Answer:
The answer is 2,475$+948= add you will get the answer rn
The weight we experience on a surface is the product of our mass and acceleration due to gravity on that surface. The mass of the astronaut in earth was 60 kg . Then his mass at moon will be 9.76 kg.
What is gravitational force ?The gravitational force is the force by which an object attracts other objects into its center of mass. Earth attracts every objects into its center. That why we are all standing on the ground.
The weight we experience on earth is due to the gravity. That is the product of mass and acceleration due to gravity of a planet is the weight on the planet.
The weight of astronaut in earth = 60 kg
acceleration due to gravity on earth = 9.8 m/s²
mass = 60/9.8 m/s² = 6.19 kg
Acceleration due gravity on moon = 1.6 m/s²
then weight on moon = 1.6 m/s² ×6.19 kg = 9.7 kg.
Therefore, the weight of the astronaut in moon will be 9.7 kg.
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Assume that, when we walk, in addition to a fluctuating vertical force, we exert a periodic lateral force of amplitude 25 NN at a frequency of about 1 HzHz. Given that the mass of the bridge is about 2000 kgkg per linear meter, how many people were walking along the 144-mm-long central span of the bridge at one time, when an oscillation amplitude of 75 mmmm was observed in that section of the bridge
Complete Question
The complete question is shown on the first uploaded image
Answer:
Explanation:
From the question we are told
The amplitude of the lateral force is [tex]F = 25 \ N[/tex]
The frequency is [tex]f = 1 \ Hz[/tex]
The mass of the bridge per unit length is [tex]\mu = 2000 \ kg /m[/tex]
The length of the central span is [tex]d = 144 m[/tex]
The oscillation amplitude of the section considered at the time considered is [tex]A = 75 \ mm = 0.075 \ m[/tex]
The time taken for the undriven oscillation to decay to [tex]\frac{1}{e}[/tex] of its original value is t = 6T
Generally the mass of the section considered is mathematically represented as
[tex]m = \mu * d[/tex]
=> [tex]m = 2000 * 144[/tex]
=> [tex]m = 288000 \ kg[/tex]
Generally the oscillation amplitude of the section after a time period t is mathematically represented as
[tex]A(t) = A_o e^{-\frac{bt}{2m} }[/tex]
Here b is the damping constant and the [tex]A_o[/tex] is the amplitude of the section when it was undriven
So from the question
[tex]\frac{A_o}{e} = A_o e^{-\frac{b6T}{2m} }[/tex]
=> [tex]\frac{1}{e} =e^{-\frac{b6T}{2m} }[/tex]
=> [tex]e^{-1} =e^{-\frac{b6T}{2m} }[/tex]
=> [tex]-\frac{3T b}{m} = -1[/tex]
=> [tex]b = \frac{m}{3T}[/tex]
Generally the amplitude of the section considered is mathematically represented as
[tex]A = \frac{n * F }{ b * 2 \pi }[/tex]
=> [tex]A = \frac{n * F }{ \frac{m}{3T} * 2 \pi }[/tex]
=> [tex]n = A * \frac{m}{3} * \frac{2\pi}{25}[/tex]
=> [tex]n = 0.075 * \frac{288000}{3} * \frac{2* 3.142 }{25}[/tex]
=> [tex]n = 1810 \ people[/tex]
What is meant by child rights?
Why did scientist struggle for thousands of years to accurately describe the solar system
Answer: C- The planets appeared to move backward in the sky occasionally.
One of the two rectangular components of a force is 20N and it makes an angle of 30
with the force. Find the magnitude of the other components.
Answer:
11.545NExplanation:
The the horizontal and vertical component be expressed as;
Fx = Fxos theta (horizontal compt)
Given Fx = 20N
theta = 30°
Get F:
20 = Fcos30
20 = 0.8660F
F = 20/0.8660
F = 23.09N
Get the magnitude of the other components. (vertical compt)
Fy = Fsin theta
Fy = 23.09sin30
Fy = 23.09(0.5)
Fy = 11.545N
Hence the magnitude of the other component is 11.545N
A missle is flying at a speed 125 m/s. if the missle has a mass of 125 kg. What’s it’s kinetic energy
Answer:
[tex]976562.5\ Joules[/tex]
Explanation:
[tex]We\ are\ given:\\The\ mass\ of\ the\ missile=125\ kg\\The\ velocity\ of\ the\ missile=125\ m/s\\Hence,\\As\ we\ know\ that,\\Kinetic\ energy\ possessed\ by\ an\ object\ with\ mass\ m\ moving\\ with\ a\ velocity\ v:\\E_k=\frac{1}{2}mv^2\\Here,\\The\ Kinetic\ Energy\ possessed\ by\ the\ missile=\frac{1}{2}*125*(125)^2=976562.5\ Joules[/tex]
A 100 kg gymnast comes to a stop after tumbling. Her feet do 5,000 J of work to stop her.
