Answer:I think it’s because both can provide heat and cold
Explanation:
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.
Does anyone know the answer to this page and the next one that it has?
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.
Which type of circuit would be best to use for lights used for decorations? Question 1 options: Series circuit. One bulb could go out and the strand will stay on. Series circuit. One bulb could go out and the rest go out. Parallel circuit. One bulb goes out and the rest go out. Parallel circuit. One bulb could go out and the strand will stay on.
Answer:
One bulb could go out and the strand will stay on.
Explanation:
In series circuit, there is only one path provided for the current to flow. So, all the lights are required to be in working condition, for the others to work. And if anyone light bulb goes out, the circuit will become incomplete and the rest of the strand will also go out. Because there is only one path for current flow which is now broken.
On the other hand, in parallel circuits, each light bulb has a separate connection with the source. Current path to each bulb is independent of the others. Therefore, if one bulb goes out, the rest of the strand will stay on.
So, the correct option is:
One bulb could go out and the strand will stay on.
A. 230 N
B. 194 N
C. 250 N
D. 340 N
Answer:
I think C 250 it is correct
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.
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]
Describe the energy transfers and transformations for a javelin, starting from the point at which an athlete picks up the javelin and ending when the javelin is stuck into the ground after being thrown.
Explanation:
When the Javelin is at rest on the ground, Potential Energy and Kinetic Energy are zero. immediately the Athlete picks the javelin up from the ground, there is an increase in the Kinetic Energy this increase continues until the javelin comes to a halt. Potential Energy also increases.
As the Athlete throws javelin, there is a decrease in the Potential Energy, the Kinetic Energy increases simultaneously until the javelin hits the ground.
after which Potential Energy and Kinetic Energy becomes zero.
The moment of inertial of the hoop yo-yo when it is: (a) rotating about its center of mass, and (b) rotating about the point where the tension force is applied
mass=332g
diamater=35.9cm
thickness.95cm
Answer:
a) The moment of inertia of the hoop yo-yo rotating about its center of mass is [tex]I_{g} = 0.0108\,kg\cdot m^{2}[/tex].
b) The moment of inertial of the hoop yo-yo rotating about its center of mass is [tex]I_{O} = 0.0216\,kg\cdot m^{2}[/tex].
Explanation:
a) The hoop yo-yo can be modelled as a tours with a minor radius [tex]a[/tex], related with the thickness, and with a major radius [tex]b[/tex], related with the diameter, and with an uniform mass. The momentum of inertia about its center of mass ([tex]I_{g}[/tex]), measured in kilogram-square meters, which is located at the geometrical center of the element, is determined by the following formula:
[tex]I_{g} = \frac{1}{4}\cdot m \cdot (4\cdot b^{2}+3\cdot a^{2})[/tex] (1)
[tex]b = 0.5\cdot D[/tex] (2)
[tex]a = 0.5\cdot t[/tex] (3)
Where:
[tex]D[/tex] - Diameter, measured in meters.
[tex]t[/tex] - Thickness, measured in meters.
[tex]m[/tex] - Mass, measured in kilograms.
If we know that [tex]m = 0.332\,kg[/tex], [tex]D = 0.359\,m[/tex] and [tex]t = 9.5\times 10^{-3}\,m[/tex], then the moment of inertia of the hoop yo-yo is:
[tex]a = 0.5\cdot (9.5\times 10^{-3}\,m)[/tex]
[tex]a = 4.75\times 10^{-3}\,m[/tex]
[tex]b = 0.5\cdot (0.359\,m)[/tex]
[tex]b = 0.180\,m[/tex]
[tex]I_{g} = \frac{1}{4}\cdot (0.332\,kg)\cdot [4\cdot (0.180\,m)^{2}+3\cdot (4.75\times 10^{-3}\,m)^{2}][/tex]
[tex]I_{g} = 0.0108\,kg\cdot m^{2}[/tex]
The moment of inertia of the hoop yo-yo rotating about its center of mass is [tex]I_{g} = 0.0108\,kg\cdot m^{2}[/tex].
b) The hoop yo-yo rotate at a point located at a distance of half diameter from the center of mass of the element, whose moment of inertia is determined by the Theorem of Parallel Axes:
[tex]I_{O} = I_{g} +m\cdot r^{2}[/tex] (4)
Where:
[tex]r[/tex] - Distance between parallel axes, measured in meters.
