acceleration due to gravity at an altitude h above the surface of the planet.
A 51.0-kg woman wearing high-heeled shoes is invited into a home in which the kitchen has vinyl floor covering. The heel on each shoe is circular and has a radius of 0.400 cm. If the woman balances on one heel, what pressure does she exert on the floor?
Answer:
Pressure = 9.94 x 10⁶ Pascals
Explanation:
given data
mass = 51 kg
radius = 0.400 cm
solution
we know Pressure that is express as here
Pressure = total force on an area ÷ the area of the area .................1
and
Force is the woman's weight so weight will be
Weight = mass × gravity .................2
put here value
Weight = 51 × 9.8 m/s²
Weight = 499.8 Newtons
and
Area of a circle of bottom of the heel = (π) × (radius)² ...................3
put here value
Area = (π) × (0.40 cm)²
Area = 0.502654 cm²
Area = 0.0000502654 m²
and
now we put value in equation 1 we get
Pressure = force ÷ area
Pressure = 499.8 ÷ 0.0000502654
Pressure = 9943221.381 N/m²
Pressure = 9.94 x 10⁶ Pascals
A 1100 kg car pushes a 2200 kg truck that has a dead battery. When the driver steps on the accelerator, the drive wheels of the car push against the ground with a force of 5000 N . Rolling friction can be neglected. You may want to review (Pages 165 - 168) . Part A What is the magnitude of the force of the car on the truck
Answer:
a) 3344 N
b) 3344 N
Explanation:
This is the complete question
1100 kg car pushes a 2200 kg truck that has a dead battery. When the driver steps on the accelerator, the drive wheels of the car push against the ground with a force of 5000 N. Rolling friction can be neglected. A. What is the magnitude of the force of the car on the truck? Express your answer to two significant figures and include the appropriate units. B. What is the magnitude of the force of the truck on the car?
Mass of the car = 1100 kg
Mass of the truck = 2200 kg
Force exerted on the ground by the car = 5000 N
The total mass in the system = 1100 + 2200 = 3300 Kg
Total force in the system = 5000 N
Recall that the force in the system = mass x acceleration
therefore,
5000 = 3300 x a
Total acceleration in the system = 5000/3300 = 1.52 m/s^2
The force on the truck individually fro the car, will be the product of this acceleration and its mass
Force on the truck = 2200 x 1.52 = 3344 N
b) Force on the car From the truck will be equal to this force but will act in the opposite direction.
Force on the car from the truck is 3344 N
power output of 87 W. At what distance will the decibel reading be 120 dB, which is noise level of a loud indoor rock concert
Answer:
r = 2.63 m
Explanation:
To find the distance at which the sound level is 120dB, you first calculate the intensity of the sound. You use the following formula:
[tex]\beta=10log(\frac{I}{I_o})[/tex] (1)
β: sound level = 120dB
I: intensity of the sound
Io: threshold of hearing = 10⁻12W/m^2
You solve the equation (1) for I and replace the values of all parameters:
[tex]\beta=log(\frac{I}{I_o})^{10}\\\\10^\beta=10^{log(\frac{I}{I_o})^{10}}=(\frac{I}{I_o})^{10}\\\\I=10^{\beta/10}I_o[/tex]
[tex]I=10^{120/10}(10^{-12}W/m^2)=1\frac{W}{m^2}[/tex]
Next, you use the following formula for the power of the sound with intensity I, and you solve for r:
[tex]I=\frac{P}{4\pi r^2}\\\\r=\sqrt{\frac{P}{4\pi I}}[/tex]
r: distance at which the sound level is 120dB
P: power of the sound = 87W
I: intensity of the sound = 1W/m^2
You replace the values of I and P for calculating r:
[tex]r=\sqrt{\frac{87W}{4\pi (1W/m^2)}}=2.63m[/tex]
The distance is at 2.63m from the source of the soundr = 2.63m
To prevent damage to floors (and to increase friction) a crutch will often have a rubber tip attached to its end. If the end of the crutch is a circle of radius 0.95 cm without the tip, and the tip is a circle of radius 2.0cm, by what factor does the tip reduce the pressure exerted by the crutch
Answer:
By a factor of about 0.23
Explanation:
Pressure is force over an area: P=F/A
Let's call the pressure without the tip P₁ and the pressure with the rubber piece P₂.
