Answers
Given data
*The given time is t = 6 seconds
*The given distance from the base of the cliif is R = 30 m
The formula for the speed is given as
[tex]\begin{gathered} R=ut \\ u=\frac{R}{t} \end{gathered}[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} u=\frac{30}{7} \\ =4.28\text{ m/s} \end{gathered}[/tex]Related Questions
A 15 V constant voltage battery is connected to 120 kΩ resistor. A 15 V constant voltage battery is connected to 120 kΩ resistor. 1) What is the current in the resistor in [mA]?Multiple Choice question2) How many electrons pass through the resistor each second?a) 7.8x1014 electronsb) 1.25x10-4 electronsc) 1.6x10-19 electrons
Answers
We will have the following:
1)
[tex]I=\frac{V}{R}\Rightarrow I=\frac{15V}{120000\Omega}\Rightarrow I=1.25\cdot10^{-4}A[/tex][tex]\Rightarrow I=0.125mA[/tex]2)
First we determine the quantity of coulombs:
[tex]Q=(1.25\cdot10^{-4}A)(1s)\Rightarrow Q=1.25\cdot10^{-4}C[/tex]Now, we know that 1 coulomb represents 6.24*10^18 electrons, so:
[tex]1.25\cdot10^{-4}C\cdot\frac{6.24\cdot10^{18}\text{electrons}}{1C}=7.8\cdot10^{14}\text{electrons}[/tex]So, approximately 7.8*10^14 electrons pass each second.
Can you compare “simple harmonic motion” & hooks law. For me?
Answers
We will have the following:
We will have that Hook's law describes systems of springs with masses atteched to them and models the nature of the force and position of such masses by the relationshipt between them. Hook's law applies only to spring systems and is linear in nature.
Simple harmonic motion models repetitive movement back and forth through an equilibrium, or central position. Hook's law is one of the multiple instances of simple harmonic motion. Simple harmonic motion not only is able to model srpings, but waves and periodic movements.
If the period of a certain wave (wavelength = 4.5 m) is 2 seconds, what is the speed of the wave?1) 0.44 m/s2) 1.1 m/s3) 9.0 m/s4) 2.3 m/s
Answers
Given data:
*The given wavelength of the wave is
[tex]\lambda=4.5\text{ m}[/tex]*The given time is t = 2 s
The formula for the speed of the wave is given as
[tex]v=\frac{\lambda}{t}[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} v=\frac{4.5}{2} \\ =2.25\text{ m/s} \\ \approx2.3\text{ m/s} \end{gathered}[/tex]Hence, the speed of the wave is v = 2.3 m/s and the correct option is (4)
Sidney's group came up with a design for the seat belt but had to figure out whattype of material to use. They picked two fabrics: one that was very rigid with littlegive and a second fabric that was stretchy. They made the two belts, got somemotorized cars and put action figures into the cars, one car per seat belt design.Then they ran the cars into a wall of the classroom. Predict which belt they choseand why.The stretch seat belt: for every action there is a reaction. The stretch belt letsyou move forward and then you move back again.The non-stretch seat belt: with no seatbelt to stop the driver with the car, thedriver flies free until stopped suddenly by impact on the steering column so thenon-stretch belt keeps the driver from flying free.The non-stretch seat belt holds the driver immobile against the seat as the carcrashed.The stretch seat belt: some stretch in the seatbelts will reduce the averageimpact force by extending the stopping distance of the passenger
Answers
The correct answer is the last one, that is, The strecth seat belt:
Read each scenario below. Then select the answer that best completes each sentence.The power from a car engine is more than the power of a bicycle because a car engine does the same amount of work in time.Raul and Katrina equally shared a frozen lunch but heated each portion in two different microwaves. Katrina’s lunch was warm in one minute while Raul’s lunch took two minutes. The power used by Raul’s microwave must be the power used by Katrina’s because his microwave took time to do the same amount of work.
