Answer: θ would equal approximately 28.7°
This is a kinematics problem, where one is only given the theta value 13.0° in regards to the range; thus, the problem is testing one's understanding of the relationships between the variables.
Range (aka x) = (v₀ sin (2θ₀))/g, where θ₀ = 13.0°
Now if we multiply the range by 2, we get:
2x = 2((v₀ sin (2θ₀))/g) → to verbalize, if range equates to (v₀ sin (2θ₀))/g, and doubling the range equals twice the product value, then:
2θ = sin⁻¹(2sin(2(13.0° )) = sin⁻¹(2(0.76255845048)) = sin⁻¹ (1.52511690096) = 57.35560850015109°/2 = θ
Thus, θ = 28.67780425
It's been awhile since I did this; though I hope it helped!
If we double the mass of an object without a change in volume, its density would be
A) half.
B) double.
C) unchanged.
Find the ratio of speeds of an electron and a negative hydrogen ion (one having an extra electron) accelerated through the same voltage, assuming non-relativistic final speeds. Take the mass of the hydrogen ion to be 6.7 X 10 -27 Kg.
Answer:
v₂ /v₁ = 2.3 10⁺²
Explanation:
The energy is conserved so the total potential energy must be transformed into kinetic energy
K = U
½ m v² = q ΔV
v = [tex]\sqrt{\frac{2q \Delta V}{m} }[/tex]
a) Let's find the speed of the electron
m = 9.1 10⁻³¹ kg
as they do not indicate the value of the power difference, we will assume that ΔV = 1 V is worth one
v = [tex]\sqrt{ \frac{2 \ 1.6 \ 10^{-19} \ 1}{9.1 \ 10^{-31}} }[/tex]
v = [tex]\sqrt {0.3516 \ 10^{12}}[/tex]
v1 = 0.593 10⁶ m / s
b) the velocity of a hydrogen ion
M = M_H + m
M = 1.673 10⁻²⁷ + 9.1 10⁻³¹
M = 1.67391 10⁻²⁷ kg
M = 1.67 10⁻²⁷ kg
v = [tex]\sqrt{ \frac { 2 \ 1.6 \ 10^{-19} \ 1}{1.67 \ 10^{-27}} }[/tex]
v = [tex]\sqrt{ 1.916167 \ 10^8 }[/tex]
v₂ = 1.38 10⁴ m / s
the relationship between these speeds is
v₂ / v₁ = 1.38 10⁴ / 0.593 10⁶
v₂ /v₁ = 2.3 10⁺²
A dog accelerates at 1.50 m/s2 to reach a velocity of 13.5 m/s while covering a distance of 49.3 m. What was his initial velocity?
Let v be the dog's initial velocity. Then
(13.5 m/s)^2 - v ^2 = 2 (1.50 m/s^2) (49.3 m)
==> v ^2 = (13.5 m/s)^2 - 2 (1.50 m/s^2) (49.3 m)
==> v = √((13.5 m/s)^2 - 2 (1.50 m/s^2) (49.3 m))
==> v ≈ 5.86 m/s
A box rests on a frozen pond, which serves as a frictionless horizontal surfaceIf a fisherman applies a horizontal force with magnitude 51.0 N to the box and produces an acceleration of magnitude 3.00 m/s2 , what is the mass of the box?
Answer:
the mass of the box is 17 kg
Explanation:
Given;
magnitude of the force applied by the fisherman, F = 51 N
magnitude of acceleration of the box, a = 3 m/s²
the mass of the box is calculated using Newton's law of motion;
F = ma
where;
m is the mass of the box
m = F / a
m = (51) / (3)
m = 17 kg
Therefore, the mass of the box is 17 kg
The current through a 3.0 Ω resistor is 0.30 A. The resistor is wired in series with a 9.0 V battery and an unknown resistor. What is the value of the unknown resistor?
Answer:
answer is 37 ohm since current e
remains same in series
The viscid silk produced by the European garden spider (Araneus diadematus) has a resilience of 0.35. If 10.0 J of work are done on the silk to stretch it out, how many Joules of work are released as thermal energy as it relaxes?