-P₁ = F/A₁= F/(πr₁²)=F/(π0.95²)
-P₂=F/A₂=F/(πr₂²)=F/(π2²)
When they ask "by what factor" it signals that we should find a ratio between the two pressures. To do this, let's divide P₁ by P₂ (I'm going to mathematical step here):
P₁/P₂=[F/(π0.95²)]x[(π2²)/F]= 2²/0.95² = 4/0.9025
So with that we can say:
P₁=(4/0.9025)P₂=4.4P₂ or
P₂=(0.9025/4)P₁=0.23P₁
What this means is that the rubber tip reduced the pressure by almost one quarter, 0.25, of what it would have been without it. Note that because we took a ratio between the two pressures that the units reduce; meaning the ratio is unitless.
By a factor of about 0.23 the tip reduces the pressure exerted by the crutch.
PressureFriction exists as the force resisting the relative motion of solid surfaces, fluid layers, and material elements sliding against each other. There exist several types of friction: Dry friction is a force that disagrees with the relative lateral motion of two solid surfaces in contact.
Pressure exists as force over an area: P=F/A
Let's name the pressure without the tip P₁ and the pressure with the rubber piece P₂.
-P₁ = F/A₁= F/(πr₁²)=F/(π0.95²)
-P₂=F/A₂=F/(πr₂²)=F/(π2²)
let's divide P₁ by P₂
P₁/P₂=[F/(π0.95²)]x[(π2²)/F]= 2²/0.95² = 4/0.9025
So with that, we can say:
P₁=(4/0.9025)P₂=4.4P₂ or
P₂=(0.9025/4)P₁=0.23P₁
Hence, By a factor of about 0.23 the tip reduces the pressure exerted by the crutch,
To learn more about Pressure refer to:
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Passengers in a carnival ride move at constant speed in a circle of radius 5.0 m, making a complete revolution in 4.0 s. As they spin, they feel their backs pressing against the wall holding them in the ride. A. What is the direction of the passengers' acceleration? a. No direction (zero acceleration) b. Directed towards center c. Directed away from center d. Directed tangentially B. What is the passengers' linear speed in m/s? C. What is the magnitude of their acceleration in m/s^2? D. What is their angular speed in rad/s?
Answer:
A. b) Directed towards center
B. [tex]v = 7.854\ m/s[/tex]
C. [tex]a_c = 12.337\ m/s^2[/tex]
D. [tex]w = 1.57\ rad/s[/tex]
Explanation:
The "force" that they feel pressing their backs against the wall is because the reaction to the centripetal acceleration .
A.
This acceleration has its direction towards the center of the circle. (option b)
B.
Their linear speed can be calculated with the equation:
[tex]v = (\theta/t)*r[/tex]
Where [tex]\theta[/tex] is the total angular position moved in radians ([tex]1\ rev = 2\pi\ radians[/tex]), 't' is the time elapsed for the angular position moved and 'r' is the radius. So we have that:
[tex]v = (2\pi/4)*5 = 7.854\ m/s[/tex]
C.
The centripetal acceleration is given by the equation:
[tex]a_c = v^2/r[/tex]
[tex]a_c = 7.854^2/5[/tex]
[tex]a_c = 12.337\ m/s^2[/tex]
D.
Their angular speed is given by the equation:
[tex]w = \theta/t = 2\pi/4 = \pi/2 = 1.57 \ rad/s[/tex]
You walk into an elevator, step onto a scale, and push the "down" button to go directly from the tenth floor to the first floor. You also recall that your normal weight is w= 635 N. If the elevator has an initial acceleration of magnitude 2.45 m/s2, what does the scale read? Express your answer in newtons.
Answer: 479. 425 N
Explanation: the calculation of a body in an elevator obeys Newton law. When it is accelerating upward, the scale reading is greater than the true weight of the person.