Answers
a)The power from a car engine is more than the power of a bicycle because a car engine does the same amount of work in less time.
b)Raul and Katrina equally shared a frozen lunch but heated each portion in two different microwaves. Katrina’s lunch was warm in one minute while Raul’s lunch took two minutes. The power used by Raul’s microwave must be the less than the power used by Katrina’s because his microwave took more time to do the same amount of work.
Explanation
work:work, in physics, measure of energy transfer that occurs when an object is moved over a distance by an external force at least part of which is applied in the direction of the displacement
so
Step 1
a)The power from a car engine is more than the power of a bicycle because a car engine does the same amount of work in less time.
power is given by:
[tex]P=\frac{\text{work done}}{time\text{ taken}}[/tex]so, we can see the time is in the denominator, so as the car engine does the same work in less time, the power of the engine is more than the power of a bicycle
Step 2
Raul and Katrina equally shared a frozen lunch but heated each portion in two different microwaves. Katrina’s lunch was warm in one minute while Raul’s lunch took two minutes. The power used by Raul’s microwave must be the less than the power used by Katrina’s because his microwave took more time to do the same amount of work.
again, as the time is in the numerator
to more times, less power , so
[tex]\begin{gathered} P_{katr\in a}=\frac{W_{katr\in a}}{\text{time taken}} \\ as\text{ the shares ase equal , the work neccesary to warm it is the same, so} \\ W_{katr\in a}=W_{raul} \\ so \\ P_{katr\in a}=\frac{W_{katr\in a}}{\text{time taken}}=\frac{W_{katr\in a}}{1} \\ Praul=\frac{W_{katr\in a}}{\text{time taken}}=\frac{W_{katr\in a}}{\text{2}} \\ P_{raul}=\frac{W_{katr\in a}}{\text{2}}=\frac{P_{katr\in a}}{2} \\ P_{raul}=\frac{P_{katr\in a}}{2} \end{gathered}[/tex]therefore,
the answer is
Raul and Katrina equally shared a frozen lunch but heated each portion in two different microwaves. Katrina’s lunch was warm in one minute while Raul’s lunch took two minutes. The power used by Raul’s microwave must be the less than the power used by Katrina’s because his microwave took more time to do the same amount of work.
I hope this helps you
Answer:
1. more than
2. less
3. less than
4. more
Explanation:
Which quantity is not required in order to calculate the mass of an ion using a mass spectrometer?O velocity vO magnetic field BO charge qO change in time t
Answers
To answer your question, lets think about the formulas needed
Centripetal force = mv^2/r
and force on a point charge = qvb
The force of the magnet is the same on the particle so you can set them equal to each other
qvb = mv^2/r
Set equal to mass
m = qbr/v
Change in time is not needed
The melting point of ethyl alcohol is -129 F. What is the Celsius reading?
Answers
We are asked to convert -129 F to Celsius. To do that we will use the following formula:
[tex]C=\frac{5}{9}(F-32)[/tex]Where:
[tex]\begin{gathered} C=\text{ reading in Celsius} \\ F=\text{ reading in Fahrenheit} \end{gathered}[/tex]Now we plug in the reading in Fahrenheit:
[tex]C=\frac{5}{9}(-129-32)[/tex]Solving the operations:
[tex]C=-89.44\text{Celsius}[/tex]Therefore, the Celsius reading is -89.44°C.
The magnetic field at point P is zero. Findthe distance, r2, from P to the second wire,1₂.I₁ = 4.32 A P------0.831 m1₂ = 7.50 Ar2r₂ = [?] m
Answers
The formula for calculating the magnetic field is
B = μo[(I1/2πR1) - (I2/2πR2)]
B = μo/2π[(I1/R1) - (I2/R2)]
where
B is the magnetic field
μo is the permeability of free space
From the information given,
B = 0
I1 = 4.32
R1 = 0.831
I2 = 7.5
R2 = ?