Answer: The energy released as thermal energy is 6.5 J
Explanation:
Energy stored by the spider when it relaxes is given by:
[tex]E_o=\text{Resilience}\times \text{Work}[/tex]
We are given:
Resilience = 0.35
Work done = 10.0 J
Putting values in above equation, we get:
[tex]E_o=0.35\times 10\\\\E_o=3.5J[/tex]
Energy released at thermal energy is the difference between the work done and the energy it takes to relaxes, which is given by the equation:
[tex]E_T=\text{Work done}-E_o[/tex]
Putting values in above equation, we get:
[tex]E_T=(10-3.5)=6.5J[/tex]
Hence, the energy released as thermal energy is 6.5 J
The energy released as thermal energy when 10 J of work is done to stretch silk will be 6.5 J
What is thermal energy?Thermal energy refers to the energy contained within a system that is responsible for its temperature. Heat is the flow of thermal energy.
Energy stored by the spider when it relaxes is given by:
[tex]\rm E_o=Resilience \ \times Work[/tex]
We are given:
Resilience = 0.35
Work done = 10.0 J
Putting values in above equation, we get:
[tex]\rm E_o=0.35\times 10[/tex]
[tex]E_o=3.5\ J[/tex]
Energy released at thermal energy is the difference between the work done and the energy it takes to relaxes, which is given by the equation:
[tex]E_T=\rm Work done -E_o[/tex]
Putting values in above equation, we get:
[tex]E_T=(10-3.5)=6.5\ J[/tex]
Hence, the energy released as thermal energy is 6.5 J
To know more about thermal energy follow
https://brainly.com/question/19666326
A 1.5 kg rock is dropped from a height of 2.0 meters onto a spring that
compresses and brings the rock to rest. (Assume no losses to thermal
energy.) How much energy is in the system before the drop.
Answer:
29.4 J
Explanation:
Before the drop, the system has only the gravitational potential energy, and this energy us given by mass×gravity×height:
1.5•9.8•2 = 29.4 J
The momentum of a falling rock is found to be 200 kg m/s. What is the mass of the rock if it falls with a velocity of 5.0 m/s
Answer:
[tex]\boxed {\boxed {\sf 40 \ kilograms}}[/tex]
Explanation:
Momentum is the product of velocity and mass. The formula is:
[tex]p=m*v[/tex]
We know the rock is falling. Its momentum is 200 kilograms meters per second and its velocity is 5 meters per second. Substitute the values into the formula.
[tex]200 \ kg \ m/s = m * 5.0 \ m/s[/tex]
We are solving for m, the mass. We must isolate the variable. It is being multiplied by 5 meters per second. The inverse of multiplication is division, so we divided both sides by 5.0 m/s.
[tex]\frac{200 \ kg \ m/s}{5.0 \ m/s}=\frac{ m* 5.0 \ m/s }{5.0 \ m/s}[/tex]
[tex]\frac{200 \ kg \ m/s}{5.0 \ m/s}=m[/tex]
The units of meters per second (m/s) cancel.
[tex]\frac{200 \ kg}{5.0 } =m[/tex]
[tex]40 \ kg = m[/tex]
The falling rock has a mass of 40 kilograms.
Some runners train with parachutes that trail behind them to provide a large drag force. These parachutes are designed to have a large drag coefficient. One model expands to a square 1.8 mm on a side, with a drag coefficient of 1.4. A runner completes a 240 mm run at 6.0 m/s with this chute trailing behind.
Required:
How much thermal energy is added to the air by the drag force?
Answer:
by the drag force, 2.4004512 × 10⁻⁵ J is added to the air.
Explanation:
Given the data in the question;
drag coefficient of Cd = 1.4
speed v = 6.0 m/s
One model expands to a square 1.8 mm on a side
Area A = 1.8 × 1.8 = 3.24 mm² = 3.24 × 10⁻⁶ m²
distance travelled s = 240 mm = 0.24 m
we know that; density of air e = 1.225 kg/m³
Now,
Dragging force F[tex]_D[/tex] = ( Cd × e × v² × A ) / 2
thermal energy = F[tex]_D[/tex] × s
so
thermal energy = ( 1.4 × 1.225 × (6)² × (3.24 × 10⁻⁶) × 0.24 ) / 2
thermal energy = ( 4.8009024 × 10⁻⁵ ) / 2
thermal energy = 2.4004512 × 10⁻⁵ J
Therefore, by the drag force, 2.4004512 × 10⁻⁵ J is added to the air.
Which statements describe using genetic factors to influence the growth of organisms? Select the three (3) that apply.
