Answer:
Positive zero error: When the jaws of the caliper are closed, the zero mark on the vernier scale is to the right of the zero mark on the main scale. This means that the caliper reads a value greater than the actual value, leading to a positive zero error.
Negative zero error: When the jaws of the caliper are closed, the zero mark on the vernier scale is to the left of the zero mark on the main scale. This means that the caliper reads a value less than the actual value, leading to a negative zero error.
Nuclear fusion in a star produces elements up to, but no heavier than, ________.
A. iron
B. lead
C. carbon
D. nitrogen
Answer:
A. iron.
In the process of nuclear fusion, lighter elements are fused together to form heavier elements. This process releases energy and is what powers the star. However, the fusion of elements heavier than iron requires energy, rather than releasing it. Therefore, once a star has produced iron in its core, it is no longer able to sustain nuclear fusion and will eventually undergo a supernova explosion.
A charge of −0.0004 C is a distance of 3 meters from a charge of 0.0003 C. What is the magnitude of the force between them?
The magnitude of the force is 3.6 x 10^-6 N, rounded to two significant figures.
The magnitude of the force between two point charges can be calculated using Coulomb's law:
F = k * (q1 * q2) /[tex]r^2[/tex]
where F is the magnitude of the force in Newtons (N), k is Coulomb's constant (9 x 10^9 N*m^2/C^2), q1 and q2 are the magnitudes of the charges in Coulombs (C), and r is the distance between the charges in meters (m).
Plugging in the given values, we get:
F =[tex](9 * 10^{9} N*m^{2}/C^2) * (-0.0004\ C) * (0.0003 C) / (3 m)^{2}[/tex]
Simplifying the expression, we get:
F = [tex]-3.6 * 10^{-6} N[/tex]
Note that the negative sign in the result indicates that the force is attractive. The magnitude of the force is [tex]3.6 * 10^{-6} N[/tex], rounded to two significant figures.
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Macmillan Learning
When a massive star reaches the end of its life, it is possible for a supernova to occur. This may result in the formation of a very
small, but very dense, neutron star, the density of which is about the same as a neutron. A neutron has a mass of 1.7 x 10-27 kg
and an approximate radius of 1.2 x 10-15 m. The mass of the sun is 2.0 x 1030 kg.
Okay, let's break this down step-by-step:
1) A neutron has a mass of 1.7 x 10-27 kg and an approximate radius of 1.2 x 10-15 m.
So we know the mass and radius of a single neutron.
2) The mass of the sun is 2.0 x 1030 kg.
So we know the total mass of the sun, which is much greater than a neutron.
3) When a massive star reaches the end of its life, it can explode as a supernova.
This supernova can form a neutron star.
4) A neutron star has a density about the same as a neutron.
So we can conclude that a neutron star has a density of:
Density = Mass / Volume
= (1.7 x 10-27 kg) / (4/3 * pi * (1.2 x 10-15 m)3)
= 1.6 x 1017 kg/m3
5) A neutron star forms from the core collapse of a massive star during supernova.
So it has a mass on the order of 1-2 times that of the sun (2 x 1030 kg),
but compressed into a sphere only about 10-20 km in radius.
So its mass would be huge, around 2 x 1030 kg, but confined to a tiny volume,
giving it an immense density, around 1.6 x 1017 kg/m3, the same as a neutron.
Does this help explain the concepts and walk through the calculations? Let me know if you have any other questions!
A car travels at a speed of 30m/s when it leaves a ramp set at an angle fo 37 degrees from the ground. How high off the ground will the car reach? What is the maximum height of the gorge could the car clear?
We can use the laws of physics to solve this problem. The key here is to recognize that the initial kinetic energy of the car (due to its speed) is converted into potential energy as it goes up the ramp, and then back into kinetic energy as it falls back down.
Using conservation of energy, we can set the initial kinetic energy equal to the final potential energy at the maximum height:
(1/2) * m * v^2 = m * g * h
where m is the mass of the car, v is its initial speed (30 m/s), g is the acceleration due to gravity (9.81 m/s^2), and h is the maximum height.
Simplifying the equation and solving for h, we get:
h = (v^2 * sin^2(theta)) / (2 * g)
where theta is the angle of the ramp (37 degrees in this case).
Plugging in the known values, we get:
h = (30^2 * sin^2(37)) / (2 * 9.81) ≈ 79.1 meters
So the car will reach a height of approximately 79.1 meters off the ground.
