Answers
We will have that the graph that describes the scenario is given by graph B.
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Suppose a force F, = (1/3)x acts on anobject. Assume at x = 0 the force pushesthe object and it starts moving to the right.What happens to the magnitude of the workdone as the object moves to the right?A. More work is done B. Less work is doneC. A constant amount of work is doneD. It is impossible to predict
Answers
We are given that a variable force acts on an object given by the function:
[tex]F_x=\frac{1}{3}x[/tex]This force will increase as the object moves since the value of "x" will increase. The work done is defined as:
[tex]W=Fd[/tex]Where "F" is the force and "d" is the total distance. This magnitude will also increase therefore, more work is done.
How do i solve this problem? Hint: The following are the three relevant equations:a = G M / r2a = v2/rT = 2πr/vEliminating a and v and solving for r gives:r = 3√(GMT2/(4π2))Be sure to plug in the period in seconds. This will provide the orbital radius from the center of the earth. Using this value and the radius of the earth, we can determine the required height above the earth's surface. Finally, we can use v = 2πr/T to find the orbital speed.
Answers
An engineer wants a satellite to orbit the Earth with a period of 48 hours.
The acceleration due to gravity at the surface of the Earth is given by
[tex]a=\frac{GM}{r^2^{}}[/tex]Where G is the gravitational constant (G = 6.67430×10⁻¹¹ Nm²/kg²), M is the mass of the Earth (m = 5.97x10²⁴ kg)
The acceleration of the satellite in a circular motion (centripetal acceleration) is given by
[tex]a=\frac{v^2}{r}[/tex]Where v is the speed of the satellite.
Equating both equations, we get
[tex]\begin{gathered} \frac{GM}{r^2}=\frac{v^2}{r} \\ \frac{GM}{r}=v^2 \end{gathered}[/tex]Substitute v = 2πr/T into the above equation
[tex]\begin{gathered} \frac{GM}{r}=(\frac{2\pi r}{T})^2 \\ \frac{GM}{r}=\frac{4\pi^2r^2}{T^2} \\ GM=\frac{4\pi^2r^2\cdot r}{T^2} \\ GM=\frac{4\pi^2r^3}{T^2} \\ r^3=\frac{GMT^2}{4\pi^2} \\ r=\sqrt[3]{\frac{GMT^2}{4\pi^2}} \end{gathered}[/tex]So, we have got the equation for radius r.
Let us first convert the period from hours to seconds
[tex]T=48\times60\times60=172800\; s[/tex]Substitute it into the above equation and find the radius (r).
[tex]\begin{gathered} r=\sqrt[3]{\frac{6.67430\times10^{-11}\cdot5.97\times10^{24}\cdot(172800)^2}{4\pi^2}} \\ r=67045443\; m \\ r=6.7045443\times10^7\; m \end{gathered}[/tex]Therefore, the satellite must orbit at an orbital radius of 67,045,443 m
(b) How far is this above the Earth's surface?
The required height above the earth's surface can be found by subtracting the radius of the earth from the orbital radius from the center of the earth that we calculated in the previous part.
The radius of Earth is 6.37×10⁶ m
[tex]\begin{gathered} h=6.7045443\times10^7-6.37\times10^6 \\ h=60675443\; m \\ h=6.0675443\times10^7\; m \end{gathered}[/tex]Therefore, the required height above the earth's surface is 60,675,443 m
(c) Speed of the satellite
The speed of the satellite is given by
[tex]\begin{gathered} v=\frac{2\pi r}{T} \\ v=\frac{2\cdot\pi\cdot67045443}{172800} \\ v=2,437.84\; \; \frac{m}{s} \end{gathered}[/tex]Therefore, the orbital speed of the satellite is 2,437.84 m/s
usWhich of the following X-Y tables agrees withthe information in this problem?A missile is moving 1810 m/s at a 20.0° angle. It needsto hit a target 29,500 m away in a 65.0° direction in9.20 s. What acceleration must its engine produce?хYB)хYC)хYA)ViVE1700619V;17006191700619<<??V?a??aa?AX12500 | 26700Ax 12500 26700Ax 12500 26700t9.209.20t9.209.20t8.653.15
Answers
The given value of the distance at which the target is present is
[tex]d=29500\text{ m}[/tex]The distance along the x-axis is,
[tex]\begin{gathered} d_x=d\cos (65^{\circ}) \\ d_x=29500\times\cos (65^{\circ}) \\ d_x=12467 \\ d_x\approx12500\text{ m} \end{gathered}[/tex]As in the given tables, the value of the distance and time 9.2 s is correct in first two only. So, the third one is not the correct option.