μo = 4π x 10^-7 Tm/A
Thus,
0 = 4π x 10^-7/2π[(4.32/0.831) - (7.5/R2)]
0 = 4 x 10^-7/2[(4.32/0.831) - (7.5/R2)]
0 = 4 x 10^-7/2(4.32/0.831) - 4 x 10^-7/2(7.5/R2)]
4 x 10^-7/2(7.5/R2)] = 4 x 10^-7/2(4.32/0.831)
7.5/R2 = 4.32/0.831
By cross multiplying,
4.32R2 = 7.5 x 0.831
R2 = (7.5 x 0.831)/4.32
R2 = 1.443 m
A box is pulled a distance of 20 meters by a force of magnitude 40N, in the direction of the force ,over a period of 10 seconds Find the power generated by the force
Answers
The formula for the power is,
[tex]\begin{gathered} P=\frac{W}{t} \\ Where, \\ P=Power \\ W=Work \\ t=Time \end{gathered}[/tex]Work done is,
[tex]\begin{gathered} W=F\times S \\ Where, \\ W=Work \\ F=Force \\ S=Displacement \\ here, \\ F=40N;\text{ }and\text{ }S=20m; \\ So, \\ W=F\times S=40\times20=800J \end{gathered}[/tex]Here time taken to do the work is 10 sec
So,
[tex]P=\frac{W}{t}=\frac{800}{10}=80\text{ watt/sec}[/tex]So power generated by the force is 80 watt/sec.
What type of electronics can be easily damaged by a static electricity discharge?
Answers
Electrostatic sensitive devices can be easily damaged by static electricity discharge.
These are mainly integrated circuits or ICs which usually operate at low voltages.
These devices can be easily damaged by static electricity discharge.
If M-16 kg, what is the tension in string 2? The angles shown are 30° for string 1 and 60° for string 2.
Answers
Since the block is at rest, i.e., equilibrium, the net force on the system must be zero: That means:
[tex]\begin{gathered} F_{1x}=F_{2x} \\ F_1\sin 30^o=F_2\sin 60^o \\ F_1=\frac{F_2\sin 60^o}{\sin 30^o}^{}\rightarrow eq(1) \end{gathered}[/tex]Also we have:
[tex]\begin{gathered} F_{1y}+F_{2y}=W \\ F_1\cos 30^o+F_2\cos 60^o=mg \\ \frac{F_2\sin60^o}{\sin30^o}\cos 30^o+F_2\cos 60^o=mg \\ F_2\sin 60^o\cot 30^o+F_2\cos 60^o=mg \\ F_2(\sin 60^o\cot 30^o+\cos 60^o)=mg \\ F_2=\frac{mg}{\sin 60^o\cot 30^o+\cos 60^o} \\ \end{gathered}[/tex]Lets say that g = 9.8 m/s², then if we plug the data we get:
[tex]\begin{gathered} F_2=\frac{16\times9.8}{(0.87)\times(1.73)+(0.50)} \\ \\ F_2=\frac{156.8}{2.01} \\ \\ F_2=78N \end{gathered}[/tex]Answer: the tension in string 2 is 78 N.
A student conducts an investigation into the motion of objects dropped from rest in the absence of air resistance. The student releases objects of various masses in a vacuum chamber and records the speed and distance fallen every 0.1 seconds for each object. What experimental evidence would lead the student to conclude that free-fall motion is constant acceleration? Select two answers. A. All objects released from rest at the same time fall together.B. The distance fallen is proportional to the square of the falling time.C. A graph of velocity vs. time is linear. D. Objects speed up as they free-fall. One of your answers can be explained using a basic equation of motion. Which answer is it and which equation is it? One of your answers can be explained using a rule about how acceleration appears on a certain graph. State the graph rule, then explain why “constant acceleration” looks the way that it does on this certain graph.
Answers
What is the power of a heater that has resistance 48 and operates at 240v
Answers
Given
Resistance of the heater, R=48 ohm
Voltage , V=240 V
To find
The power of the heater
Explanation
The power is given by
[tex]P=\frac{V^2}{R}[/tex]Putting the values,
[tex]P=\frac{240^2}{48}=1200\text{ W}[/tex]Conclusion
The power is 1200 W
What is the main purpose for learning about significant figures in science and/or technology courses? Give a detailed answer without plagiarism.