Which of the following was the girl's velocity when she began to stop?
Answer:10 m/s
Explanation:
A 100 kg gymnast comes to a stop after tumbling. Her feet do 5,000 J of work to stop her. The girl's velocity when she began to stop is 10 m/sec.
What is velocity?When an item is moving, its velocity is the rate at which its direction is changing as seen from a certain point of view and as measured by a specific unit of time. Velocity is vector quantity.
Uniform motion an object is said to have uniform motion when object cover equal distance in equal interval of time within exact fixed direction. For a body in uniform motion, the magnitude of its velocity remains constant over time. Here in the question velocity is changing by using work energy theorem we have,
work done = change in kinetic energy
5000 = (.5).m.v² , where v is velocity.
5000 = (.5).100.v²
v = 10 m/sec
A 100 kg gymnast comes to a stop after tumbling. Her feet do 5,000 J of work to stop her. The girl's velocity when she began to stop is 10 m/sec.
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1. Derive the equation of the trajectory of a projectile.
(3mks)
2. A ball is thrown with an initial speed uof 30 m/s at an angle & above thehorizontal, where
sin 0 = 4/5 and cos 0 = 3/5.
(5mks)
(a) decompose the vector u into itsx and y components.
(b) When t= 2 s, find the position of the ball andthe magnitude and direction of its
velocityu.
(c) Determine
the value
of the highest point of the ball's trajectory.
(d) calculate how much time has elapsed forthe ball to reach the highest point.
(e) Calculate the values of the total time of theball's flight T and the horizontal range R.
ASAN
3. In the model of the hydrogen atom proposed by Niels Bohr an electron circulatesa
stationary proton in a circle of radius 7' = 5.28 x 10-11 m with a speedı= 2.18 x 106m's
(a) Find the magnitude of the electron sradial acceleration in this model.
(1mk)
(b) Determine the period of the motion,
(Imk)
As an astronaunt travels from the surface of the earth to a postion that is four times
as far away from the center of the earth, the astronaut's
a) mass decreases
b)
mass remains the same
c) weight increases
d) weight remains the same
Answer: mass remains the same
Explanation:
My teacher gave me the answer
As an astronaut travels from the surface of the earth to a position that is four times as far away from the center of the earth, the astronaut's:
B. Mass remains the sameAccording to the given question, we can see that an astronaut is on a journey to a new location which is four times further from the center of the earth. We need to find out if there is a change to the mass of the astronaut.
As a result of this, we can see that the mass of the astronaut remains the same as he travels to a position which is four times further from the center of the earth, because the gravitational forces are the same.
Therefore, the correct answer is option B
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100 points!! word bank!
⬇️Article⬇️
Have you ever cut an apple in half and looked at the layers inside? When you cut something in half, the resulting view is called a cross section. When you look at the cross section of an apple, you see several layers: the skin, the pulp, the core, and the seeds. Much like the apple, Earth is made up of layers, too. If you could look at a cross section of our planet, you would see the crust. the mantle, the outer core, and the inner core. If Earth were an apple, the crust would be the apple’s skin. The mantle would be the apple’s pulp, making up most of the inside. Earth also has a central core, similar to an apple’s core, though Earth’s core does not contain any seeds! Of course, scientists cannot cut the whole planet in half to see a cross section the way you can with an apple. How do you think scientists know about Earth’s internal layers?
Scientists divide Earth’s interior into distinct layers.