From the given question, the value of the acceleration is to be found.
The question mark arrows are on the acceleration is only on the b table.
Hence, Table B agrees with the information in the given situation.
A rock tied to the end of string swings at a constant angular rate. If you are told thatthe string can support a total of 210 N of force before breaking, what is the maximumangular velocity the rock can rotate if the rock has a mass of 0.23 kg, and the length ofthe string is 0.35 m? Give your answer in units of radian per second.
Answers
Given data
The magnitude of the total force is F = 210 N
The mass of the rock is m = 0.23 kg
The length of the string is L = 0.35 m
The expression for the maximum angualr velocity is given as:
[tex]\begin{gathered} F=m\omega^2L \\ \omega=\sqrt{\frac{F}{mL}} \end{gathered}[/tex]Substitute the value in the above equation.
[tex]\begin{gathered} \omega=\sqrt[]{\frac{210\text{ N}}{0.23\text{ kg }\times\text{0.35 m}}} \\ \omega=51.07\text{ rad/s} \end{gathered}[/tex]Thus, the maximum angular velocity of the rock is 51.07 rad/s.
B) what is the frequency of the sound wave;i.e, the tuning fork? C) the water continues to leak out the bottom of the tube When the tube next resonates with tuning fork, what is length of the air column?
Answers
Answer:
60 cm
Explanation:
The wavelength of the sound wave in the air column closed at one end is given by
[tex]\lambda=\frac{4}{n}L[/tex]where n = 1, 3, 5, 7,9, 11, 13, 15 are all odd numbers.
Now in our case, we know that the length of the pipe is 225 or 2.25 m. The 8th resonance corresponds to n = 15; therefore, the wavelength of the sound wave is
[tex]\lambda=\frac{4}{15}\times225[/tex][tex]\boxed{\lambda=60\; cm\text{.}}[/tex]Hence, the wavelength of the sound wave is 60 cm.
A blue car weighing 1,302 kg is accelerating forward at a rate of 4 m/s². What is the forward force of the car?
Answers
Answer:
5208N
Explanation:
force=mass×acceleration
mass(kg)=1,302
acceleration(m/s²)=4
f= m × a
f= 1302 × 4
f=5208N
D. O499.917 JE. O 1358.602 J4. An object of mass 5 kg is sliding downa friction less inclined plane of length3 m that makes an angle of10 deg. Calculate its speed just before it hits the ground. (1 point)A. 03.195 m/sB. O 1.958 m/sC. O5.58 m/sD. O2.525 m/sE. 04.225 m/s5. A roller coaster extends to the groundfrom a height of 39 m (point A) and then risesAn obiect of mass 12 kg
Answers
Given
Mass of the object, m=5 kg
Length of the inclined plane, s=3m
Angle of inclination,
[tex]\theta=10^o[/tex]To find
Calculate its speed just before it hits the ground.
Explanation
Here the acceleration is
[tex]\begin{gathered} a=gsin\theta \\ \Rightarrow a=9.8sin10^o \end{gathered}[/tex]The initial velocity is zero.
So the final velocity is
[tex]\begin{gathered} v=\sqrt{2as} \\ \Rightarrow v=\sqrt{2\times9.8sin10^o\times3} \\ \Rightarrow v=3.19\text{ m/s} \end{gathered}[/tex]Conclusion
The velocity before it hit the ground, A.3.19 m/s
A small moon is in a circular orbit around an exoplanet. It orbits at an altitude of 700 km. This exoplanet is the same size as Earth, but is only 1/2 as dense.Does this exoplanet exert a torque on its moon? Why or why not?No, because the force exerted by the planet on the meteoroid has a negligible magnitude.Yes, because the meteoroid's direction of motion is constantly changing.Yes, because the planet exerts a centripetal force on the meteoroid.No, because the planet exerts a force on the meteoroid parallel to its position vector relative to the center of mass of the planet.