Answers
The significant figure of a value or measurement is the number of important digits that it contains
The main purpose for learning about significant figures in science and/or technology courses is precision of measurements.
In science and technology experiments and reseaches, we always want to ensure that measured values are as close to the true values as possible. This is because any deviation from the true value can invalidate the result of the experiment. Also, errors due to repeated approximations can have cummulative effects on the experimental results.
Therefore, the knowledge of significant figures is useful in science and technology to be able to record values of experimental measurements as close as possible to the actual value.
A magnet that retains its magnetism and does not need to be “recharged” is called a(n) ____.
a. electromagnet b. permanent magnet c. temporary magnet d. voltage regulation
Answers
Answer:
B
Explanation:
permanent magnet
how much work is required to stop an electron which is moving with a speed of 1.10x10^6m/s
Answers
Work required to stop an electron which is moving with a speed of 1.10x10^6m/s is [tex]-\frac{3.971}{5^{19}\cdot \:1048576}[/tex]
What is electron?There are three main types of particles in an atom: protons, neutrons, and an electron that is bonded to an atom. For a variety of reasons, electrons are different from other particles. They possess both wave-like and particle-like properties, exist outside of the nucleus, have a substantially lower mass than protons. A particle that is not composed of smaller parts is an electron, which is also an elementary particle. They are not elementary particles because quarks are assumed to make up protons and neutrons.
We can solve the problem by using the work-energy theorem: in fact, the work done on the electron must be equal to its change in kinetic energy, therefore
[tex]W=K_{f}-K_{i}[/tex]
[tex]W=\frac{1}{2} mv^{2} - \frac{1}{2} mu^{2}[/tex]
where,
m=9.11 × 10⁻³¹kg is the mass of the electron
v is the final velocity of the electron
u is the initial velocity of the electron
In the problem, we are told that the electron is initially moving at,
v = 1.90×10⁶ m/s
Therefore, the work required to stop it as
W = -1/2 mv²
W = -1/2 (1.10×10⁻³¹)(1.90×10⁶)²
[tex]W= -\frac{3.971}{5^{19}\cdot \:1048576}\\[/tex]
and the work is negative since it is in the opposite direction to the motion of the electron.
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The internal energy of an expanding gas changes by 2000j. How much work is done if the process is done adiabatically?
Answers
Given
The internal energy of an expanding gas changes by U=2000J.
To find
The work done
Explanation
Since the work done is adiabatic so the change in heat is zero.
By first law of thermodynamics,
[tex]\begin{gathered} U=Q+W \\ \Rightarrow2000+0+W \\ \Rightarrow W=2000J \end{gathered}[/tex]Conclusion
The work done is 2000J
Two equal, but oppositely charged particles are attracted to each other electrically. The size of the force of attraction is 33.55 N when they are separated by 55.75 cm. What is the magnitude of the charges in microCoulombs ?
Answers
Given
Two equal, but oppositely charged particles are attracted to each other electrically.
Force of attraction is F=33.55 N
Distance between them, d=55.75 cm=0.5575 m
To find
What is the magnitude of the charges in microCoulombs ?
Explanation
Let the charge be q.
We know the force of attraction is given by
[tex]\begin{gathered} F=k\frac{q^2}{d^2} \\ \Rightarrow33.55=9\times10^9\times\frac{q^2}{(0.5575)^2} \\ \Rightarrow q=3.39\times(10)^{-5}C \\ \Rightarrow q=\pm33.9\mu C \end{gathered}[/tex]Conclusion
The charges are:
[tex]+33.9\mu C,-33.9\mu C[/tex]The magnitude of equal charges are:
[tex]\lvert{q}\rvert=33.9\mu C[/tex]A force of 585 N is exerted on a 407 kg mass a distance of 13660 km above the surface of a planet having a mass of 7.9E24 kg. Determine the average density of the planet in kg/cubic meter. Derive and express algebraic solution in terms of givens: F, m, mp, alt and G.
