Answer:
127.53 J
Explanation:
Given that
Height of the floor, h = 1.3 m
Acceleration due to gravity, g = 9.81 m/s²
Mass of the box, m = 10 kg
To find the work done on the box by the janitor, we use the formula for potential energy. Remember that potential energy is actually the energy stored in an object as a result of its position.
P.E = mgh, substituting each parameters for its value, we then have
P.E = 10 * 9.81 * 1.3
P.E = 127.53 J
Therefore, we can conclude that the amount of work done on the box by the janitor is 127.53 J
Characteristics of good home insulation :
Answer:
Explanation:
all
did sans from Undertale really die or was he just trolling
Answer:
I think he was just trolling.
Explanation:
They don't give out any good reasons or evidence about him dying. I think they are just pretending. You tell me tho.... Many people think he did die, while others think that he didnt.
What do you think?!
3. Ohm's law is represented by the equation I = V/R. Explain how the current would change if the amount of resistance decreased and the voltage stayed the same.
Answer:
Current will increase
Explanation:
In Ohm's law the equation for current is current = voltage / resistance.
In order to explain how current is effected when resistance decreases while voltage stays the same, lets represents the situations with some possible inputs.
Let's compare 3 different closed circuits all with a voltage of 10V.
In circuit 1, the resistance is 5 ohm.
In circuit 2, the resistance is 2 ohms.
In circuit 3, the resistance is 1 ohms.
The current of:
Circuit 1 = Voltage / Resistance = 10 V / 5 ohms = 2 Amps
Circuit 2 = Voltage / Resistance = 10V / 2 ohms = 5 Amps
Circuit 3 = Voltage / Resistance = 10V / 1ohm = 10 Amps
As you can see in this representation, as the resistance in a circuit increases while the voltage is constant, the total current is increased.
As the resistance decreases, the amount of current increases.
What is Ohm's law?Ohm's law states that the current through a conductor between two points is directly proportional to the voltage across the two points.
Ohm's law is represented by the equation
I = V/R
Ohm's Law tells us that the electrical current in a circuit can be calculated by dividing the voltage by the resistance.
As the current change if the resistance is decreased,
In this case, there is a inverse relationship between the two variables. As the resistance increases, the current decreases, provided all other factors are kept constant.
In other words, the current is directly proportional to the voltage and inversely proportional to the resistance.
As the current is directly proportional to the voltage and inversely proportional to the resistance.
So,
an increase in the voltage will increase the current as long as the resistance is held constant.
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The Mars Curiosity rover was required to land on the surface of Mars with a velocity of 1 m/s. Given the mass of the landing vehicle and parachute is 2270 kg, the drag coefficient is effectively 0.5, the atmosphere density is 0.71 that of Earth (take Earth atmosphere density as 1.2 kg/m3), and the Martian gravitational acceleration is 3.689 m/s2, find the required total frontal area (in m2) of the lander plus a parachute to land at the given velocity. Assume the landing vehicle has achieved terminal velocity as it falls through the Martian atmosphere.
Answer:
The value is [tex]A = 39315 \ m^2[/tex]
Explanation:
From the question we are told that
The velocity which the rover is suppose to land with is [tex]v = 1 \ m/s[/tex]
The mass of the rover and the parachute is [tex]m = 2270 \ kg[/tex]
The drag coefficient is [tex]C__{D}} = 0.5[/tex]
The atmospheric density of Earth is [tex]\rho = 1.2 \ kg/m^3[/tex]
The acceleration due to gravity in Mars is [tex]g_m = 3.689 \ m/s^2[/tex]
Generally the Mars atmosphere density is mathematically represented as
[tex]\rho_m = 0.71 * \rho[/tex]
=> [tex]\rho_m = 0.71 * 1.2[/tex]
=> [tex]\rho_m = 0.852 \ kg/m^3[/tex]
Generally the drag force on the rover and the parachute is mathematically represented as
[tex]F__{D}} = m * g_{m}[/tex]
=> [tex]F__{D}} = 2270 * 3.689[/tex]
=> [tex]F__{D}} = 8374 \ N[/tex]
Gnerally this drag force is mathematically represented as
[tex]F__{D}} = C__{D}} * A * \frac{\rho_m * v^2 }{2}[/tex]
Here A is the frontal area
So
[tex]A = \frac{2 * F__D }{ C__D} * \rho_m * v^2 }[/tex]
=> [tex]A = \frac{2 * 8374 }{ 0.5 * 0.852 * 1 ^2 }[/tex]
=> [tex]A = 39315 \ m^2[/tex]
If the forces acting on an object are balanced, what must be true about the motion of the object? (DOK 1, AKS 8b) A. It must be changing direction. B. It must be accelerating. C. It must be moving sporadically. (slowly and then quickly) D. It must have a constant velocity.