Answers
According to kepler's 2nd law a planet sweeps equal area in equal time around a planet.
here the angular momentum of planet is constant
if angular momentum is constant then according to conservation of angular momentum the net torque on the moon due to the exoplanet will be equal to 0.
So here the net torque = 0.
That's why last option is correct.
You are wanting to walk on an icy road without falling over. You have a ruler, some cotton wool and some salt and vinegar crisps. Devise a plan to increase the friction so that you can walk safely.
Answers
Using the concept of friction, we got salt > vinegar crisps >cotton wool > ruler is the order of friction which they provide.
We know very well that the frictional force is directly proportional to the surface of the area and the surface of the icy road is very smooth. So the contact area of man's foot with the surface of the icy road is very small and the fraction between man's foot and the surface of the icy road is also small so this is the reason why man can not walk on the icy road properly and safely.
So in order to walk safely on icy roads we have to increase friction so in order to increase friction we have to first use salt then vinegar crisps then cotton wool and then a ruler.
Hence the order of friction which they provide is salt > vinegar crisps >cotton wool > ruler
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Bob drops a rock from the roof of a building (5m tall) onto Joe's head (1.5m tall). The mass of the rock is 0.5kg.a) What is the speed of the rock as it his Joe's head?b) If the rock is only travelling 8.0 m/s, how much work was done by air resistance?
Answers
Given:
The initial height of the rock was,
[tex]h_i=5\text{ m}[/tex]The final height of the rock is,
[tex]h_f=1.5\text{ m}[/tex]The mass of the rock is,
[tex]m=0.5\text{ kg}[/tex]The travelling speed of the rock is,
[tex]v^{\prime}=8.0\text{ m/s}[/tex]To find:
a) The speed of the rock at Joe's head
b) how much work was done by air resistance
Explanation:
The displacement of the rock is,
[tex]\begin{gathered} h=h_i-h_f \\ =5-1.5 \\ =3.5\text{ m} \end{gathered}[/tex]The final speed at Joe's head is,
[tex]\begin{gathered} v=\sqrt{u^2+2gh} \\ =\sqrt{0^2+2\times9.8\times3.5} \\ =\sqrt{68.6} \\ =8.28\text{ m/s} \end{gathered}[/tex]Hence, the speed of the rock at Joe's head is 8.28 m/s.
b)
The speed at Joe's head was 8.0 m/s, and the loss of kinetic energy is,
[tex]\begin{gathered} \frac{1}{2}\times m[(8.23)^2-(8.0)^2] \\ =\frac{1}{2}\times0.5\times[3.73] \\ =0.93\text{ J} \end{gathered}[/tex]This loss of energy is the work done by the air resistance.
Hence, the work done by the air resistance is 0.93 J.
Draw the following components of a circuitCellBatteryLight Bulb Fuse
Answers
ANSWER and EXPLANATION
We want to draw the circuit diagram of the given components.
The fuse will be placed in between the battery (and cell) and the load
To do this, use the appropriate symbols as shown below:
That is the answer.
Amanda's Coffee Shop makes a blend that is a mixture of two types of coffee. A coffee costs Amanda $5.60 per pound, and type B coffee costs $4.55 per pound. This month, Amanda made 141 pounds of the blend, for a total cost of $728.70. How many pounds of type B coffee did she use?
Answers
We are given that Amanda made 141 pounds of coffee. If "x" is the pounds of type A coffee and "y" is the amount of type b coffee then we can write this mathematically as:
[tex]x+y=141,(1)[/tex]We are also given that the cost of type A is $5.6 per pound and that the cost of type B is $4.55 per pound and that the total cost is $728.70, this can be written mathematically as:
[tex]5.6x+4.55y=728.7,(2)[/tex]Now, we solve for "x", first by subtracting "y" from both sides:
[tex]y=141-x[/tex]Now, we substitute the value of "y" in equation (2):
[tex]5.6x+4.55(141-x)=728.7[/tex]Now, we apply the distributive property in the parenthesis:
[tex]5.6x+641.55-4.55x=728.7[/tex]Now, we add like terms:
[tex]1.05x+641.55=728.7[/tex]Now, we subtract 641.55 from both sides:
[tex]\begin{gathered} 1.05x=728.7-641.55 \\ 1.05x=87.15 \end{gathered}[/tex]Now, we divide both sides by 1.05:
[tex]x=\frac{87.15}{1.05}[/tex]solving the operations:
[tex]x=83[/tex]Now, we substitute the value of "x" in equation (1):
[tex]\begin{gathered} y=141-83 \\ y=58 \end{gathered}[/tex]Therefore, the amount of coffee type A is 83 pounds and type B is 58 pounds.