Answers
The average density of a planet is given by:
[tex]\rho=\frac{m}{V}[/tex]where m is the mass of the planet and V is its volume. We know the mass of the planet but we don't know its volume, to find it we will need find its radius.
To find the radius of the planet we can use Newton's Law of gravitation:
[tex]F=G\frac{mM}{d^2}[/tex]where G is the gravitational constant, m is the mass of the object, M is the mass of the planet and d is the distance between the planet and the object. Let r be the radius of the planet, and x be the distance from the surface of the planet to the object (x=13660 in this case); then we have:
[tex]\begin{gathered} F=G\frac{mM}{(r+x)^2} \\ (r+x)^2=\frac{GmM}{F} \\ r+x=\pm\sqrt[]{\frac{GmM}{F}} \\ r=-x\pm\sqrt[]{\frac{GmM}{F}} \end{gathered}[/tex]Plugging the values given we have:
[tex]\begin{gathered} r=-13660\times10^3\pm\sqrt[]{\frac{(6.67\times10^{-11})(407)(7.9\times10^{24})}{585}} \\ \text{ Using the positive root we have:} \\ r=5.49\times10^6 \\ \text{ Using the negative root we have:} \\ r=-3.28\times10^7 \end{gathered}[/tex]Since the radius of the planet has to be positive we choose the positive solution.
Now, that we know the radius of the planet we can calculate its volume; assuming the planet is spherical we have that:
[tex]V=\frac{4}{3}\pi r^3[/tex]then we have:
[tex]\begin{gathered} V=\frac{4}{3}\pi(5.49\times10^6)^3 \\ V=6.92\times10^{20} \end{gathered}[/tex]Finally we can calculate the density:
[tex]\begin{gathered} \rho=\frac{7.9\times10^{24}}{6.92\times10^{20}} \\ \rho=11398 \end{gathered}[/tex]Therefore, the average density is 11398 kg/m^3
Numerical problem on Coulomb's law 1. Find the magnitude of product of two charges kept 1 metre aparat when a active force of 9.10^9 newton is acting between them.
Answers
In order to calculate the product of charges, we can use the formula below for the force acting on two charges (Coulomb law):
[tex]|F|=k_e\frac{|q_1q_2|_{}}{r^2}[/tex]Where ke is the Coulomb constant (ke = 9 * 10^9), q1 and q2 are the charges and r is the distance between the charges.
So, for r = 1 and F = 9 * 10^9, we have:
[tex]\begin{gathered} 9\cdot10^9=9\cdot10^9\frac{|q_1q_2|}{1^2} \\ |q_1q_2|=1 \end{gathered}[/tex]Therefore the product of the charges is 1 Coulomb.
Point p, to me, doesn't look like a conjunction point where I can apply Kirchhoff's first law, which is the sum of the current in is equal to the sum of the current out. And how do I explain using charges that flow through?
Answers
ANSWER
a) The current at P is 1.0A and it is moving towards the node.
b) A total of 1C charges flow through P in 1.0 seconds
EXPLANATION
a) To find this, we have to apply Kirchoff's node law which states that:
Therefore, we have that:
[tex]4.0+3.0-8.0+P=0[/tex]Solve for P:
[tex]\begin{gathered} 4.0+3.0-8.0+P=0 \\ -1.0+P=0 \\ \Rightarrow P=1.0A \end{gathered}[/tex]This implies that the current at point P is 1A and it is moving towards the node (in the left direction).
b) To find the electric charge that flows through P in 1.0s, we have to use the formula for current:
[tex]I=\frac{q}{t}[/tex]where I = current, q = electric charge and t = time
Therefore, we have to find q:
[tex]\begin{gathered} q=I\cdot t \\ \Rightarrow q=1.0\cdot1.0 \\ q=1.0C \end{gathered}[/tex]Therefore, a total of 1C charges flow through P in 1.0 seconds.
convert 675 nm to meters
Answers
Given,
The number needed to be converted is 675 nm
A nanometer is equal to one billionth of a meter.
Mathematically,
[tex]1nm^{}=1\times10^{-9}\text{ m}[/tex]Thus to convert any number from nanometer to meter, we need to multiply the number by 10⁻⁹ m.