To find the maximum height of the gorge that the car could clear, we need to look at the horizontal distance that the car travels while in the air. This distance is given by:
d = v * t
where t is the time that the car spends in the air. We can find t by setting the initial potential energy (at the maximum height) equal to the final kinetic energy (at the end of the jump):
m * g * h = (1/2) * m * (v_f^2)
where v_f is the final velocity of the car at the end of the jump (when it lands back on the ground). We can solve for v_f using:
v_f = sqrt(2gh)
where h is the maximum height (found earlier). Plugging this into the conservation of energy equation, we get:
t = sqrt(2h/g)
Now we can plug in the known values for v and theta to get:
d = 30 * sqrt(2h/g) * cos(theta)
Plugging in the values for h and theta, we get:
d = 30 * sqrt(2*79.1/9.81) * cos(37) ≈ 187.2 meters
So the maximum height of the gorge that the car could clear is approximately 187.2 meters.
1. A rich merchant finds himself stranded in the middle of a frozen lake. Don't ask how he got there. Think Twilight Zone... The surface
is perfectly frictionless. All he has is his clothing and a large bag of gold coins. How can he save himself? Write down your answer.
2. You are hammering a nail into a hard piece of wood. You are using one of your little brothers light, toy hammers and getting nowhere fast. Finally, you grab a hammer with a heavier head and your task goes much easier. Which one of Newtons laws did you finally remember? explain.
The rich merchant can save himself by throwing some gold coins in one direction.
You finally remembered Newton's second law of motion, which states that the acceleration of an object is directly proportional to the net force applied to it and inversely proportional to its mass.
How to understand Newton's laws?Newton's third rule of motion states that for every action, there is an equal and opposite reaction. When he tosses the gold coins in one direction, the flung coins cause a rebound force in the opposing direction. This recoil force will provide him with a little amount of velocity, which he may then employ to go in the opposite direction he tossed the coins. He can slowly make his way to the coast by continuing this method.
It is because the heavier hammer has more mass than the toy hammer, when the same force is applied to both hammers, the heavier hammer suffers less acceleration than the lighter hammer. This implies that the heavier hammer applies more power to the nail, making it easier to drive into the hard wood.
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if the battery is 4.0V, the voltmeter reading across R is 2.0V and the resistance per unit length of wire AX is 2 ohm per metre, calculate the current in the circuit when /AP/ is 40.0cm. (neglect the internal resistance of the battery)
The current in the circuit when AP is 40.0 cm is 2.5 A.
We can use Ohm's law to calculate the current in the circuit:
V = IR
where V is the voltage, I is the current, and R is the resistance.
The resistance of the wire AX can be calculated as:
R = ρ ([tex]\frac{L}{A}[/tex])
where ρ is the resistivity of the wire material, L is the length of the wire, and A is the cross-sectional area of the wire.
Since the resistance per unit length of wire AX is given as 2 [tex]\frac{ohm}{m}[/tex] ,we can calculate the resistance of a 40 cm length of wire AX as:
R = (2 [tex]\frac{ohm}{m}[/tex]) × ([tex]\frac{40 cm}{100 \frac{cm}{m}}[/tex]) = 0.8 ohm
Using the voltmeter reading across R as 2.0V and the battery voltage as 4.0V, we can calculate the voltage across the wire AX as:
V = 4.0V - 2.0V = 2.0V
Therefore, we can calculate the current in the circuit as:
I = [tex]\frac{V}{R}[/tex] = [tex]\frac{2.0V }{0.8 ohm}[/tex] = 2.5 A
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A magnifying glass has a converging lens of focal length of 10.3 cm. At what distance from a nickel should you hold this lens to get an image with a magnification of +1.53?
The magnifying glass should be held 16.4 cm away from the nickel to obtain an image with a magnification of +1.53.
Using the thin lens equation, 1/f = 1/o + 1/i, where f is the focal length, o is the object distance, and i is the image distance, and the magnification equation, M = -i/o, where M is the magnification, we can solve for the object distance.
First, solve for the image distance i:
M = -i/o
1.53 = -i/o
i = -1.53o
Then, substitute i into the thin lens equation:
1/0.103 = 1/o + 1/(-1.53o)
Solving for o, we get:
o = 16.4 cm
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A thin uniform rod has a length of 0.480 m and is rotating in a circle on a frictionless table. The axis of rotation is perpendicular to the length of the rod at one end and is stationary. The rod has an angular velocity of 0.37 rad/s and a moment of inertia about the axis of 3.10×10−3 kg⋅m2. A bug initially standing on the rod at the axis of rotation decides to crawl out to the other end of the rod. When the bug has reached the end of the rod and sits there, its tangential speed is 0.132 m/s. The bug can be treated as a point mass.