Answer:
D
Explanation:
If the forces on an object are balanced (or if there are no forces acting on it), this is what happens: a stationary object stays still a moving object continues to move at the same speed and in the same direction Remember that an object can be moving, even if there are no forces acting on it.
The forces acting on an object are balanced, then it must have a constant velocity. Hence, option D is correct.
What is Force?A strain is a phenomenon that can change an object's velocity in physics. A force can change or increase an excess object's velocity. It makes logical sense to use pushing or pulling to express force. Being vector quantities, energies have both size and direction. Utilizing the SI system, it is calculated in newtons (N). The letter F stands for force.
When forces acting on an item are balanced (or absent), the following occurs: a static object remains still while a moving object maintains its direction and speed. Do not forget that an item can be moving even in the absence of external forces.
When the forces are in equilibrium then it will move in a particular direction also it will move at a constant speed.
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Assuming perfect optics, if the smallest feature you can resolve when observing around 660 nm is of angular size 0.04 arcsec, what would the size (in arcsec) of the finest feature you can resolve if you make your observation around wavelength 1320 nm, assuming everything else is the same
Answer:
The value is [tex]\theta _2 = 0.08 \ arcsec[/tex]
Explanation:
From the question we are told that
The first wavelength is [tex]\lambda_1 = 660 \ nm = 660 *10^{-9 }[/tex]
The first angular size is [tex]\theta_1 = 0.04 \ arcsec[/tex]
The second wavelength is [tex]\lambda _2 = 1320 \ nm = 1320 *10^{-9 } \ m[/tex]
Generally according to Rayleigh Criterion we have that
[tex]\theta= 1.22 * \frac{\lambda }{D}[/tex]
given every other thing remains constant we have that
[tex]\theta = k * \lambda[/tex]
Here k represented constant so
[tex]k = \frac{\theta }{\lambda}[/tex]
=> [tex]\frac{\theta_1}{ \lambda_1} = \frac{\theta_2}{ \lambda_2}[/tex]
=> [tex]\frac{\theta_1}{ \theta_2} = \frac{\lambda_1}{ \lambda_2}[/tex]
So
[tex]\frac{ 0.04}{ \theta_2} =0.5[/tex]
=> [tex]\theta _2 = 0.08 \ arcsec[/tex]
the capacity of computer of performing more than one task at the same time is called ............of computer.
Answer:
We conclude that the capacity of the computer of performing more than one task at the same time is called the versatility of the computer.
Explanation:
We know that when a PC or laptop is capable of doing multiple tasks at the same time, this particular characteristic of a computer is termed as the versatility of the computer.
In other words, the versatility of the computer makes sure the computer or PC can do multitasking at the same time.
The versatility of a computer makes sure the machine can perform different types of works completed at the same time.
For example, using a computer, we can listen to music while playing games at the same time.
Therefore, we conclude that the capacity of the computer of performing more than one task at the same time is called the versatility of the computer.
School bus drivers are extra careful when they cross railroad tracks. They stop before they cross railroad tracks to be sure that there is no train coming. Yet, trains typically move more slowly through towns than cars do. If trains usually move more slowly than cars, why can they do more damage if they hit something in their path?
Answer:
because trains are bigger and heavier than cars.