If we know that [tex]I_{g} = 0.0108\,kg\cdot m^{2}[/tex], [tex]m = 0.332\,kg[/tex] and [tex]r = 0.180\,m[/tex], then the moment of inertial of the hoop yo-yo rotating about the point where the tension force is applied is:
[tex]I_{O} = 0.0108\,kg\cdot m^{2}+(0.332\,kg)\cdot (0.180\,m)^{2}[/tex]
[tex]I_{O} = 0.0216\,kg\cdot m^{2}[/tex]
The moment of inertial of the hoop yo-yo rotating about its center of mass is [tex]I_{O} = 0.0216\,kg\cdot m^{2}[/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.
A food processor draws 8.47 A of current when connected to a potential difference of 110 V.
How much electrical energy is consumed by this food processor monthly (30 days) if it is used on average of 10.0 min every day?
Answer:
27.95[kW*min]
Explanation:
We must remember that the power can be determined by the product of the current by the voltage.
[tex]P=V*I[/tex]
where:
P = power [W]
V = voltage [volt]
I = amperage [Amp]
Now replacing:
[tex]P=110*8.47\\P=931.7[W][/tex]
Now the energy consumed can be obtained mediate the multiplication of the power by the amount of time in operation, we must obtain an amount in Kw per hour [kW-min]
[tex]Energy = 931.7[kW]*30[days]*10[\frac{min}{1day} ]=279510[W*min]or 27.95[kW*min][/tex]
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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A ball of clay is thrown at a wall at a velocity of 28.0 m/s. After it hits the wall, it
takes 0.020 s to stop. What is the acceleration of the clay when it hits the wall?
Answer:
Explanation:
1400 m/s2
The acceleration of the ball when it hits the vertical wall is -1,400 m/s²
The given expression:
velocity of the ball, v = 28 m/s
time of motion of the ball, t = 0.02 s
To find:
the acceleration of the ball when it hits the vertical wall.The acceleration of the ball is calculated as follows:
[tex]acceleration = \frac{\Delta velocity}{\Delta time} = \frac{Final \ velocity\ - \ initial \ velocity }{time} \\\\final \ velocity \ of \ the \ ball \ when \ it \ hits \ the \ wall = 0\\\\acceleration = \frac{0 - 28}{0.02} \\\\acceleration = - \ 1,400 \ m/s^2[/tex]
Thus, the acceleration of the ball when it hits the vertical wall is -1,400 m/s²
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Car A travelled 300k in 5 hours, what was its average speed?
Speed =distance/time
Speed=300/5=60 km/h
mark brainliest please
A block of wood of length L = 21.0 cm, width w = 9.53 cm, and height h = 5.92 cm is just barely immersed in water by placing a mass m on the top of the block. The density of the wood is rho = 0.390 g/cm^3. The value of m is:______
a. 0.72 kg
b. 1.2 kg
c. 1.6 kg
d. 7.1 kg
e. 0.36 kg
Answer:
0.462kg
Explanation:
Density = Mass/Volume
Given
Density of the wood = 0.390 g/cm^3
Volume of the wood = Length * Width * Height
Volume = 21 * 9.53 * 5.92
Volume = 1,184.7696cm³
Get the mass m;
mass = density * volume
mass = 0.390 * 1,184.7696
mass = 462.060144g
The mass in kg is 0.462kg
What is black body radiation? Explain in detail.
An object that absorbs all radiation falling on it, at all wavelengths, is called a black body. When a black body is at a uniform temperature, its emission has a characteristic frequency distribution that depends on the temperature. Its emission is called black-body radiation
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An object that floats in water weighs 20 N in air.
a. What is the weight of the object in water?
b.What is the Upthrust acting on the object in water?
c. What is the weight of the water displaced by the object?
Answer:
a. Weight of Object in Water = 20 N
b. Up thrust = 20 N
c. Weight of Water Displaced = 20 N
Explanation:
a.
The weight of the object remains same in the water as well. Because, the same force of gravity is acting there as well. Hence,
Weight of Object in Water = 20 N
b.
Since, the object floats on the water. Therefore, according to Archimedes' principle the up thrust force acting on the object must be equal to the weight of object:
Up thrust = Weight of object
Up thrust = 20 N
c.
From Archimedes' Principle, we know that the up thrust or the Buoyant force is equal to the weight of the water displaced by the object. therefore:
Weight of Water Displaced = Up thrust
Weight of Water Displaced = 20 N
A sound wave enters a new medium where sound travels faster. How does this affect the frequency and wavelength of the sound?