(a) What is the mass of the rod?
(b) What is the mass of the bug?
(a) To solve for the mass of the rod, we can use the formula for rotational kinetic energy:
KE = (1/2) * I * w^2
where KE is the rotational kinetic energy, I is the moment of inertia, and w is the angular velocity.
At the beginning, when the bug is at the axis of rotation, the rotational kinetic energy of the rod is:
KE1 = (1/2) * I * w^2 = (1/2) * (3.10×10−3 kg⋅m^2) * (0.37 rad/s)^2 = 0.036 J
When the bug reaches the other end of the rod, the rotational kinetic energy is:
KE2 = (1/2) * I * w^2 = (1/2) * (3.10×10−3 kg⋅m^2) * (0.132 m/s)^2 / (0.480 m)^2 = 3.62×10^-5 J
The change in kinetic energy is due to the work done by the bug as it crawls along the rod. The work done by the bug can be calculated as the product of the force it exerts on the rod and the distance it crawls:
W = F * d
The force the bug exerts on the rod can be calculated using Newton's second law for rotational motion:
τ = I * α
where τ is the torque, α is the angular acceleration, and I is the moment of inertia.
Since the rod is rotating with a constant angular velocity, its angular acceleration is zero, so the net torque on the rod must be zero. This means that the torque exerted by the bug on the rod must be equal and opposite to the torque due to the angular momentum of the rod:
τ_bug = τ_rod
F * d = I * w
F = I * w / d
Substituting the values given, we get:
F = (3.10×10−3 kg⋅m^2) * (0.37 rad/s) / (0.480 m) = 0.0237 N
Now we can use the work-energy principle to find the mass of the rod:
W = KE2 - KE1 = F * d = (1/2) * m * v^2
where m is the mass of the rod and v is the tangential velocity of the bug at the end of the rod.
Substituting the values given, we get:
(1/2) * m * (0.132 m/s)^2 = 3.62×10^-5 J - 0.036 J
Simplifying and solving for m, we get:
m = 0.221 kg
Therefore, the mass of the rod is 0.221 kg.
(b) To find the mass of the bug, we can use the same equation we used to calculate the force it exerted on the rod:
F = I * α / d
where α is the angular acceleration of the rod caused by the bug crawling along it.
Since the rod is a thin uniform rod, its moment of inertia can be calculated as:
I = (1/3) * m * L^2
where L is the length of the rod.
Substituting the values given, we get:
I = (1/3) * (0.221 kg) * (0.480 m)^2 = 0.0202 kg⋅m^2
Now we can calculate the angular acceleration caused by the bug crawling along the rod. Since the rod is rotating with a constant angular velocity, its angular acceleration is given by:
α
Fade
finish
α = (w_f^2 - w_i^2) / (2 * θ)
where w_i is the initial angular velocity of the rod, w_f is the final angular velocity of the rod after the bug has crawled to the end, and θ is the angle through which the bug crawls, which is equal to the length of the rod:
θ = L = 0.480 m
Substituting the values given, we get:
α = (0.132 m/s)^2 / (2 * 0.480 m) = 0.072 rad/s^2
Now we can calculate the force exerted by the bug on the rod:
F = I * α / d = (0.0202 kg⋅m^2) * (0.072 rad/s^2) / (0.480 m) = 0.00303 N
Finally, we can use Newton's second law to find the mass of the bug:
F = m * a
where a is the tangential acceleration of the bug, given by:
a = r * α
where r is the distance from the bug to the axis of rotation, which is equal to the length of the rod minus the distance the bug has crawled, or:
r = L - d = 0.480 m - 0.480 m = 0 m
Therefore, the tangential acceleration of the bug is zero, and its mass is:
m = F / a = 0.00303 N / 0 m/s^2 = undefined
This means that the force exerted by the bug on the rod is not enough to cause any tangential acceleration, and therefore the mass of the bug is negligible compared to the mass of the rod. We can assume that the bug has zero mass for practical purposes.
(a) Calculate the force (in N) the woman in the figure below exerts to do a push-up at constant speed, taking all data to be known to three digits. (You may need to use torque methods from a later chapter.) 401.15
(b)How much work (in J) does she do if her center of mass rises 0.260 m?