Explanation:
An observer stands 150 meters from a fireworks display rocket, which is fired directly upward. When the rocket reaches a height of 200 meters, it is traveling at a speed of 12 meters/second. At what rate is the angle of elevation formed with the observer increasing at that instant
Answer:
[tex]-0.288\ \text{rad/s}[/tex]
Explanation:
x = Distance of observer from the initial location of the rocket = 150 m
y = Vertical displacement of the rocket from the ground = 200 m
r = Distance between observer and rocket
[tex]\dfrac{dy}{dt}[/tex] = Rate of change in height of rocket = 12 m/s
[tex]\dfrac{dx}{dt}[/tex] = Rate of change in x = 0
Distance between observer and rocket at y = 200 m
[tex]r=\sqrt{x^2+y^2}\\\Rightarrow r=\sqrt{150^2+200^2}\\\Rightarrow r=250\ \text{m}[/tex]
[tex]\tan\theta=\dfrac{y}{x}[/tex]
Differentiating with respect to time
[tex]\sec^2\theta\dfrac{d\theta}{dt}=\dfrac{y\dfrac{dx}{dt}-x\dfrac{dy}{dt}}{x^2}\\\Rightarrow \dfrac{d\theta}{dt}=\dfrac{\dfrac{y\dfrac{dx}{dt}-x\dfrac{dy}{dt}}{x^2}}{\sec^2\theta}\\\Rightarrow \dfrac{d\theta}{dt}=\dfrac{\dfrac{0-150\times 12}{150^2}}{(\dfrac{250}{150})^2}\\\Rightarrow \dfrac{d\theta}{dt}=-0.288\ \text{rad/s}[/tex]
The rate of change of the angle of elevation is [tex]-0.288\ \text{rad/s}[/tex].
Two copper spheres are currently 1.2 meters apart. One sphere has a charge of +2.2 x 10-4 C and the other has a charge of -8.9 x 10-4C.
What is the force between the charged spheres?
Is the force attractive or repulsive?
Answer:
The force between the charged spheres is 1,223.75 N
The force is attractive because opposite charges mutually attract.
Explanation:
Coulomb's Law
The electrical force between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between the two objects.
Expressing the above as a formula:
[tex]\displaystyle F=k\frac{q_1q_2}{d^2}[/tex]
Where:
[tex]k=9\cdot 10^9\ N.m^2/c^2[/tex]
q1, q2 = the objects' charge
d= The distance between the objects
The copper spheres have charges of:
[tex]q_1=+2.2\cdot 10^{-4}\ c[/tex]
[tex]q_2=-8.9\cdot 10^{-4}\ c[/tex]
Note the charges have opposite signs.
They are separated by d=1.2 m
Applying Coulomb's formula:
[tex]\displaystyle F=9\cdot 10^9\frac{2.2\cdot 10^{-4}\cdot 8.9\cdot 10^{-4}}{1.2^2}[/tex]
[tex]\displaystyle F=9\cdot 10^9\frac{1.958\cdot 10^{-7}}{1.44}[/tex]
F = 1,223.75 N
The force between the charged spheres is 1,223.75 N
The force is attractive because opposite charges mutually attract.
A 3000-kg truck moving with a velocity of 25 m/s hits a 1000-kg parked car. The impact causes the 1000-kg car to be set in motion at 7 m/s. What is the velocity of the truck after the collision?
Answer:
v₃ = 22.67 [m/s]
Explanation:
In order to solve this problem, we must use the principle of conservation of the quantity of linear momentum. Where momentum is conserved before and after the collision, i.e. remains the same.
The terms on the left of the equation represent the amount of linear momentum before the collision and the members on the right represent the momentum after the collision.