An astronaut has a mass of 59 kilograms. What will her gravitational force be on the Moon? The
gravitational attraction on the Moon is 1.62 (1 point)
O 578.2 N
O 60.62 N
O 95.58 N
O 36.42 N
Answers
Taking into account the Newton's Second Law, the correct answer is the third option: the gravitational force of the astronaut on the Moon is 95.58 N.
Newton's second lawAcceleration in a body occurs when a force acts on a body. There are two factors that influence the acceleration of an object: the net force acting on it and the mass of the body.
Newton's second law defines the relationship between force and acceleration mathematically. This law says that the acceleration of an object is directly proportional to the sum of all the forces acting on it and inversely proportional to the mass of the object.
Mathematically, Newton's second law is expressed as:
F= m×a
where:
F = Force [N]m = Mass [kg]a = Acceleration [m/s²]Gravitational force be on the MoonIn this case, you know:
F= ?m= 59 kga= gravitational attraction on the Moon= 1.62 m/s²Replacing in Newton's second law:
F= 59 kg× 1.62 m/s²
Solving:
F= 95.58 N
Finally, gravitational force on the Moon is 95.58 N.
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What is force acting on an object having a mass of 560 g on Earth?
Answers
Given,
[tex]{ \blue{ \tt{m = 560g}}}[/tex]
[tex]{ \blue{ \tt{g = 9.8 m/s²}}}[/tex]
[tex]{ \blue{ \tt{W = ?}}}[/tex]
[tex]{ \purple{ \tt{W = mg}}}[/tex]
[tex]{ \purple{ \sf{W= 560 × 9.8}}}[/tex]
[tex]{ \boxed{ \red{ \sf{W = 5488N}}}}[/tex]
Suppose a magnetic field exists with the north pole at the top of the computer monitor and the south pole at the bottom of the monitor screen. If a positively charged particle entered the field moving from your face to the other side of the monitor screen, which way would the path of the particle bend? Select one:a.leftb.rightc.upd.downe.none of the above
Answers
Given:
The direction of magnetic field is from the top to bottom.
The direction of current is into the screen.
To find the direction of particle.
Explanation:
According to Fleming's left hand rule,
First finger indicates the direction of magnetic field,
Central finger indicates the direction of current,
Thumb indicates the direction of motion.
On applying Fleming left hand rule, the direction of particle bend will be towards left.
Thus, the correct option is a
On July 19, 1969, the lunar orbit of Apollo 11 was adjusted to an average height of 122 kilometers above the Moon's surface. The radius of the Moon is 1840 kilometers, and the mass of the Moon is 7.3 x 1022 kilograms. How long did it take to orbit once? Include units in your answer. Answer must be in 3 significant digits.
Answers
Given data:
* The mass of the Moon is,
[tex]m=7.3\times10^{22}\text{ kg}[/tex]* The radius of the Moon is,
[tex]\begin{gathered} R=1840\text{ km} \\ R=1840\times10^3\text{ m} \end{gathered}[/tex]* The height of the Apollo 11 is,
[tex]\begin{gathered} h=122\operatorname{km} \\ h=122\times10^3\text{ m} \end{gathered}[/tex]Solution:
The period of revolution of the Apollo 11 around the Moon is,
[tex]T=2\pi\sqrt[]{\frac{(R+h)^3}{Gm}}[/tex]where G is the gravitational constant,
Substituting the known values,
[tex]\begin{gathered} T=2\pi\sqrt[]{\frac{(1840\times10^3+122\times10^3)^3}{6.67\times10^{-11}\times7.3\times10^{22}}} \\ T=2\pi\times\sqrt[]{\frac{(1962\times10^3)^3}{48.69\times10^{11}}} \\ T=2\pi\times1245.46 \end{gathered}[/tex]Thus, the value of time period is,
[tex]\begin{gathered} T=7825.46\text{ s} \\ T=\frac{7825.46}{60\times60}\text{ hr} \\ T=2.17\text{ hr} \end{gathered}[/tex]Thus, Apollo 11 takes 2.17 hours to complete orbit once around the Moon.