(c) What is her useful power output (in W) if she does 30 push-ups in 1 min? (Should work done lowering her body be included? See the discussion of useful work in Work, Energy, and Power in Humans.)
The force is 400.2 N
The work done is 120 J
The power is 48W
What is Force?Force is a physical concept that describes the influence that one object has on another object, causing it to accelerate or deform. Force can be defined as any influence that changes the motion of an object, such as a push or a pull.
How to solve:
under equilibrium condition
F * 1.45 m =68 kg * 9.81 m/s^2 *0.87 m
F =400.2 N
b)
work done = m*g*h =68 kg*9.81 m/s^2*0.180 m =120.0744 J =120 J
c)
power =120.0744 J *(24 /60 s) =48.02976 W = 48 W
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A 21 g block of ice is cooled to −77 ◦C. It is added to 593 g of water in an 92 g copper calorimeter at a temperature of 28◦C. Find the final temperature. The specific heat of copper is 387 J/kg · ◦C and of ice is 2090 J/kg · ◦C . The latent heat of fusion of water is 3.33 × 105 J/kg and its specific heat is 4186 J/kg · ◦C . Answer in units of ◦C. ( 2 significant digits pls)
Answer:m 1= 36 g=0.036 kg - the mass of ice T1= −77 ◦C. - temperature of ice m2=589 g=0.589 kg - mass of water C1=2090 J/kg · ◦C - specific heat of ice λ = 3.33 × 10^5 J/kg - latent heat of fusion of water T2=26◦C. - temperature of water C2= 4186 J/kg · ◦C . - - specific heat of water m3=74 g=0.074kg - mass of copper T3=26◦C. - temperature of copper C3=387 J/kg ·◦C - specific heat of copper is and of ice is T - ? - final temperature 1 1 ( 0 − 1 ) − ℎ 0 1 − 1 2 ( − 0 ) − ℎ 2 2 ( − 2 ) − 3 3 ( − 3 ) − m 1 C 1 (0−T 1 )−ice heating to 0 o C m1λ−ice melting m 1 C 2 (T−0)−melted water heating m 2 C 2 (T−T 2 )−water cooling m 3 C 3 (T−T 3 )−copper cooling 1 1 ( 0 − 1 ) + 1 + 1 2 ( − 0 ) + 2 2 ( − 2 ) + 3 3 ( − 3 ) = 0 m 1 C 1 (0−T 1 )+m1λ+m 1 C 2 (T−0)+m 2 C 2 (T−T 2 )+m 3 C 3 (T−T 3 )=0 0.036 ⋅ 2090 ⋅ 77 + 0.036 ⋅ 3.33 ⋅ 1 0 5 + 0.036 ⋅ 4186 ⋅ + 0.589 ⋅ 2090 ⋅ ( − 26 ) + 0.074 ⋅ 387 ⋅ ( − 26 ) = 0 0.036⋅2090⋅77+0.036⋅3.33⋅10 5 +0.036⋅4186⋅T+0.589⋅2090⋅(T−26)+0.074⋅387⋅(T−26)=0 1410.344 ⋅ = 14969.368 = 10.6 1 1410.344⋅T=14969.368 T=10.61 o C Answer: = 10.6 1 T=10.61 o C
Explanation:
The brass pipe pictured to the right has few of the trumpet’s attributes. There is no mouthpiece,
no bell, and no valves, but it is a brass tube that you could buzz your lips into. What is the lowest
note that would be possible to blow on this pipe?
10 cm
Answer: the lowest note that would be possible to blow on this brass pipe is approximately 1700 Hz
Explanation: f = c / (2L)
where c is the speed of sound in brass, which is around 340 m/s.
Changing over the length of the pipe to meters, we have:
L = 10 cm = 0.1 m
Stopping within the values, we get:
f = 340 / (2 x 0.1) = 1700 Hz
1. Which characteristic of a substance is constant?
a phase
b mass
Ospecific heat
d kinetic energy
The end of Hubbard Glacier in Alaska advances by an average of 105 feet per year. What is the speed of advance of the glacier in m/s? ( 1 ft = 0.305 m, 1 yr = 3.156 × 10's )
1.01 x 10-6 m/sec is the speed of advance of the glacier in m/s
1 foot = .3048 meters
105 ft = 105(.3048) = 32.004 m
(365.25 days)(24 hrs/day)(60 min/hr)(60 sec/min)
= 31,557,600 sec/yr
m/sec = 32.004/31557600 = 1.0141455624 x 10-6 m/sec
Round to 1.01 x 10-6 m/sec
Hubbard Glacier has altered through time, how?Hubbard Glacier in Alaska has been slowly thickening and moving towards Disenchantment Bay since observations began in 1895. The progress contrasts with several neighbouring thinning and receding glaciers in Alaska and elsewhere in the world.