[tex](m_{1}*v_{1})+(m_{2}*v_{2})=(m_{1}*v_{3})+(m_{2}*v_{4})[/tex]
where:
m₁ = mass of the truck = 3000 [kg]
m₂ = mass of the car = 1000 [kg]
v₁ = velocity of the truck before the coliision = 25 [m/s]
v₂ = velocity of the car parked = 0 (without movement)
v₃ = velocity of the truck after the collision [m/s]
v₄ = velocity of the car after the collision = 7 [m/s]
Now replacing:
[tex](3000*25)+(1000*0)=(3000*v_{3})+(1000*7)\\75000-7000 = 3000*v_{3}\\v_{3}=22.67 [m/s][/tex]
An 5-kg animal accelerates from rest to 10 m/s in 10 m and encounters an average drag force of 12 N. What is the average power output in Watts if this takes 2.0 seconds
Answer:
60 watts
Explanation:
Given that
Mass of the animal, m = 5 kg
Velocity of the animal, v = 10 m/s
Distance moved by the animal, d = 10 m
Drag force, F(d) = 12 N
Time taken, t = 2 seconds
To start with, we need to find the power output in itself before proceeding to find the average power output. And as such, we have
Power = Force * Distance/Time
But Distance/Time is velocity, so
Power = Force * Velocity
Power = 12 * 10
Power = 120 W.
We then use this power gotten to find the average power output.
Power(avg) = P/t
Power(avg) = 120 / 2
Power(avg) = 60 Watts.
Therefore, the average power output is found to be 60 Watts
Alexandra is dragging her 37.8 kg golden retriever cross the wooden floor by applying a horizontal force. What force must be applied to move a dog with a constant speed of 1 m/s? The coefficient of kinetic friction between the dog in the floor is one point
Answer:
F = 385.56 N
Explanation:
Given that,
Mass, m = 37.8 kg
He applies a horizontal force and cross the wooden floor.
We need to find the force that must be applied to move a dog with a constant speed of 1 m/s.
The coefficient of kinetic friction between the dog in the floor is 1.02.
The net force acting on it is given by :
F = f- μmg
f is force due to constant speed, f = 0 (since, a = 0)
F = μmg
= 1.02 × 37.8 × 10
= 385.56 N
Hence, the required force is 385.56 N.
When calculating speed what goes into the calculator first
A. Time
B. Distance
C. Speed
Answer:
B. Distance
Explanation:
When calculating speed, the value of the given distance is first entered on the calculator because it is in the numerator.
Speed is the rate of change of distance with time;
Speed = [tex]\frac{distance}{time}[/tex]
The value of the distance is inputted first before that of the time is entered.
This way the division sign evaluates for the speed.
A boy on a bicycle approaches a brick wall as he sounds his horn at a frequency 400 Hz. The sound he hears reflected back from the wall is at a frequency 408 Hz. At what speed is the boy riding his bicycle toward the wall
Answer:
6.8 m/s
Explanation:
To solve this, we would use the Doppler's Effect. Doppler's effect is represented by the formula
f = [(g + vr)/(g + vs)].fo, where
f = observed frequency, 408 Hz
g = speed of sound in air, 340 m/s
vr = velocity of the receiver
vs = velocity of the source, 0 m/s
fo = source frequency, 400 Hz
Now, applying the values to the equation, we have
408 = [(340 + vr)/(340 + 0) * 400
408/400 = (340 + vr)/340
1.02 * 340 = 340 + vr
346.8 = 340 + vr
vr = 346.8 - 340
vr = 6.8 m/s
Therefore the speed at which the boy is riding his bicycle towards the wall is 6.8 m/s
As the fragment falls through the atmosphere it is heated and some of the material is vaporized. Explain how you could determine the composition of this hot vaporized material from the light it emits.
Answer:
by determining the wavelength of the light emitted
Explanation:
The composition of the hot vaporized materials of an asteroid can be find out by the wavelength of the light emitted when the fragment material is burnt out.
Some of the asteroids that are very large do not burn completely in the atmosphere and they fall on the surface of the earth. Thus there is a possibility of radioactivity. An atom is radioactive or unstable when the forces that constitutes the nucleus are not balanced and then the nucleus have excess of energy. Thus these unstable and highly energetic nucleus will try to attain stability by ejecting protons and neutrons, or by releasing gamma rays, beta particles or positrons, etc. Thus they will radiate energy and will cause radioactivity.
a 0.8 ,^3 insulated rigid tank contains 1.54 kg of carbon dioxide at 100 kPa. Now paddle wheel work is done on the system until pressure in the tank rises to 135 kpa assuming the ideal gas model and negligible kinetic and potential energy effects determine the paddle wheel work dine during the process and the energychange during this process
Answer:
The answer is "[tex]W= 100.44 \ KJ\ \ \ \W_{win}=7.23 \ KJ[/tex]"
Explanation:
Please find the complete question in the attached file.