A car moves with a constant speed of 220 km/h. Express this speed in m/s and calculate the distance covered in 20 seconds
Answers
Hello..!
We first apply the data to the problem.
Data:
V = 220km/hT = 20sD = ?Now, we convert the "km/h" to "m/s".
Conversion:
220km/h • (1000m/1km) • (1h/3600s)V = 61.1m/sThen, we apply the formula that is.
Formula:
D = V • TFinally we develop the problem.
Developing:
D = (61.1m/s) • (20s)D = 1222mThe distance covered is 1222 meters.
Answer:
Formula:
D = v • tResolver:D = (61.1m/s) • (20s)D = 1222mHope this helped
A graduated cylinder measures ________, usually in units of ___________.
Answers
Answer:
A graduated cylinder measures in units of volume, usually in units of milliliters.
Two pieces of window glass are separated by a distance, d. If a beam of light of wavelength l=469 nm passes through the first piece of glass. What is the minimum distance, in nm, such that the light intensity transmitted through the right is a maximum?
Answers
The distance needs to be optimized to cause constructive interference, which can be seen on the following drawing:
So the distance needs to be exactly of a single wavelength.
Then, our final answer is d=469nm
Please do number three for me step-by-step and explain the east and west thank you
Answers
the Given that the distance from mall to home is 1000 m.
The distance from home to the library is 1200 m.
So the total displacement is 1000 +1200 = 2200 m.
The displacement is from the library to the mall, so the displacement is towards the west.
So the displacement is 2200 m towards west.
Rashad and Carlos measure the length, width and height of a textbook at 6 cm, 4 cm, and 3respectively. What is the volume of the textbook?
Answers
Answer:
72 cm³
Explanation:
The volume of a textbook can be calculated as
Volume = Length x Width x Height
Replacing the length by 6 cm, the width by 4 cm, and the height by 3 cm, we get:
Volume = 6 cm x 4 cm x 3 cm
Volume = 72 cm³
Therefore, the volume of the textbook is 72 cm³
What are the two most important factors in determining if a force is doing physics work?
Answers
The important factors in determining if a force is doing physics work include
1) The amount or magnitude of the force being applied
2) The displacement caused by the applied force
Hence,
Work = force x displacement
Create a table comparing and contrasting constructive and destructive interference .
Answers
Constructive interference:-
1. The constructive interference takes place, when the amplitude of the two or more waves combines to form the resultant wave with more amplitude.
2. The phase difference between the waves during the constructive interference is,
[tex]\varnothing=\frac{n\pi}{2}[/tex]where n is the even number.
3. In the constructive interefernce, the bright fringes are observed.
Destructive interference:-
1. The destructive interference takes place when the ampltude of two or more wave combines to form the resultant wave with less amplitude.
2. The phase difference between the waves during the destructive interference is,
[tex]\varnothing=\frac{n\pi}{2}[/tex]where n is the odd number.
3. In the destructive interference, the dark fringes are formed.
Why is the sand on a dry beach not a fluid but when a wave comes in and out the sand does become a fluid and your feet sink when standing on it?
Answers
When the sand is mixed with the water, it will form a quicksand. It behaves like a solution, and when the water is removed from the sand, it will act as a solute.
When our feet sink into the sand, our movements will cause you to dig yourself deeper into it.
What is the acceleration of a car that starts at rest and achieves a final velocity of 33.5 m/s in a time of 12 seconds?
Answers
Variables we are given and that we know:
vi: initial velocity; vi = 0 m/s
vf: final velocity; vf = 33.5 m/s
t: time; t = 12 s
We need to find a: acceleration
We can use the formula:
vf = vi + a*t
Let's substitute the values we know:
33.5 = 0 + a*12
33.5 = 12a
Let's solve for a:
a = 33.5/12
a = 2.7917 m/s^2
Point charges create equipotential lines that are circular around the charge (in the plane of the paper). What is the potential energy, in nJ, of a 1 nC charge located 2.82 m from a 2 nC charge ?