Thus,
[tex]675\text{ nm}=675\times10^{-9}\text{ m}[/tex]Therefore 675 nm is equal to 675×10⁻⁹ m
An astronaut drops two pieces of paper from the door of a lunar landing module. One piece of paper is
crumpled, and the other piece is folded into an airplane. Why do the two pieces of paper land on the Moon's
surface at the same time? (1 point)
O The Moon's gravity is much weaker than Earth's.
O The mass of the paper folded into an airplane must be greater than the mass of the crumpled paper.
O The pieces of paper were not dropped from a sufficient height for air resistance to affect their falls.
O The Moon has practically no atmosphere, so there is no air resistance.
Answers
The moon has practically no atmosphere so there is no air resistance.
What is Lunar Landing Module?Grumman created the two-stage Apollo Lunar Module (LM) to transport two men from lunar orbit to the lunar surface and back. The ascending rocket engine, equipment bays, and pressurized crew quarters made up the higher ascent stage.
The landing gear, descent rocket engine, and lunar surface experiments were all located on the lower descent stage.
For a second unmanned Earth-orbit test trip, LM 2 was constructed. A second unmanned LM test mission was deemed unnecessary because the LM 1 test flight, which was carried out as part of the Apollo 5 mission, was so successful.
Therefore, The moon has practically no atmosphere so there is no air resistance.
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What happens to parallel light rays that pass through a thin convex lens?They remain parallel. ⊝They all bend, deflecting in the same direction. ⊝They converge at the focal point on the opposite side. ⊝They diverge as though originating from the same-side focal point.
Answers
Convex and concave lenses deviate the trajectory of light rays that pass through them, as the following diagram shows:
Therefore, the answer is:
[tex]\text{They converge at the focal point on the opposite side.}[/tex]The half-life of tungsten 188 is 69.4 days. Initially, there are 0.725 kg of this isotope. How much of the isotope will remain after 147 days?(a) 0.104 kg(b) 0.167 kg(c) 0.237 kg(d) 0.255 kg
Answers
In order to determine the amount of tungsten after 147 days, use the following formula for the radioactive decay:
[tex]A=A_oe^{-\lambda t}[/tex]where
A: amount of tungsten after t days
Ao: initial amount of tungsten = 0.725 kg
t: time = 147 days
λ: decay constant
Then, it is necessary to find the value of λ. Use the following formula:
[tex]\lambda=\frac{\ln 2}{t_{\frac{1}{2}}}[/tex]where t1/2 is the half-life of tungsten (69.4 days)
[tex]\lambda=\frac{\ln 2}{69.4}=0.00998[/tex]next, replace the previous result and the values of the other parameters into the formula for A:
[tex]A=(0.725kg)e^{-(0.00998)(147)}=(0.725kg)(0.23)=0.167\operatorname{kg}[/tex]Hence, after 147 days, there are 0.167 kg of tungsten 188
Consider a satellite in a circular geostationary orbit about the Earth, meaning it orbits above a fixed point on the Earth's equator.
You may use the following values in your calculations:
- Mass of the Earth, M is 6.0x1024kg;
- Gravitational constant G, is 6.67 x 10-11 N m2 kg-2
- Radius of the Earth, r is 6400 km or 6.4 x 106 m. At what distance from the centre of the Earth does the satellite orbit?
Answers
Answer:
m V^2 / R = G M m / R^2 balancing forces
R = G M / V^2 (I)
V = 2 π R / (24 * 3600) to complete 1 rev in 24 hrs
V = π R / (12 * 3600) = R * 7.27E-5
V^2 = R^2 * 5.29E-9
R = G M / (R^2 * 5.29E-9) using (I)
R = (G M / 5.29E-9)^1/3
R = (6.67E-11 * 6.0E24 / 5.29E-9)^1/3
R = (66.7 * 6.0 / 5.29)^1/3 E7
R = 4.22E7 meters = 42200 km about 7 earth radii
An object is spun around in a circle of radius 6.0m with a frequency of50 Hz.a. What is the period of its rotation?b. What is its velocity?c. What is its acceleration?
