Let's state the Newton's Third Law of Motion.
The Newton's Third Law of Motion states that when two bodies interact, the forces they apply to each other are equal in magnitude and opposite in direction.
Therefore, since the forces the bodies exert equal forces in opposite direction on each other, it means that forces come in pairs.
The forces can be said to be action-reaction force pairs
Therefore, Newton's third Law of Motion states that all forces come in pairs.
ANSWER:
All forces come in pairs.
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.
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
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
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
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
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.
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
The correct answer is the last one, that is, The strecth seat belt:
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.
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]_7. What happens to the total resistance of the resistors if they are connected inparallel? A. increases C. decreases B. remains the same D. increases then decreases
In order to find what happens to the total resistance when resistors are connected in parallel, let's analyze the formula below for the total resistance between two resistors in parallel:
[tex]\begin{gathered} \frac{1}{RT}=\frac{1}{R_1}+\frac{1}{R_2}\\ \\ RT=\frac{R_1R_2}{R_1+R_2} \end{gathered}[/tex]When two resistors are in parallel, they have the same voltage.
That means more current will flow, because there will be current flowing from each resistor and the voltage is the same.
Calculating the total resistance, we will find a smaller resistance than the two resistors used, because with the same voltage, the current is greater.
Therefore the correct option is C.
What type of electronics can be easily damaged by a static electricity discharge?
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.
Can you compare “simple harmonic motion” & hooks law. For me?
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
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)
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.
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:
how much work is required to stop an electron which is moving with a speed of 1.10x10^6m/s
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 melting point of ethyl alcohol is -129 F. What is the Celsius reading?
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.
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
Answer:
B
Explanation:
permanent magnet
What is physics and explain elements of physics
Physics is the branch of science that deals with mass , time , space, energy and their inter-relations .
The application of
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.
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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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
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.
which Vector goes from 5, 5 to 4, 0
The required vector is î + 5ĵ
Those physical quantities that have magnitude, as well as direction, are represented by a Vector. Some examples of Vector quantities are Force, Acceleration, Velocity, and Torque.
According to the given information, we represent the points in vector form,
The 1st vector is [tex]R_{1}[/tex] = 5î + 5ĵ
And the Second Vector is [tex]R_{2}[/tex] = 4î
Now, it is given that the required vector goes from 5,5 to 4,0.
So, the required vector is R,
Thus, R = (5-4)î + (5-0)ĵ
R = î + 5ĵ
Hence, the required vector is î + 5ĵ
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Answer:
Vector B
Explanation:
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
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
Which statement describes the sign of gravitational force?(1 point)
A Gravitational force is negative because it is attractive.
B Gravitational force is negative because it is repulsive.
C Gravitational force is positive because it is attractive.
D Gravitational force is positive because it is repulsive
The statement that describes the sign of gravitational force is as follows: Gravitational force is negative because it is attractive (option A).
What is gravitational force?Gravitational force is a very long-range, but relatively weak fundamental force of attraction that acts between all particles that have mass. It is believed to be mediated by gravitons.
According to the Newton’s law of gravitation, which explains the gravitational force, every particle in the universe attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
The negative sign in the gravitational force equation indicates that the gravitational force interaction is always attractive.
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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?
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
The internal energy of an expanding gas changes by 2000j. How much work is done if the process is done adiabatically?
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
What is the main purpose for learning about significant figures in science and/or technology courses? Give a detailed answer without plagiarism.
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 cord is wrapped around the rim of a wheel 0.230 m in radius, and a steady pull of 37.0 N is exerted on the cord. The wheel is mounted on frictionless bearings on a horizontal shaft through its center. The moment of inertia of the wheel about this shaft is 4.60 kg⋅m2.
The angular acceleration of the wheel is 1.85m/s²
Acceleration is the rate at which speed and direction of velocity vary over time. Something is said to be accelerating when it starts to move faster or slower.
We are given that,
The pull force of the wheel = τ = 37.0N
The radius of the wheel = r = 0.230m
The moment of inertia = I = 4.60kg-m²
Here the angular acceleration = α = ?
The formula for angular acceleration can be given as,
α = τ /I
α = [(37.0N)×( 0.230m)]/4.60kg-m²
α = 1.85 m/s²
Hence , the angular acceleration of the wheel would be 1.85m/s²
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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
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.
What is the power of a heater that has resistance 48 and operates at 240v
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
An object is placed 3.00 cm in front of a concave lens with a focal length of 5.00 cm. Find the image distance.