Scientists can distinguish different layers in Earth depending on the properties used to identify each layer. For example, scientists identify the crust, mantle, and core based on each layer’s basic chemical composition. In other words, the crust, mantle, and core are each made up of different chemical elements. The crust and mantle are composed primarily of the elements oxygen and silicon. These are known as silicates. Silicates of the mantle contain heavier elements. This makes them denser than those found in the crust. The core is composed of the densest materials, primarily iron and nickel. Earth’s layers can also be identified by using each layer’s physical state of matter. The crust and the uppermost part of the mantle, together, make up a layer called the lithosphere. The lithosphere is the cool, rigid, outermost layer of Earth. The lithosphere is in the solid
In a typical silicate molecule, a silicon atom is bonded to four oxygen atoms. The atoms form a crystal structure.
Scientists use models to represent the different layers of Earth’s interior.
When constructing a model of the layers of Earth, scientists need to consider the chemical composition, state of matter, and thickness of each layer. Just like the skin of an apple, Earth’s crust is very thin compared to the other layers. It is about 25–70 km thick beneath the continents. Under the oceans, the crust is only about 5–7 km thick; however, it is much denser. The mantle is much thicker than the crust is, taking up most of Earth’s volume. The mantle begins directly beneath Earth’s crust. It reaches all the way to the outer core, about 2,900 km below the planet’s surface. In other words, Earth is about 1% crust, 83% mantle, and 16% core.
The crust and uppermost mantle are solid. The rest of the mantle is solid with plasticity. The core is made of very dense iron and nickel. The outer core is liquid, because it is hot enough to melt the iron and nickel. The inner core is solid. Even though it is as hot as the outer core, there is so much pressure at the very center of Earth that the iron and nickel stay in a solid state.
Looking to the Future: Exploring Earth’s Interior
Despite what you may have read in stories or seen in movies, scientists have never journeyed to the center of Earth. In fact, scientists have never made it through Earth’s crust! However, this has not stopped them from trying. The crust at the bottom of the oceans is much thinner than the crust of the continents. Therefore, drilling through the oceanic crust is the best chance that scientists have to make it to the mantle.
Answer:
Except in the crust, the interior of the Earth cannot be studied by drilling holes to take samples. Instead, scientists map the interior by watching how seismic waves from earthquakes are bent, reflected, sped up, or delayed by the various layers.
Explanation:
A waterbed has a force of 1300N on the floor. It exerts 347 Pa of pressure. What is the area of the waterbed?
Answer:
Pressure = Force/Area
347 = 1300/Area
Area = 1300/347
area= 3.47m2
Answer:
I hope the above pic will help you
I will always help you understanding your assingments have a great day
#Captainpower
What is the water cycle ?
Answer:
The water cycle shows the continuous movement of water within the Earth and atmosphere. ... Liquid water evaporates into water vapor, condenses to form clouds, and precipitates back to earth in the form of rain and snow. Water in different phases moves through the atmosphere (transportation).
Explanation:
It's the water cycle.
19. What energy transformation occurs in a battery?
A. chemical energy to thermal energy
C. electric energy to chemical energy
B. chemical energy to electric energy
D. thermal energy to chemical energy
i think correct answer is b
Answer:
B.chemical energy changes to electric energy.Explanation:
I hope l helped you. ❤❤i just got hit on the head by a basketball.
story of my life.
Season Leaders
Go to Stats
Points Per Game
1.
Talen Horton-Tucker
26.0
2.
Nikola Jokic
26.0
3.
Zion Williamson
26.0
4.
Giannis Antetokounmpo
24.5
5.
Patty Mills
24.0
Rebounds Per Game
1.
Rudy Gobert
14.0
2.
Harry Giles III
13.5
3.
Clint Capela
13.0
4.
Nikola Vucevic
13.0
5.
Giannis Antetokounmpo
12.0
Assists Per Game
1.
Ja Morant
9.0
2.
Bam Adebayo
8.0
3.
T.J. McConnell
7.5
4.
DeMar DeRozan
7.0
5.
Spencer Dinwiddie
7.0
More Leaders
Hustle Leaders
Advanced Leaders
All Time Leaders
Look at this model of an atom. Using a periodic table, which element does it represent?
Answer:
Nitrogen
Explanation:
10) The coil of transformer is coated by enamel why
Answer:
Enamel is used to coat the wires, it is the thinnest possible insulator. The coils are made up of large number of turns and enamel makes it possible to cram a lot of wires (coils) in much smaller space.