-increasing use of hybrid crops
-altering genes in DNA to create new plants
-increasing human population
-increasing climate change
-developing disease or pest resistant crops
Answer:
- increasing use of hybrid crops
- altering genes in DNA to create new plants
- developing disease or pest resistant crops
Explanation:
The use of genetic factors to influence the growth of a plant encompasses manipulating the genetic constituent (gene) of such plant.
For example,
- Increasing use of hybrid crops entails mating two pure bred plants based on a gene of interest responsible for a particular trait, to form a hybrid.
- Altering genes in DNA to create new plants is also a genetic factor as it has to with gene modification.
- developing disease or pest resistant crops means that the genetic make up of such plant has been modified to be resistant to pest/disease.
20 points, im begging for help‼️
How much capacitance is needed to
store 0.00100 J of energy when the
charge on the capacitor is
4.86 x 10-5 C?
[?] x 10?!F
Answer:
Capacitance= 1.18×10^-6
Answer: 1.18*10^-6
Explanation:
. A car increases velocity from 20 m/s to 60 m/s in a time of 10 seconds. What was the acceleration of the car?
Answer:
0.3333
Explanation:
Acceleration = change in velocity/time
a = 20 m/s / 60 m/s
a = 0.3333 m/s^2
Water with a volume flow rate of 0.001 m3/s, flows inside a horizontal pipe with diameter of 1.2 m. If the pipe length is 10m and we assume fully developed internal flow, find the pressure drop across this pipe length.
Answer:
[tex]\triangle P=1.95*10^{-4}[/tex]
Explanation:
Mass [tex]m=0.001[/tex]
Diameter [tex]d=1.2m[/tex]
Length [tex]l=10m[/tex]
Generally the equation for Volume flow rate is mathematically given by
[tex]Q=AV[/tex]
[tex]V=\frac{Q}{\pi/4D^2}[/tex]
[tex]V=\frac{0.001}{\pi/4(1.2)^2}[/tex]
[tex]V=8.84*10^{-4}[/tex]
Generally the equation for Friction factor is mathematically given by
[tex]F=\frac{64}{Re}[/tex]
Where Re
Re=Reynolds Number
[tex]Re=\frac{pVD}{\mu}[/tex]
[tex]Re=\frac{1000*8.84*10^{-4}*1.2}{1.002*10^{-3}}[/tex]
[tex]Re=1040[/tex]
Therefore
[tex]F=\frac{64}{Re}[/tex]
[tex]F=\frac{64}{1040}[/tex]
[tex]F=0.06[/tex]
Generally the equation for Friction factor is mathematically given by
[tex]Head loss=\frac{fLv^2}{2dg}[/tex]
[tex]H=\frac{0.06*10*(8.9*10^-4)^2}{2*1.2*9.81}[/tex]
[tex]H=19.9*10^{-9}[/tex]
Where
[tex]H=\frac{\triangle P}{\rho g}[/tex]
[tex]\triangle P=\frac{19.9*10^{-9}}{10^3*(9.81)}[/tex]
[tex]\triangle P=H*\rho g[/tex]
[tex]\triangle P=1.95*10^{-4}[/tex]
A light-emitting diode (LED) connected to a 3.0 V power supply emits 440 nm blue light. The current in the LED is 11 mA , and the LED is 51 % efficient at converting electric power input into light power output. How many photons per second does the LED emit?
Answer:
3.73 * 10^16 photons/sec
Explanation:
power supply = 3.0 V
Emits 440 nm blue light
current in LED = 11 mA
efficiency of LED = 51%
Calculate the number of photons per second the LED will emit
first step : calculate the energy of the Photon
E = hc / λ
=( 6.62 * 10^-34 * 3 * 10^8 ) / 440 * 10^-9
= 0.0451 * 10^-17 J
Next :
Number of Photon =( power supply * efficiency * current ) / energy of photon
= ( 3 * 0.51 * 11 * 10^-3 ) / 0.0451 * 10^-17
= 3.73 * 10^16 photons/sec
what happened on march 21 every year in the northern hemisphere
Answer:
B. The Spring equinox
Explanation:
The vernal equinox marks the moment the sun crosses the celestial equator. The vernal equinox happens on March 19, 20, or 21 every year in the Northern Hemisphere. In the Southern Hemisphere, this same event marks the beginning of fall. (Source: What Exactly Is The Spring Equinox? - Dictionary.com)
Hopefully this helps.
Boron is one position to the left of carbon on the periodic table. The atomic number of carbon is 6. Given its position on the periodic table what is the atomic number of boron?
Answer:
5
Explanation:
A professional quarterback throws a 0.40 kg football. what is the force of weight?