Answers
The potential energy U of a system formed by two point charges separated a distance r is:
[tex]U=k\frac{q_1q_2}{r}[/tex]Where k is Coulomb's Constant:
[tex]k=8.98755\times10^9N\frac{m^2}{C^2}[/tex]Repace q1=1nC, q2=2nC and r=2.82m to find the potential energy of the system:
[tex]\begin{gathered} U=k\frac{q_1q_2}{r} \\ \\ =(8.98755\times10^9N\frac{m^2}{C^2})\times\frac{(1\times10^{-9}C)(2\times10^{-9}C)}{2.82m} \\ \\ =6.3741489...\times10^{-9}Nm \\ \\ \approx6.37nJ \end{gathered}[/tex]Therefore, the potential energy of the system is approximately 6.37 nanoJoules.
One force that has always on you is the gravitational force on you by earth what is the newtons third law paired to this forth think opposites
Answers
c. The normal force on you by earth
The Newton's third law states that, for every action there is equal and opposite reactionhe
In which of the following processes does the Sun play little to no role inproviding energy?A. Plate tectonicsB. Ocean currentsC. SeasonsD. Trophic levels
Answers
Plate tectonics are influenced by the earth's mantle and its variations, therefore, the role of the sun is less in comparison with the other processes described.
A box sits on a table. The box weighs 80 N. If I attach a rope to the box and lift in the upward direction with a 25 N force, what is the normal force acting on the box
Answers
If I tie a rope to an 80N box that is resting on a table and push it upward with a 25N force, the normal force operating on the box will be (80N-25N) = 55N force.
What is force?A force is just an effect that can alter an object's motion according to physics. An object with mass can change its velocity, or accelerate, as a result of a force. A pushing or a pulling motion is a straightforward way to describe force. A force is a vector quantity since it has either magnitude and direction. A force or energy that has been applied, exerted, or brought about to move or transform the natural forces. In science, the term "force" has a particular definition. At this level, calling a force a pushing or a pull is entirely appropriate. An object doesn't "have a force in it" or "contain a force." A force is applied to one thing by another. The idea of a force encompasses both living and non-living things.
How do you calculate force and why force unit is Newton?The SI unit measuring force is the newton, denoted by the letter N. The definition of a force formula is given by Newton's second law of motion: An object's forces are equal towards its mass times its acceleration. F = m ⨉ a. You must use SI units of force (newtons), mass (kg), and acceleration in order to apply this formula (meters per second squared).
The force that causes a mass of one kilogram to accelerate by one metre per second per second is known as 1 kgm/s₂. In terms of its contribution to classical mechanics, especially Newton's second law of motion, it is named after Isaac Newton.
F = m × a
F = Force
m = mass of an object
a = acceleration
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Two equally charged, 4.982 g spheres are placed with 3.173 cm between their centers. When released, each begins to accelerate at 258.312 m/s2. What is the magnitude of the charge, in micro-Coulombs, on each sphere?
Answers
Given:
The mass of the spheres is m = 4.982 g
The distance between the center of spheres is d = 3.173 cm
The acceleration is a = 258.312 m/s^2.
To find the magnitude of charge in micro Coulomb on each sphere.
Explanation:
According to Newton's second law, the force will be
[tex]F\text{ =ma}[/tex]According to Coulomb's law, the force will be
[tex]F=\frac{kq^2}{r^2}[/tex]Here, k is the Coulomb's constant whose value is
[tex]k=9\times10^9\text{ N m}^2\text{ /C}^2[/tex]On equating the forces, the charge will be
[tex]\begin{gathered} ma=\frac{kq^2}{r^2} \\ q=\sqrt{\frac{mar^2}{k}} \end{gathered}[/tex]On substituting the values, the magnitude of charge will be
[tex]\begin{gathered} q=\sqrt{\frac{(4.982\times10^{-3})\times258.312\times(3.173\times10^{-2})^2}{9\times10^9}} \\ =3.79\text{ }\times10^{-7}\text{ C} \\ =0.379\text{ }\times10^{-6}\text{ C} \\ =0.379\text{ }\mu C \end{gathered}[/tex]The magnitude of the charge of each sphere is 0.379 microCoulomb