An object is moving diagonally (down and to the left). You want it to stop moving. In what direction (or
directions) should you exert a force to get the object to stop?
a. Force direction(s):
b. Explain your answer:
A satellite orbits the Earth (mass = 5.98 x 1024 kg) once every = 43200 s. At what radius does the satellite orbit?
Answer:
26621 km
Explanation:
We are given;
Mass: m = 5.98 x 10^(24) kg
Period; T = 43200 s
Formula for The velocity(v) of the satellite is:
v = 2πR/T
Where R is the radius
Formula for centripetal acceleration is;
a_c = v²/R
Thus; a_c = (2πR/T)²/R = 4π²R/T²
Formula for gravitational acceleration is:
a_g = Gm/R²
Where G is gravitational constant = 6.674 × 10^(-11) m³/kg.s²
Now the centripetal acceleration of the satellite is caused by its gravitational acceleration. Thus;
Centripetal acceleration = gravitational acceleration.
Thus;
4π²R/T² = Gm/R²
Making R the subject gives;
R = ∛(GmT²/4π²)
Plugging in the relevant values;
R = ∛((6.674 × 10^(-11) × 5.98 x 10^(24) × 43200²)/(4 × π²))
R = 26.621 × 10^(6) m
Converting to km, we have;
R = 26621 km
Answer:
2.66 *10^7
Explanation:
Sorry I don't know the process if you were interested in that :/
A high-resistance material is used as an insulator between the conductors of a length of coaxial cable. The resistance material, which forms a hollow tube, has an inner radius a and an outer radius b, and the insulator provides a resistance R between the conductors. If a second insulator, made of the same material and having the same length, is made with double both the inner radius and the outer radius of the first, what resistance would it provide between the conductors
This question is incomplete, the complete question is;
A high-resistance material is used as an insulator between the conductors of a length of coaxial cable. The resistance material, which forms a hollow tube, has an inner radius a and an outer radius b, and the insulator provides a resistance R between the conductors. If a second insulator, made of the same material and having the same length, is made with double both the inner radius and the outer radius of the first, what resistance would it provide between the conductors
a) (In2)R
b) 4R
c) R/(In2)
d) 2R
e) R
Answer: Option e) R is the correct answer.
Explanation:
Given that;
Inner radius = a
Outer radius = b
Conical Cylinder
∫dR = ∫(edr/2πrL)
R = e/2πL In e |ᵇₐ
R = e/2πL In(b/a) ------------- let this be equation 1
Taking a look at the second cone
a' = 2a
b' = 2a
R' = e/2πL In(2b/2a)
{L = L'}
R' = e/2πL In(b/a) -------let this be equation 2
now lets compare the two equation
R = e/2πL In(b/a)
R' = e/2πL In(b/a)
so R' = R
Therefore Option e) R is the correct answer.
a 2 kg block is attached to a horizontal ideal spring with a spring constant 200N/m. when the
how do you determine half life
Answer:
T= In(2)/___ =__ In(2)
Explanation:
the equation pices arnt on the keyboard so i put the blanks
sorry
jope it helps some. Have a good day ir night! :)
Two spheres of equal mass, A and B, are projected off the edge of a 1.0 m bench. Sphere A has a horizontal velocity of 10 m/s and sphere B has a horizontal velocity of 5 m/s.
__ 5. If both spheres leave the edge of the table at the same instant, sphere A will land
a. at some time after sphere B.
b. at the same time as sphere B.
c. at some time before sphere B.
d. There is not enough information to decide.
__ 6. If both spheres leave the edge of the table at the same instant, sphere A hits the floor at the spot marked X. Sphere B will hit the floor
a. at some point between the edge of the table and X.
b. at some point past X.
c. at the same distance from the table as X.
d. there is not enough information to decide.
Answer:
c. because A will land first becuase its heavier
and D.
Explanation:
A pail in a water well is hoisted by means of a frictionless winch, which consists of a spool and a hand crank. When Jill turns the winch at her fastest water-fetching rate, she can lift the pail the 28.0 m to the top in 11.0 s. Calculate the average power supplied by Jill's muscles during the upward ascent. Assume the pail of water when full has a mass of 7.30 kg.