Answer:
3.92N
Explanation:
Force= mass×accelerarion due gravity
But mass= 0.40kg
acceleration due to gravity = 9.8 m/s^2
Force = 0.40×9.8
Force=3.92N
What is the primary function of the lower
respiratory system?
O to move nutrients to the cells throughout the
body
O to move blood to the cells throughout the body
O to extract oxygen from the air that the body breathes in
O to extract oxygen from the blood that the body makes
Answer:
C
Explanation:
100%
Select the correct answer.
If you increase the frequency of a sound wave four times, what will happen to its speed?
A.
The speed will increase four times.
B.
The speed will decrease four times.
C.
The speed will remain the same.
D.
The speed will increase twice.
E.
The speed will decrease twice.
Answer:
A. The speed of the wave increases four times.
Answer:
A. The speed will increase four times.
Explanation:
If you increase the frequency of a sound wave four times, the speed will increase four times. So, option (A) is correct.
A spring, with a spring constant of 4000 N/m, is oriented horizontally, and compressed by 10cm. When released, the spring launches a block of mass 1.0 kg along a 5.0-m horizontal section of track, where the coefficient of friction between the block and track is 0.20. The block then goes up a frictionless ramp angled at 60o with the horizontal. How high up the ramp does the block go before it starts to slide back down
Answer:
[tex]d=1.2m[/tex]
Explanation:
From the question we are told that:
Spring constant [tex]k= 4000 N/m[/tex]
Compressed [tex]l_d= 10cm=>0.10[/tex]
Mass [tex]m=1.0kg[/tex]
Length of horizontal section [tex]l=5.0-m[/tex]
Coefficient of friction [tex]\mu=0.20[/tex]
Angle [tex]\theta=60 \textdegree[/tex]
Generally the equation for Kinetic Energy K.E is mathematically given by
[tex]K.E=\mu mgL+mgdsin\theta[/tex]
[tex]\frac{1}{2}k*l_d^2=\mu mgL+mgdsin\theta[/tex]
[tex]\frac{1}{2}(4000)*0.1^2=0.2*1*9.8*5+1*9.8*d*sin60[/tex]
[tex]d=1.2m[/tex]
Speeding up
Slowing down
Standing still
Holding at a constant non-zero velocity
Answer:
speeding up is the answer
Explanation:
from the graph it can be seen that as the time (horizontal axis) increases the speed of vehicle (vertical axis) increases
At 20 ◦C a copper wire has a resistance of 4×10−3 Ω and a temperature coefficient of resistivity of 3.9×10−3 (C◦)−1, its resistance at 100 ◦C is
A.
52.5 × 10-3 Ω
B.
5.25 × 10-3 Ω
C.
5.25 × 10-4 Ω
D.
5.25 × 10-2 Ω
E.
25.5 × 10-3 Ω
Answer:
[tex]R _{t} = R _{0}( \alpha t + 1) \\ = 4 \times {10}^{ - 3} (3.9 \times {10}^{ - 3} \times 20 + 1) \\ = 4 \times {10}^{ - 3} (1.078) \\ = 4.312 \times {10}^{ - 3} \: Ω[/tex]
the magnitude of the electrical force acting between a +2.4x10-8c charge and 1+1.8x10-6 charge that are separated by 1.008m is
Answer:
3.83×10¯⁴ N
Explanation:
From the question given above, the following data were obtained:
Charge 1 (q₁) = +2.4x10¯⁸ C
Charge 2 (q₂) = +1.8x10¯⁶ C
Distance apart (r) = 1.008 m
Electrical constant (K) = 9×10⁹ Nm²/C²
Force (F) =?
The magnitude of the electrical force acting between the two charges can be obtained as follow:
F = Kq₁q₂ / r²
F = 9×10⁹ × 2.4x10¯⁸ × 1.8x10¯⁶ / (1.008)²
F = 0.0003888 / 1.016064
F = 3.83×10¯⁴ N
Thus the magnitude of the electrical force acting between the two charges is 3.83×10¯⁴ N
Vectors 퐴, 퐵and 퐶are added together. 퐴has a magnitude of 20.0 units and makes an angle of 60.0° counterclockwise from the negativex-axis. 퐵has a magnitude of 40.0 units and makes an angle of 30.0° counterclockwise from the positive x-axis.퐶has a magnitude of 35.0 units and makes an angle of 60.0° clockwise from the negative y-axis. Determine the magnitude of the resultant vector 퐴+퐵+퐶and its direction as an angle measured counterclockwise from the positive x-axis.