It is given by N= m(g+a)
When it is accelerating downward, the scale reading is less than the true weight.
It so given by N = m(g-a)
The answer to the above questions is in the attached photo
Answer:
the scale will read 476.414 N
Explanation:
Weight = 635 N
mass = (weight) ÷ (acceleration due to gravity 9.81 m/^2)
mass m = 635 ÷ 9.81 = 64.729 kg
initial acceleration of the elevator a = 2.45 m/s^2
the force produced by the acceleration of the elevator downwards = ma
your body inertia force try to counteract this force, by a force equal and opposite to the direction of this force, leading to an apparent weight loss
apparent weight = weight - ma
apparent weight = 635 - (64.729 x 2.45)
apparent weight = 635 - 158.586 = 476.414 N
What is meant civilized?
Answer:
at an advanced stage of social and cultural development. "a civilized society"
Explanation:
polite and well-mannered "I went to talk to them and we had a very civilized conversation" hope this helps you :)
discuss any three factors that affect speed of sound
The inductor in a radio receiver carries a current of amplitude 200 mA when a voltage of amplitude 2.40 V is across it at a frequency of 1400 kHz. What is the value of the inductance
Answer:92
Explanation:
A 30 W engine generates 3600 J of energy. How long did it run for?
Answer:
so the time taken will be 120 seconds
Explanation:
power=30W
work done=3600J
time=?
as we know that
[tex]power=\frac{work done}{time taken}[/tex]
evaluating the formula
power×time taken=work done
[tex]time taken=\frac{work done}{power}[/tex]
[tex]time taken=\frac{3600J}{30W}[/tex]
[tex]Time taken=120seconds[/tex]
i hope this will help you :)
Two long, straight, parallel wires are carrying a current I in the same direction. Select all that apply to the force felt by one wire due to the other.
a. It is parallel to the current in the other wire.
b. It is perpendicular to the field produced by the other wire.
c. It is perpendicular to the current in the other wire.
d. It is parallel to the field produced by the other wire.
Answer: Option C.
It is perpendicular to the field produced by the other wire
Explanation
It is perpendicular to the field produced by the other wire because the force between the two parallel line is attractive in nature because the current is flowing in the same direction.
The two parallel wires carrying current in the same direction attract each other and their magnetic field will be attracted to each other.
f NASA wants to put a satellite in a circular orbit around the sun so it will make 2.0 orbits per year, at what distance (in astronomical units, AU) from the sun should that satellite orbit
Answer:
r = 0.63 AU
Explanation:
In this question, we are interested in calculating the distance from the sun that the satellite should orbit.
From the question;
given time period of satellite is T = 0.5 year
Mathematically;
T is directy proportional to r^3/2
Hence;
T = constant * r^3/2
0.5/1 = r^1.5 / 1^1.5
( 1 is the the distance of the earth’s orbit to the sun in AU)
r = 0.63 AU
A thick copper wire connected to a voltmeter surrounds a region of time-varying magnetic flux, and the voltmeter reads 6 volts. If instead of a single wire we use a coil of thick copper wire containing 20 turns, what does the voltmeter read
Answer:
The voltmeter will read 120 volts
Explanation:
For a given magnetic field strength, undergoing a change in flux linkage with time. If a single wire has on it, due to this field an induced EMF of
E = -dΦ/dt
then, increasing the area of the single wire loop by using a wire of the same material containing N number of loops loops will increase the induced EMF to
E = -NdΦ/dt
where E is the induced EMF
N is the number of turns
dΦ/dt is the rate of change of the magnetic flux
From this, we can see that if the voltage reading due to the single thick copper wire is 6 volts, using a coil of thick copper wire containing 20 turns will give a reading of
==> E = 6 x 20 = 120 volts
A car travels at 23.27 m/s for 6.08 s. How far did it travel?
Answer:
As S= Vt
S = (23.27)(6.08)
S = 141.48 m
Explanation:
When the pivot point of a balance is not at the center of mass of the balance, how is the net torque on the balance calculated? When a force is applied directly to the pivot point of a balance, what is the torque due to that force?