Hubbard Glacier's base ice is around 400 years old since it takes that long for ice to travel the glacier's entire length. Frequently, icebergs the size of a ten-story structure are calved off by the glacier.
Sea levels are rising as a result of glaciers, such those that cover Greenland, melting as a result of climate change.
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A simple machine is able to move a 400 N load a distance of 20 cm when a force of 20 N is moved through a distance of 5.0 m. Calculate: a)the work input b)the work output c)the actual mechanical advantage
(a) The work input on the machine is 10 J.
(b) The work output of the machine is 80 J.
(b) The Machanical advantage of the machine is 20.
What is a machine?Machine is any device that enables work to be done easily.
(a) To calculate the work input of the machine, we use the formula below
Formula:
W' = Ed'........................... Equation 1Where:
W' = Work inputE = Effortd' = Distance moved by effortFrom the question,
Given:
E = 20 Nd' = 0.5 mSubstitute these values into equation 1
W' = 20×0.5W' = 10 J(b) Also, for work output, we use the formula below
W = Ld................. Equation 2Where:
W = Work outputL = Load = 400 Nd = 20 cm = 0.2 mSubstitute these values into equation 2
W = 400×0.2W = 80 J(c) To calculate the mechanical advantage of the machine, we use the formula below
M.A = L/E..................... Equation 3Given:
L = 400E = 20 NSubstitute into equation 3
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Which of the following was one of Hubble's conclusions due to red shift?
A. There are millions of galaxies in the universe, not just ours.
B. The universe is contracting.
C. Background radiation shows the Big Bang occurring.
D. The universe is expanding.
D. The universe is expanding. Hubble's observation of red shift in the light from distant galaxies led him to conclude that these galaxies were moving away from us and that the universe was expanding.
What was Hubble's observation?Hubble's most famous observation was his discovery of the relationship between the redshift of light from distant galaxies and their distance from Earth. He observed that the light from distant galaxies was shifted toward longer, redder wavelengths, which indicated that the galaxies were moving away from us. By analyzing the degree of redshift, Hubble was able to calculate the distance of these galaxies from Earth and found that they were much farther away than previously thought. This led him to conclude that the universe was expanding and laid the foundation for the Big Bang theory.
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11.1. Is ammeter A1 connected in series or in parallel in this diagram?
Ammeter is always connected in series with the circuit in which the current is to be measured.
An ammeter is connected, to gauge the current flowing through a part or circuit. Since current in a series connection stays constant and an ammeter's resistance is relatively low, the current being measured is unaffected. For the purpose of measuring current, an ammeter is linked in series.
A shunt running parallel to the metre carries the majority of the current at high current values, hence an ammeter can measure a wide range of current values. An ammeter is represented by a circle with a capital A inside it on circuit diagrams.
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NO USE OF AI BOTS TO answer question pls answer thanks
Answer:
19.2 centimeters or 0.192 meters. Both are the same.
Explanation:
T = 2π√(L/g)
T = 60.0 s / 14 ≈ 4.29 s
L = (gT²) / (4π²) ≈ 0.384 m
h + 0.384 m = height above the field
H = h + 0.192 m
H = h
h ≈ 0.192 m or 19.2 cm.
A runner covers a distance of 500m in 14.5minutes. Calculate the average speed
Answer:
The formula to calculate average speed is:
average speed = total distance / total time
In this case, the distance covered is 500 meters and the time taken is 14.5 minutes. However, we need to convert the time to seconds to get the answer in meters per second.
14.5 minutes = 14.5 x 60 seconds = 870 seconds
So, the average speed is:
average speed = total distance / total time
average speed = 500 meters / 870 seconds
average speed = 0.57 meters per second
Therefore, the average speed of the runner is 0.57 meters per second.
An RLC series circuit has a 1.00 kn resistor, a 155 mH inductor, and a 25.0 nF capacitor.
(a) Find the circuit's impedance (in ) at 485 Hz.
(b) Find the circuit's impedante (in 2) at 7.50 kHz.