Answer:
182.28 W
Explanation:
Here ,
m = 7.30 Kg
distance , d= 28.0 m
time , t = 11.0 s
average power supplied = change in potential energy/time
average power supplied = m×g×d/time
average power supplied = 7.30×9.81×28/11
average power supplied = 182.28 W
the average power supplied is 182.28 W
a steel girder weighting 700 lb is hoisted from gund level to the roof of a 70-ft building using a chain that weighs 4lb/running foot. Find the work done
Answer:
66.475kJ
Explanation:
Work done is the product of the force and perpendicular distance in direction of the force.
Work done = Force × distance
Given
Weight = 700lb
Distance = 70ft
1lb force = 4.448N
700lb force = 700(4.448) = 3113.6N
Distance = 70(0.305) = 21.35m
Work done = 3113.6×21.35
Work done ≈ 66,475Joules
Work done = 66.475kJ
Calculate the potential energy of a coffee cup that is resting on a 24.5 meter ledge, and
weighs 5.4 Newtons
Answer:
E_{pot} = 132.3 [J]
Explanation:
In order to solve this problem, we must use the definition of the potential energy which can be calculated by means of the following formula.
[tex]E_{pot}=W*h[/tex]
where:
Epot = potential energy [J]
W = weight = 5.4 [N]
h = elevation = 24.5 [m]
[tex]E_{pot}=5.4*24.5\\E_{pot}=132.3[J][/tex]
.
Calculate the potential energy of a 6 kg bowling ball suspended 3 m above the surface of the Earth.
Answer: 176.4 J
Explanation:
Explain
applications
MRI
X-ray
Ultrasound
Infrared Radiation
Answer and Explanation:
--> MRI (Magnetic Resonance Imaging) is an imaging technique that involves the use of radio waves and magnetic field to generate high detailed images of the human body. It's clinical applications includes:
• Bone and joint MRI: these are done to check for bone and joint problems such as bone marrow disorders, arthritis, or bone tumours.
• A Spine MRI: This is used to investigate the spinal disc and innervations for disorders such as spinal tumours.
• Chest MRI: This is used to detect abnormalities of the heart, heart valves and coronary arteries.
--> X-RAY: These are electromagnetic waves of short wavelength with high penetrating power. They are produced when fast moving electrons strike a mass of heavy atoms such as those of metals. It's applications includes:
• They are used to detect hidden cracks in materials.
• They are used to show broken bones in human body.
• They are used in the study of internal structures of crystals.
• In agriculture, x-rays are used in killing germs.
--> ULTRASOUND: This makes use of high frequency sounds to detect abnormalities in the human body. An ultrasound machine transmits sound waves into the body which are reflected at the surfaces between the tissues of different density. It can also be used in different disciplines such as imaging, cleaning, mixing, navigation and communication It's applications includes:
•. Detection of Cracks: when applied on metallic surface under investigation, high frequency sound wave reflects back which and be predicted and recorded.
--> INFRARED RADIATION: This is a type of electromagnetic radiation with wavelength longer than visible light. It is visible to the human eye and most heated surfaces transmits infrared radiation. It's applications includes:
• it is used for the purpose of sensing and detection. For example the night vision goggles and infrared cameras. Remote control makes use of infrared light waves to change channels in the television.
A double-slit interference pattern is created by two narrow slits spaced 0.21 mm apart. The distance between the first and the fifth minimum on a screen 61 cm behind the slits is 6.2 mm What is the wavelength of the light used in this experiment?
Answer:
[tex]\lambda =533.6 nm [/tex]
Explanation:
the slits spacing, d = 0.21 mm
distance of screen, D = 61 cm
The condition for minima is given as
[tex]dsin(\theta) = \left ( n+\frac{1}{2} \right )\lambda[/tex]
So, first minima, n = 0
[tex]dsin(\theta_1) = \frac{1}{2}\lambda[/tex]
fifth minima, n = 4
[tex]dsin(\theta_5) = \frac{9}{2}\lambda[/tex]
[tex]d(sin(\theta_5) -sin(\theta_1))= 4\lambda[/tex]
For small angle
[tex]d(tan(\theta_5) -tan(\theta_1))= 4\lambda[/tex]
From the figure:
[tex]d(\frac{y_5}{D}-\frac{y_1}{D})= 4\lambda[/tex]
[tex]\frac{d}{D} (y_5-y_1) = 4\lambda[/tex]
[tex]\lambda = \frac{d}{4D} (y_5-y_1)[/tex]
[tex]\lambda = \frac{0.021}{4(61)} (6.2 \times 10^{-3})[/tex]
[tex]\lambda =533.6 nm[/tex]
"533.5 nm" would be the wavelength of the light used in this experiment.