Answer:
Explanation:
Net torque is calculated by multiplying the force with distance from the point of application of force to the point of pivot .
If more than 2 forces are present, then we either subtract the product of forces with their respective distances from pivot point or we add them . It depends on whether they both are present on opposite sides of pivot or on same side of pivot .
When a force is applied directly to the pivot point of balance, then the torque on due that force = 0 (zero) .
It is so because the torque is defined as the product of force and perpendicular distance from the pivot point but here the distance is 0 , therefore torque is zero.
A vertical spring-mass system undergoes damped oscillations due to air resistance. The spring constant is 2.15 ✕ 104 N/mand the mass at the end of the spring is 10.7 kg.
(a) If the damping coefficient is b = 4.50 N · s/m, what is the frequency of the oscillator?Hz
(b) Determine the fractional decrease in the amplitude of the oscillation after 7 cycles.
Answer:
7.13Hz
Explanation:pls see attached file
in a _system supply and demand forces affect the production and consumption decisions. There is little to no _control in such a system
Answer:
in a free market system supply and demand forces affect the production and consumption decisions. There is little to no government control in such a system .
Explanation:
A free market is an economic system in which prices are based on competition between private actors and are not affected by other factors besides supply and demand, that is, where there are no external variables that condition the market.
Free market economy systems are characterized by limited government intervention, which characterizes democratic, liberal states and the modern global economy, in which the market in its private face makes most of the economic decisions, leaving the government a minimum amount of necessary regulations.
A 1500-kg car moving at 20 m/s strikes a truck waiting at a traffic light, hooking bumpers. The two continue to move together at 5 m/s. What was the mass of the truck
Answer:
The mass of the truck is 6000kg
The electric field between square the plates of a parallel-plate capacitor has magnitude E. The potential across the plates is maintained with constant voltage by a battery as they are pulled apart to twice their original separation, which is small compared to the dimensions of the plates. The magnitude of the electric field between the plates is now equal to
Answer:
The magnitude of the electric field between the plates is half its initial value.
Explanation:
We know the electric field E = V/d where V = voltage applied and d = separation between plates.
Since V is constant and V = Ed,
So, E₁d₁ = E₂d₂ where E₁ = initial electric field at separation d₁, d₁ = initial separation of plates, E₂ = final electric field at separation d₂ and d₂ = final separation of plates.
So, E₂ = E₁d₁/d₂
Now, the distance between the plates is twice their original separation. Thus, d₂ = 2d₁
So, E₂ = E₁d₁/2d₁ = E₁/2
So, E₂ = E₁/2
Thus, the magnitude of the electric field between the plates is half its initial value.
On Apollo missions to the Moon, the command module orbited at an altitude of 160 km above the lunar surface. How long did it take for the command module to complete one orbit?
Answer:
T = 2.06h
Explanation:
In order to calculate the time that the Apollo takes to complete an orbit around the moon, you use the following formula, which is one of the Kepler's law:
[tex]T=\frac{2\pi r^{3/2}}{\sqrt{GM_m}}[/tex] (1)
T: time for a complete orbit = ?
r: radius of the orbit
G: Cavendish's constant = 6.674*10^-11 m^3.kg^-1.s^-2
Mm: mass of the moon = 7.34*10^22 kg
The radius of the orbit is equal to the radius of the moon plus the distance from the surface to the Apollo:
[tex]r=R_m+160km\\\\[/tex]
Rm: radius of the moon = 1737.1 km
[tex]r=1737.1km+160km=1897.1km=1897.1*10^3 m[/tex]
Then, you replace all values of the parameters in the equation (1):
[tex]T=\frac{2\pi (1897.1*10^3m)^{3/2}}{\sqrt{(6.674*10^{-11}m^3/kgs^2)(7.34*10^22kg)}}\\\\T=7417.78s[/tex]
In hours you obtain:
[tex]T=7417.78s*\frac{1h}{3600s}=2.06h[/tex]
The time that the Apollo takes to complete an orbit around the moon is 2.06h