From of the ideal gas relation that initial and the last temperatures were determined:
[tex]T_1 = \frac{P_1 V}{m R}[/tex]
[tex]= \frac{100 \times 0.8}{1.54 \times 0.1889} \\\\ = 275 \ K[/tex]
[tex]T_2 = \frac{P_2 V}{m R}[/tex]
[tex]= \frac{135 \times 0.8}{1.54 \times 0.1889} \\\\ = 317 \ K[/tex]
In the initial and final states, the internal energies for given temperatures are described from A-20 by means of intelmpolation and divided by the carlxon molar mass.
[tex]u_1 = 141.56 \frac{KJ}{kg}\\\\u_2 = 206.78 \frac{KJ}{kg}[/tex]
The real job is just the difference between internal energies:
[tex]W = m(u_2 - u_1) \\\\[/tex]
[tex]= 1.54(206.78 -141.56) \ kJ \\\\ =100.44 \ kJ[/tex]
In the initial and final states, the zero entries are as determined as internal energies:
[tex]S_1^{\circ} =4.788 \frac{KJ}{kg K}\\\\S_2^{\circ} =5.0478 \frac{KJ}{kg K}[/tex]
From its energy increase, the minimum work required is determined:
[tex]W_{min} = m(u_2-u_1 - T_0(s_2 -S_1))\\\\=W-mT_0(S_2^{\circ}- S_1^{\circ} -R \In \frac{P_2}{P_1})\\\\= 100.44kJ -1.54 \times 298( 5.0478-4.788-0.1889 \In \frac{135}{100})\\\\=7.23\ KJ\\[/tex]
Through which material would you expect sound waves to move fastest? O A. Iron O B. Water O c. Air O D. Milk
Answer:
Solids
Is this what your looking for, It might tell you the answer?
which button allow you to move the text enter to a higher heading level
Answer:
The entry level is the basic part of any text or power point which needs to be pushed to the higher levels for reading the entire content
Three runners compete in a 400m race on an oval-shaped running track. At a certain time, one runner is 15m away from the finish line. Which term describes the finish line?
Answer:
Reference point.
Explanation:
Motion can be defined as a change in location (position) with respect to a reference point.
This ultimately implies that, motion would occur as a result of a change in location (position) of an object with respect to a reference point or frame of reference i.e where it was standing before the effect of an external force.
In this scenario, three runners compete in a 400m race on an oval-shaped running track. At a certain time, one runner is 15m away from the finish line. Thus, the term which best describes the finish line is a reference point.
A reference point can be defined as a fixed point that typically marks the position from which a body starts its motion or change its location.
WILL GIVE BRAINLESS IF U ANSER ALL CORECTLY
1 What does wind do as it goes up and over a mountain range?
2 What is the climate on the windward side of a mountain range like?
3 What is the climate on the leeward side of a mountain like?
4 What are rain shadow deserts?
5 What are the characteristics (wind, precipitation, vegetation) seen on the windward side of the Sierra Nevada Mountains?
6 What are the characteristics (wind, precipitation, vegetation) seen on the leeward side of the Sierra Nevada Mountains?
Answer:
1-As winds rise up the windward side of a mountain range, the air cools and precipitation falls.
2-Mountains and mountain ranges can cast a rain shadow. As winds rise up the windward side of a mountain range, the air cools and precipitation falls.
3-Mountains and mountain ranges can cast a rain shadow. As winds rise up the windward side of a mountain range, the air cools and precipitation falls. On the other side of the range, the leeward side, the air is dry, and it sinks.
4-Rain shadow deserts are formed because tall mountain ranges prevent moisture-rich clouds from reaching areas on the lee, or protected side, of the range.