Answers
From the information given,
radius = 6
frequency = 50
a) Period, T is the reciprocal of frequency. This means that
T = 1/frequency
Thus,
T = 1/50 = 0.02
Period = 0.02 s
b) The formula for calculating velocity for an object in circular motion, V is expressed as
V = 2pi x radius/T
By substituting pi = 3.14, radius = 6 and T = 0.02 into the formula,
V = 2 x 3.14 x 6/0.02
V = 1884
Velocity is 1884 rad/s
c) The formula for calculating centripetal acceleration is expressed as
A = V^2/radius
Substituting V = 1884 and radius = 6 into the formula,
A = 1884^2/6 = 591576
Acceleration = 591576 rad/s^2
Which of the following choices correctly describes the orbital relationship between Earth and the sun?
a.
The sun orbits Earth in a perfect circle
c.
The sun orbits Earth in an ellipse, with Earth at one focus
b.
Earth orbits the sun in a perfect circle
d.
Earth orbits the sun in an ellipse, with the sun at one focus
Answers
The correct statement which describes the relationship between earth and the sun is that earth orbits the sun in an ellipse, with the sun at one focus.
What is an Orbit?An orbit in celestial mechanics is the curved path taken by an object, such as the path taken by a planet around with a star, a celestial body around a planet, or a manufactured satellite around an object or location in space, such as a planet, satellite, meteorite, or Lagrange point.
Ordinarily, the physics term "orbit" implies a trajectory that repeats itself with a regular basis, though it can also denote a non-repeating trajectory.
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A book of mass m = 10kg is balanced on a table Determine the weight of the book and the component of the reaction of the table in each of the following cases: 1)The book is simply placed on the table2) Someone puts their hand on the book by exerting a vertical force of 40N3)Someone is pulling this book vertically upwards with a force of 40N
Answers
Given,
The mass of the book, m=10 kg
The weight of an object is given by the product of its mass and the acceleration due to gravity.
Thus the weight of the book is given by,
[tex]W=mg[/tex]Where g is the acceleration due to gravity.
On substituting the known values,
[tex]\begin{gathered} W=10\times9.8 \\ =98\text{ N} \end{gathered}[/tex]Therefore the weight of the book is 98 N
1)
When the book is placed on the table reaction of the table will be equal to the weight of the book. Thus, the reaction of the table, in this case, will be 98 N
2)
The vertically downward force applied on the book, F=40 N
The reaction of the table when there is a vertically downward force is equal to the sum of the weight and the applied force.
Thus the reaction of the table, in this case, is given by,
[tex]N=W+F[/tex]On substituting the known values,
[tex]\begin{gathered} N=98+40 \\ =138\text{ N} \end{gathered}[/tex]Thus, the reaction of the table when a vertically downward force is acting on it is 138 N
3)
The vertically upward force acting on the book, F=40 N
As the book is in equilibrium, the net force acting on the book must be equal to zero. That is total upward forces acting on the book must be equal to the total downward forces. Thus,
[tex]\begin{gathered} N+F=W \\ \Rightarrow N=W-F \end{gathered}[/tex]Where N is the normal force.
On substituting the known values,
[tex]\begin{gathered} N=98-40 \\ =58\text{ N} \end{gathered}[/tex]Thus the reaction of the table, in this case, is 58 N
Which of the following is a graph of the velocity of an object as it falls fromrest if drag is not ignored? Explain your choice
Answers
The air drag is a force that depends on the speed of an object relative to the wind. Under certain conditions, it can be modeled as:
[tex]F=-bv[/tex]Where b is a constant.
As a falling object reaches a speed so that its weight is cancelled out by the air drag, the object will reach a maximum velocity.
In a speed vs time gaph, the speed would approach the maximum speed like an asymptote.
On the other hand, since the object falls from rest, the initial speed on the graph must be zero.
Taking these considerations into account, the correct graph for the movement of an object that falls from rest if air drag is not ignored, is option B.