(c) If the voltage source has V
rms
=
___mA (at 485 Hz)
___mA (at 7.50 kHz)
(d) What is the resonant frequency (in kHz) of the circuit?
___kHz
408 V, what is Irms (in mA) at each frequency?
(e) What is Irms (in mA) at resonance?
___mA
(a) The impedance of the circuit at 485 Hz is approximately 1253.53 Ω.
(b) The impedance of the circuit at 7.50 kHz is approximately 1256.04 Ω.
(c) The current (Irms) at 485 Hz is approximately 326 mA, and the current at 7.50 kHz is approximately 325 mA.
(d) The resonant frequency of the circuit is approximately 25.74 kHz.
(e) The current (Irms) at resonance is approximately 408 mA.
What is the impedance of the circuit?a) The impedance (Z) of an RLC series circuit can be calculated using the formula:
Z = √(R^2 + (ωL - 1/ωC)^2)
where;
R is the resistance,L is the inductance, C is the capacitance, and ω is the angular frequency in radians per second.Given:
R = 1.00 kΩ = 1000 Ω
L = 155 mH = 0.155 H
C = 25.0 nF = 25.0 × 10^(-9) F
f = 485 Hz
First, we need to convert the frequency from Hz to radians per second:
ω = 2πf
Substituting the given values into the impedance formula:
Z = √((1000)^2 + ((2π × 485) × 0.155 - 1/(2π × 485 × 25.0 × 10^(-9)))^2)
Calculating Z:
Z = √((1000)^2 + (481.663)^2) ≈ 1253.53 Ω
(b) Given:
f = 7.50 kHz = 7500 Hz
ω = 2πf
Substituting the given values into the impedance formula:
Z = √((1000)^2 + ((2π × 7500) × 0.155 - 1/(2π × 7500 × 25.0 × 10^(-9)))^2)
Calculating Z:
Z = √((1000)^2 + (568.28)^2) ≈ 1256.04 Ω
So, the impedance of the circuit at 7.50 kHz is approximately 1256.04 Ω.
(c) Given:
Vrms = 408 V
At 485 Hz:
Irms = Vrms / Z (using the impedance calculated in part (a))
Irms = 408 / 1253.53 ≈ 0.326 A = 326 mA
At 7.50 kHz:
Irms = Vrms / Z (using the impedance calculated in part (b))
Irms = 408 / 1256.04 ≈ 0.325 A = 325 mA
(d) The resonant frequency of an RLC circuit can be calculated using the formula:
f_resonant = 1 / (2π√(LC))
Given:
L = 155 mH = 0.155 H
C = 25.0 nF = 25.0 × 10^(-9) F
Substituting the given values into the resonant frequency formula:
f_resonant = 1 / (2π√(0.155 × 25.0 × 10^(-9)))
Calculating f_resonant:
f_resonant ≈ 25.74 kHz
(e) At resonance, the impedance of the inductor (ωL) and the capacitor (1/(ωC)) cancel each other out, resulting in the minimum impedance of the circuit.
Therefore, at resonance, the impedance (Z) of the circuit is equal to the resistance (R) only.
Given:
R = 1.00 kΩ = 1000 Ω
Vrms = 408 V
Using Ohm's law, we can calculate the current (Irms) at resonance:
Irms = Vrms / R
Irms = 408 / 1000 = 0.408 A = 408 mA
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Given vectors A → = 2.0 x ^ − 3.2 y ^ and B → = − 1.2 x ^ + 2.9 y ^ what is the angle between the vector D → = A → − B → and the positive x-axis?
Okay, let's solve this step-by-step:
1) Given:
A → = 2.0 x ^ − 3.2 y ^
B → = − 1.2 x ^ + 2.9 y ^
Find: Angle between D → = A → − B → and positive x-axis
2) Subtract the vectors to find D →:
D → = 2.0 x ^ − 3.2 y ^ - (− 1.2 x ^ + 2.9 y ^ )
= 2.0 x ^ − 3.2 y ^ + 1.2 x ^ - 2.9 y ^
= 3.2 x ^ − 6.1 y ^
3) To find the angle, we use the inverse tangent ratio of y/x:
Angle = tan−1(−6.1/3.2) = -61.97°
4) Since the angle is in the 4th quadrant, it is measured clockwise from the positive x-axis. So the final angle is 180 - 61.97 = 118.03°
Therefore, the angle between the vector D → = A → − B → and the positive x-axis is 118.03°.
Let me know if you have any other questions!