5-Mountains and mountain ranges can cast a rain shadow. As winds rise up the windward side of a mountain range, the air cools and precipitation falls. On the other side of the range, the leeward side, the air is dry, and it sinks. So there is very little precipitation on the leeward side of a mountain range.
6-Mountains and mountain ranges can cast a rain shadow. As winds rise up the windward side of a mountain range, the air cools and precipitation falls. On the other side of the range, the leeward side, the air is dry, and it sinks. So there is very little precipitation on the leeward side of a mountain range.
Explanation:
#6 and 5 are the same
What is the force required to stretch a spring whose constant value is 35 N/m by an amount of 0.25 m?
Answer:
or whatever then look it up it
Explanation:
NASA's Ames Research Center has a large centrifuge used for astronaut training. The centrifuge consists of a 3880-kg, 18.0-m-long tubular structure, which rotates about its center. Find the centrifuge's rotational inertia when two 105-kg seats are mounted at either end of the tube, 7.92 m from the rotation axis, and both are occupied by 72.6-kg astronauts. Treat the tube as a thin rod and the astronauts and seats as point masses.
Answer:
I = 1.27×105 kg⋅m2
Explanation:
How can magnets cause objects to have kinetic energy?
Answer:
If there is a system of magnets being held in place, there is potential energy. When you let go the potential energy makes to kinetic energy and the magnets move.
Magnets cause objects to have kinetic energy as a result of the magnetic
force present in it.
Magnets is a material which has a strong metallic field which attracts
ferromagnetic substances such as steel, iron etc.
When the magnets are put near these substances , the magnets attract them and causes them to move towards its direction . The magnets causes the conversion of the potential energy into kinetic energy through the strong metallic field.
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A clock is designed that uses a mass on the end of a spring as a timing mechanism. If the oscillation time needed is exactly one second, what is the spring constant required if the mass is 1.3 kg
Answer:
The value is [tex]k = 51.34 \ N/ m[/tex]
Explanation:
From the question we are told that
The mass is [tex]m = 1.3 \ kg[/tex]
The needed oscillation time is [tex]T = 1 \ s[/tex]
Generally the spring constant is mathematically represented as
[tex]k = \frac{4 \pi^2 * m }{ T^2}[/tex]
=> [tex]k = \frac{4* 3.142^2 * 1.3 }{ 1^2}[/tex]
=> [tex]k = 51.34 \ N/ m[/tex]
Does anyone know the answer to this page and the next one that it has?
Use of lubricants increase efficency of machine
Answer:
Lubrication reduces friction and allows moving machine parts to slide smoothly past each other. Selecting the appropriate lubrication solution can help reduce premature bearing failures and increase machine uptime, productivity and energy efficiency.
Explanation:
When the plutonium bomb was tested in New Mexico in 1945, approximately 1 gram of matter was converted into energy. Suppose another bomb is tested, and 1.5 grams of matter are converted into energy. How many joules of energy are released by the explosion?
Answer:
The value is [tex]E = 1.35 *10^{14} \ J[/tex]
Explanation:
From the question we are told that
The mass of matter converted to energy on first test is [tex]m = 1 \ g = 0.001 \ kg[/tex]
The mass of matter converted to energy on second test [tex]m_1 = 1.5 \ g = 1.5 *10^{-3} \ kg[/tex]
Generally the amount of energy that was released by the explosion is mathematically represented as
[tex]E = m * c^2[/tex]
=> [tex]E = 1.5 *10^{-3} * [ 3.0 *10^{8}]^2[/tex]
=> [tex]E = 1.35 *10^{14} \ J[/tex]
What are some negative impacts of land pollution on humans?
Answer:
Land pollution touches essentially every area of the living world, including: Water that isn't safe to drink. Polluted soil, which leads to a loss of fertile land for agriculture. Climate change, which causes an onslaught of disastrous problems, including flash floods and irregular rainfalls.
Explanation:
HOPE THIS HELPS!!!!!!