А. The frequency increases and the wavelength decreases.
B. The frequency decreases and the wavelength increases.
C. The frequency stays the same and the wavelength increases.
D.The frequency stays the same and the wavelength decreases.
E. Neither the frequency nor the wavelength is affected.
Answer:
The frequency stays the same and the wavelength decreases.
Explanation:
When sound wave enters a new medium where sound travels faster, its frequency will remain same because it depends only on the source.
The relation between wavelength and speed is inverse, it means when the speed of sound increases, its wavelength will decrease.
So, the frequency stays the same and the wavelength decreases. Hence, the correct option is (d).
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:
write the characteristic of aluminum
Answer:
see below
Explanation:
alluminium is,
1.lustrous
2.highly malleable
3.reactive metal
4.some times acts as non metal
5.good conductor
6.obtained from bauxite
7.Atomic number 13 so electrons and protons are 13
8.neurons are 14
9.ductile
10.melting point is very high approximately 660 degree celsius.
Answer:
It has low density, is non-toxic, has a high thermal conductivity, has excellent corrosion resistance and can be easily cast, machined and formed.
Explanation:
An arrow is shot straight up in the air at an initial speed of 15 m/s. After how much time is the arrow heading downward at a speed of 8 m/s?
Answer:
2.35 seconds
Explanation:
Remark
This is a question where direction matters. Let us call down + and up minus. It won't matter. The answer will be the same.
Formula
a = (vf - vi)/t
Givens
a = 9.8 m/s^2
vi = - 15 m/s
vf = 8m/s
Solution
9.8 = (8 - - 15)/t Multiply both sides by t
t * 9.8 = 23 Divide by 9.8
t = 23/9.8
t = 2.35 s
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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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:
A golfer shoots a ball horizontally off a cliff with a speed of 80.0 m/s, and it lands 471m from the base of the cliff.
1. How long is the golf ball in the air?
2. What is the height of the cliff?
3. What is the golf ball/s speed when it lands?
Answer:
1. t = 5.89 s
2. h = 170 m
3. Vf = 57.8 m/s
Explanation:
1.
First, we analyze the horizontal motion of the golf ball. Assuming the air friction to be negligible, the horizontal motion will be uniform. So, e can use the following equation:
[tex]s = vt[/tex]
where,
s = horizontal distance covered by the golf ball = 471 m
v = horizontal speed of golf ball = 80 m/s
t = time taken by the golf ball in air = ?
Therefore,
[tex]471\ m = (80\ m/s)t\\\\t = \frac{471\ m}{80\ m/s}\\\\[/tex]
t = 5.89 s
2.
Now, we analyze the vertical motion. Using 2nd equation of motion:
[tex]h = v_{i}t + \frac{1}{2}gt^2[/tex]
where,
h = height of cliff = ?
vi = vertical component of initial speed of ball = 0 m/s(ball was shot horizontally)
g = acceleration due to gravity = 9.81 m/s²
t = time of flight = 5.89 s
Therefore,
[tex]h = (0\ m/s)(5.89\ s) + \frac{1}{2}(9.81\ m/s^2)(5.89\ s)^2[/tex]
h = 170 m
3.
Now, we can use 1st equation of motion:
[tex]v_{f} = v_{i} + gt\\v_{f} = 0 m/s + (9.81\ m/s^2)(5.89\ s)\\[/tex]
Vf = 57.8 m/s
Young's double slit experiment is one of the quintessential experiments in physics. The availability of low cost lasers in recent years allows us to perform the double slit experiment rather easily in class. Your professor shines a green laser (568 nm) on a double slit with a separation of 0.107 mm. The diffraction pattern shines on the classroom wall 3.5 m away. Calculate the fringe separation between the third order and central fringe.
Answer:
18.58 mm
Explanation:
Formula for the fringe width is;
β = λL/d
We are given;
λ = 568 nm = 568 × 10^(-9) m
L = 3.5 m
d = 0.107 mm = 0.107 × 10^(-3) mm
β = (568 × 10^(-9) × 3.5)/(0.107 × 10^(-3))
β = 0.01858 m = 18.58 mm
a 6.5 kg box is pulled across a floor by a horizontal 12.0 N tension force. If there is a 5 N friction exerted on the box, what is its acceleration
Explanation:
Fnet = 12.0N - 5.0N = 7.0N
a = Fnet/m = 7.0N/6.5kg = 1.07m/s².
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]
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
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].
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.