Answer:
Magnitude = 15.86 units
direction = 69 degree below negative X axis
Explanation:
A = 20 units at 60.0° counterclockwise from the negative x - axis
B = 40 units at 30.0° counterclockwise from the positive x - axis
C = 35 units at 60.0° clockwise from the negative y - axis
Write the vectors in the vector form
[tex]\overrightarrow{A} =20 (- cos 60 \widehat{i} - sin 60 \widehat{j})=- 10\widehat{i} - 17.3 \widehat{j}\\\\\overrightarrow{B} =40 (cos 30 \widehat{i} + sin 30 \widehat{j})= 34.6\widehat{i} +20 \widehat{j}\\\\\overrightarrow{C} =35 (- sin 60 \widehat{i} - cos 60 \widehat{j})=- 30.3\widehat{i} - 17.5 \widehat{j}\\\\Now\\\\overrightarrow{A} + \overrightarrow{B} + \overrightarrow{C} = (- 10 + 34.6 - 30.3) \widehat{i} + (-17.3 + 20-17.5)\widehat{j}\\\\[/tex]
[tex]\\\overrightarrow{A} + \overrightarrow{B} + \overrightarrow{C} = - 5.7\widehat{i} -14.8\widehat{j}[/tex]
The magnitude is given by
[tex]= \sqrt{5.7^2 + 14.8^2} = 15.86 units[/tex]
The direction is given by
[tex]tan\theta = \frac{- 14.8}{- 5.7}\\\\\theta= 69^o[/tex]
below negative X axis.
A 4.00 kg ball is swung in a circle on the edge of a 1.50 m rope. The time it takes for the ball to complete one rotation is 3.40 s. Please show all work and equation.
a) What is the velocity of the ball?
b) What is the acceleration of the ball?
c) What is the force on the ball?
Answer:
The answer is below
Explanation:
The length of the rope is equal to the radius of the circle formed by the complete rotation of the rope. Therefore the radius = 1.50 m.
a) The distance covered by the rope when completing one rotation is the same as the perimeter of the circle. Hence:
Distance covered in one rotation = 2π * radius = 2π * 1.5 = 3π meters
The velocity of the ball = Distance / time = 3π meters / 3.4 seconds = 2.77 m/s
b) The initial velocity (u) is 0 m/s, the final velocity is 2.77 m/s during time (t) = 3.4 s. Hence acceleration (a):
v = u + at
2.77 = 3.4a
a = 0.82 m/s²
c) Force on ball = mass * acceleration = 4 * 0.82 = 3.28 N
. How many meters away is a cliff if an echo is heard 0.5 s after the original sound? ( Assume that sound travels at 343 m/s on that day
Answer:
171.5 m
Explanation:
To find the distance, speed x time
342 x 0.5
171.5 m
Hope this helped!
The drag force Fd, imposed by the surrounding air on a vehicle moving with velocity V is given by:
Fd =C dA 2 pV2
where Cd is a constant called the drag coefficient, A is the projected frontal area of the vehicle, and \rhorho is the air density.
Determine the power, in hp, required to overcome aerodynamic drag for an automobile moving at
(a) 25 miles per hour,
(b) 70 miles per hour.
Assume Cd=0.28,
A= 25ft2
and p=0.075Ib/ft2
Answer:
Explanation:
a)
Given that:
V = 25 mi/hr
To ft/sec, we have:
[tex]V = 25 \times \dfrac{5280}{3600} ft/s[/tex]
[tex]V = \dfrac{110}{3} ft/s[/tex]
[tex]\rho = 0.075 \ lb/ft^3[/tex]
[tex]\rho = 0.075 \times \dfrac{1 \ lbf s^2/ft}{32.174 \ lbm}[/tex]
[tex]\rho = \dfrac{0.075}{32.174 } lbf.s^2/ft^4[/tex]
[tex]C_d = 0.28[/tex]
A = 25ft²
Recall that:
The drag force [tex]F_d =\dfrac{C_dA \rho V^2}{2}[/tex]
[tex]F_d =\dfrac{1}{2}\times 0.28 \times 25\times \dfrac{0.075}{32.174}\times (\dfrac{110}{3})^2[/tex]
[tex]F_d =10.967 \ lbf[/tex]
[tex]P = F_dV \\ \\ P = 10.97 \times (\dfrac{110}{3}} \\ \\ P = 402.3 \ hp[/tex]
For 70 miles per hour, we have:
[tex]V = 70 \times \dfrac{5280}{3600} ft/s[/tex]
[tex]V = \dfrac{308}{3} ft/s[/tex]
The drag force [tex]F_d =\dfrac{C_dA \rho V^2}{2}[/tex]
[tex]F_d =\dfrac{1}{2}\times 0.28 \times 25\times \dfrac{0.075}{32.174}\times (\dfrac{308}{3})^2[/tex]
[tex]F_d =85.99 \ lbf[/tex]
[tex]P = F_dV \\ \\ P = 85.99 \times (\dfrac{308}{3}}) \\ \\ P = 8828.2 \ hp[/tex]
A motorist travels due North at 90 km/h for 2 hours. She changes direction and travels West at 60 km/for 1 hour.