Answer:
Look at any ecosystem and there could be multiple forms of contamination—streams full of toxic chemicals from industrial processes, rivers overloaded with nutrients from farms, trash blowing away from landfills, city skies covered in smog. Even landscapes that appear pristine can experience the effects of pollution sources located hundreds or thousands of miles away.
Pollution may muddy landscapes, poison soils and waterways, or kill plants and animals. Humans are also regularly harmed by pollution. Long-term exposure to air pollution, for example, can lead to chronic respiratory disease, lung cancer and other diseases. Toxic chemicals that accumulate in top predators can make some species unsafe to eat. More than one billion people lack access to clean water and 2.4 billion don’t have adequate sanitation, putting them at risk of contracting deadly diseases.
Air pollution brings to mind visions of smokestacks billowing black clouds into the sky, but this pollution comes in many forms. The burning of fossil fuels, in both energy plants and vehicles, releases massive amounts of carbon dioxide into the atmosphere, causing climate change. Industrial processes also emit particulate matter, such as sulfur dioxide, carbon monoxide and other noxious gases. Indoor areas can become polluted by emissions from smoking and cooking. Some of these chemicals, when released into the air, contribute to smog and acid rain. Short term exposure to air pollution can irritate the eyes, nose and throat and cause upper respiratory infections, headaches, nausea and allergic reactions. Long-term exposures can lead to chronic respiratory disease, lung cancer, and heart disease. Long-term exposures also can lead to significant climatic changes that can have far reaching negative impacts on food, water and ecosystems. #SAVE OUR WORLD
I would appreciate some help for this worksheet question
Answer:
a) [tex]v(2.5\,s) = 40\,\frac{m}{s}[/tex], b) [tex]v(3.5\,s) = -60\,\frac{m}{s}[/tex], c) [tex]v(4.5\,m) = 0\,\frac{m}{s}[/tex], d) [tex]v(7.5\,s) = 60\,\frac{m}{s}[/tex]
Explanation:
Mathematically speaking, the instantaneous velocity is the slope of the curve, which can be estimated by means of the definition of secant line, equivalent to the definition of average velocity:
[tex]v(t) = \frac{x(t+0.5\,s)-x(x-0.5\,s)}{(t+0.5\,s)-(t-0.5\,s)}[/tex] (1)
Where:
[tex]x(t+0.5\,s)[/tex], [tex]x(t-0.5\,s)[/tex] - Position of the particle at [tex]t +0.5\,s[/tex] and [tex]t-0.5\,s[/tex], measured in meters.
[tex]t[/tex] - Time, measured in seconds.
Now we proceed to calculate each instantaneous velocity:
a) [tex]t = 2.5\,s[/tex]
[tex]v(2.5\,s) = \frac{x(3\,s)-x(2\,s)}{3\,s-2\,s}[/tex]
[tex]v(2.5\,s) = \frac{120\,m-80\,m}{1\,s}[/tex]
[tex]v(2.5\,s) = 40\,\frac{m}{s}[/tex]
b) [tex]t = 3.5\,s[/tex]
[tex]v(3.5\,s) = \frac{x(4\,s)-x(3\,s)}{4\,s-3\,s}[/tex]
[tex]v(3.5\,s) = \frac{60\,m-120\,m}{1\,s}[/tex]
[tex]v(3.5\,s) = -60\,\frac{m}{s}[/tex]
c) [tex]t = 4.5\,s[/tex]
[tex]v(4.5\,s) = \frac{x(5\,s)-x(4\,s)}{5\,s-4\,s}[/tex]
[tex]v(4.5\,s) = \frac{60\,m - 60\,m}{1\,s}[/tex]
[tex]v(4.5\,m) = 0\,\frac{m}{s}[/tex]
d) [tex]t = 7.5\,s[/tex]
[tex]v(7.5\,s) = \frac{x(8\,s)-x(7\,s)}{8\,s-7\,s}[/tex]
[tex]v(7.5\,s) = \frac{0\,m - (-60\,m)}{1\,s}[/tex]
[tex]v(7.5\,s) = 60\,\frac{m}{s}[/tex]