According to the question,
Slits spacing,
d = 0.21 mmDistance of screen,
D = 61 cmThe condition for minima is given as:
→ [tex]d (sin \Theta) = (n+\frac{1}{2} ) \lambda[/tex]
So,
1st minima, n = 0
→ [tex]d sin(\Theta_1)= \frac{1}{2} \lambda[/tex]
5th minima, n = 4
→ [tex]d (sin (\Theta_5) - sin (\Theta_1) = 4 \lambda[/tex]
For small angle,
→ [tex]d (tan (\Theta_5) - tan (\Theta_1) = 4 \lambda[/tex]
From the figure, we get
→ [tex]d(\frac{y_5}{D} - \frac{y_1}{D} )= 4 \lambda[/tex]
[tex]\frac{d}{D} (y_5 -y_1) = 4 \lambda[/tex]
[tex]\lambda = \frac{d}{4D}(y_5-y_1)[/tex]
[tex]= \frac{0.021}{4\times 61} (6.2\times 10^{-3})[/tex]
[tex]= 533.6 \ nm[/tex]
Thus the above answer is correct.
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Use the following information to answer questions 6 and 7:
A61.0 kg grocery cart is sitting at rest. In order to change the cart's state of motion, Xavier must apply 72.1
Newtons of force on the cart.
6. How much magnitude of force does the cart exert back on Xavier's hand?
A. 4400 N
B. 1.20 N
C. 72.1 N
D. 0.00 N
After 10 seconds, the dirt has achieved a velocity of +3.62 m/s. Xavier is still applying 72.1 Not force on the
cart.
7. Xavier's sister Maria claims that the cart is in dynamic equilibrium. Is Maria correct?
A Maria is correct because the cart is traveling at a constant velocity while experiencing balanced
forces.
B. Maria is correct because the cart's velocity has changed.
C. Maria is incorrect because the cart is traveling at a constant velocity while experiencing
unbalanced forces.
D. Maria is incorrect because the cart's velocity has changed
Question 6
C. 72.1 N
Its simply the newton Third law, whenever two objects interact, they exert equal and opposite forces on each other.
Question 7
D. Maria is incorrect because the cart's velocity has changed
The car is accelerating so, It is not in dynamic equilibrium. Remember that by the Newton Second Law:
F = ma
The cart is in dynamic equilibrium. The correct options for the two question are:
6. C
7. A
6.) Given that a 61.0 kg grocery cart is sitting at rest. And Xavier applied 72.1 Newtons of force on the cart.
According to Newton's third law of motion which state that, in every action, there will be equal and opposite reaction. The magnitude of force That the cart exert back on Xavier's hand will be opposite and equal to the force applied. which is negative 72.1 N
7.) If the grocery cart is moving in a constant velocity, then it will achieve a dynamic equilibrium. In this scenario, the air resistance force balances with the applied force.
Therefore, Maria is correct because the cart is traveling at a constant velocity while experiencing balanced forces.
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If forces acting on an object are unbalanced, the object could experience a change in ________.
mass, speed, or direction
negative acceleration or mass
direction, mass, or both
speed, direction, or both
D) Speed, direction, or both
Mass doesn’t just CHANGE like that.
A street bridge is 5.5m long if the linear expansion of steel is 0.00001 oc How much will it expand when temperatures is by 10oc? Give answer in Cm
Answer:
[tex]l_o=550.055\ cm[/tex]
Explanation:
Given that,
Length of a street bridge, l = 5.5 m
The coefficient of bridge, [tex]\alpha =0.00001 ^0 C[/tex]
We need to find how much will it expand when temperatures is by 10°C.
The change in length per unit original length is given by :
[tex]\dfrac{\Delta l}{l}=\alpha \Delta T\\\\\Delta l = l\alpha \Delta T\\\\=5.5\times 0.00001 \times 10\\\\\Delta l=0.00055\\\\(l_o-l)=0.00055\\\\l_o=0.00055+5.5\\\\=5.50055\ m\\\\l_o=550.055\ cm[/tex]
Hence, the length will expanded 550.055 cm.