ANSWER:
-1.88 cm
STEP-BY-STEP EXPLANATION:
The situation would be the following:
Given:
Object distance (u): -3 cm
Focal distance (f): -5 cm
Using the lens formula, we calculate the distance of the image:
[tex]\begin{gathered} \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \\ \\ \text{We replacing} \\ \\ \frac{1}{v}-\frac{1}{-3}=\frac{1}{-5} \\ \\ \frac{1}{v}\cdot \:15v-\frac{1}{-3}\cdot \:15v=\frac{1}{-5}\cdot \:15v \\ \\ 15+5v=-3v \\ \\ 5v=-3v-15 \\ \\ 8v=-15 \\ \\ v=-\frac{15}{8} \\ \\ v=-1.88\text{ cm} \end{gathered}[/tex]The image distance is -1.88 cm
Answer:
Explanation:
yfdyrdrd
If M-16 kg, what is the tension in string 2? The angles shown are 30° for string 1 and 60° for string 2.
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.
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?
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
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
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
A girl holds an arrow in her hands, ready to shoot the arrow at a dragon who is just outside her arrow’s range of 45 meters. The girl then slides down a cliffside, giving her a velocity of 8.5 m/s at an angle of 45 below the horizontal. She points her bow 30 degrees above the horizontal and shoots, the arrow leaves the bow with a velocity of 45 m/s. The girl is 55 meters away from the dragon horizontally. How far above the girl is the dragon? (Assuming the arrow hits.)I already figured out how long it takes for the arrow to reach the dragon (4.72 seconds) but I don’t know how to calculate the distance between the girl and the dragon. Please help!
First, assume that the arrow leaves the bow with a velocity of 45 m/s above the horizontal with respect to the bow.
Since the bow is moving at 8.5 m/s 45º below the horizontal, find the initial velocity of the arrow with respect to the ground:
[tex]\vec{v}_{AG}=\vec{v}_{AB}+\vec{v}_{BG}[/tex]This equation reads:
The velocity of the arrow with respect to the ground is equal to the velocity of the arrow with respect to the bow plus the velocity of the bow with respect to the gound.
Notice that this is a vector equation. Then, the vertical and horizontal components of the velocities must be added separately:
[tex]\begin{gathered} v_{AG-x}=v_{AB-x}+v_{BG-x} \\ v_{AG-y}=v_{AB-y}+v_{BG-y} \end{gathered}[/tex]Find the vertical and horizontal components of the velocity of the arrow with respect to the bow and the velocity of the bow with respect to the ground:
[tex]\begin{gathered} v_{AB-x}=v_{AB}\cos (\theta) \\ =45\frac{m}{s}\cdot\cos (30º) \\ =38.97\frac{m}{s} \end{gathered}[/tex][tex]\begin{gathered} v_{AB-y}=v_{AB}\sin (\theta) \\ =45\frac{m}{s}\sin (30º) \\ =22.5\frac{m}{s} \end{gathered}[/tex]Similarly, for the velocity of the bow with respect to the ground:
[tex]\begin{gathered} v_{BG-x}=6.01\frac{m}{s} \\ v_{BG-y}=-6.01\frac{m}{s} \end{gathered}[/tex]Then, the vertical and horizontal components of the initial velocity of the arrow with respect to the ground, are:
[tex]\begin{gathered} v_{AG-x}=38.97\frac{m}{s}+6.01\frac{m}{s}=44.98\frac{m}{s} \\ \\ v_{AG-y}=22.5\frac{m}{s}-6.01\frac{m}{s}=16.49\frac{m}{s} \end{gathered}[/tex]Use the horizontal component of the velocity to find how long it takes for the arrow to travel a horizontal distance x of 55 meters. Then, use that time to find the vertical position of the arrow.
Since the horizontal movement of the arrow is uniform, then:
[tex]v_{AG-x}=\frac{x}{t}_{}[/tex]Isolate t and substitute x=55m, v_{AG-x}=44.98 m/s:
[tex]\begin{gathered} t=\frac{x}{v_{AG-x}} \\ =\frac{55m}{44.98\frac{m}{s}} \\ =1.2227s \end{gathered}[/tex]The vertical motion of the arrow is a uniformly accelerated motion. Then, the vertical position is given by:
[tex]y=v_{AG-y}t-\frac{1}{2}gt^2[/tex]Replace v_{AG-y}=16.49 m/s, t=1.2227s and g=9.81 m/s^2 to find the vertical position of the arrow when the horizontal position is 55 meters. This matches the elevation of the dragon with respect to the girl when the girl shoots:
[tex]\begin{gathered} y=(16.49\frac{m}{s})(1.2227s)-\frac{1}{2}(9.81\frac{m}{s^2})(1.2227s)^2 \\ =12.829\ldots m \\ \approx12.8m \end{gathered}[/tex]Therefore, the dragon is 12.8 meters above the girl when the arrow is shoot.