a) Calculate the average speed of the motorist [4]
b) Calculate the average velocity of the motorist.
Answer:
a) S = 63.2 km/h
b) V = 63.2 km/h*(-0.316 , 0.949)
Explanation:
Let's define:
North as the positive y-axis
East as the positive x-axis.
Also, remember the relation:
Distance = Time*Speed
Let's assume that she starts at the position (0km, 0km)
Then she travels due North at 90km/h for two hours, then the displacement is
90km/h*2h = 180km to the north
Then the new position is:
(0km, 180km)
Then she travels West at 60km/h for one hour.
Then the distance traveled to the West (negative x-axis) is:
60km/h*1h = 60km to the west
Then the new position is:
(-60km, 180km).
a) The average speed is defined as the quotient between the displacement and the time.
We know that the total time traveled is 3 hours.
And the displacement is the difference between the final position and the initial position.
this is:
D = √( -60km - 0km)^2 + (180km - 0km)^2)=
D = √( (60km)^2 + (180km)^2) = 189.7 km
Then the average speed is:
S = (189.7 km)/(3 h) = 63.2 km/h
b) Now we want to find the average velocity, this will be equal to the average speed times a versor that points from the origin to the direction of the final position.
So, if the final position is (-60km, 180km)
We need to find a vector that represents the same angle, but that is on the unit circle.
Then, if the module of the final position is 189.7 km (as we found above), then the versor is just given by:
(-60km/ 189.7 km, 180km/ 189.7 km)
(-60/189.7 , 180/189.7)
We can just check that the module of the above versor is 1.
[tex]module = \sqrt{(\frac{-60}{189.7} )^2 + (\frac{180}{189.7} )^2} = \frac{1}{189.7}* \sqrt{(-60 )^2 + (180 )^2} = 1[/tex]
Then the average velocity is:
V = 63.2 km/h*(-60/189.7 , 180/189.7)
We can simplify our versor so the velocity equation is easier to read:
V = 63.2 km/h*(-0.316 , 0.949)
A vessel at rest at the origin of an xy coordinate system explodes into three pieces. Just after the explosion, one piece, of mass m, moves with velocity (-21 m/s) and a second piece, also of mass m, moves with velocity (-21 m/s) . The third piece has mass 3m. Just after the explosion, what are the (a) magnitude and (b) direction (as an angle relative to the x axis) of the velocity of the third piece
Answer:
25
Explanation:
magnitude and (b) direction (as an angle relative to the x axis) of the velocity
Vector A= 3.7 i + 1.0 j and vector B = 3.0 i + 6.5 j. What is vector
(A-B).A?
Answer:
(A - B).A = -2.91
Explanation:
First, let's define the sum and dot product of vectors.
For two vectors V = (x₁, y₁) and W = (x₂, y₂) we have:
sum (or subtraction):
V + W = (x₁, y₁) + (x₂, y₂) = (x₁ + x₂, y₁ + y₂)
dot product:
V.W = (x₁, y₁).(x₂, y₂) = x₁*x₂ + y₁*y₂
Here remember the notation:
V = x₁*i + y₁*j = (x₁, y₁)
Now let's solve our problem, we have:
A = (3.7, 1.0)
B = (3.0, 6.5)
Then:
(A - B).A = ( (3.7, 1.0) - (3.0, 6.5) ).(3.7, 1.0)
= (3.7 - 3.0, 1.0 - 6.5).(3.7, 1.0)
= (0.7, -5.5).(3.7, 1.0) = (0.7*3.7) + (-5.5)*(1.0) = -2.91