Answer:
When in free fall, the only force acting upon your body is the force of gravity - a non-contact force. Since the force of gravity cannot be felt without any other opposing forces, you would have no sensation of it. You would feel weightless when in a state of free fall.
What were the physical activities in your childhood that you still do today? Do you spend more time now in doing these activities as compared before?
What is the velocity of a plane to travel 3000 miles from New York to California in 5.0 hours
Answer:
10 miles per minute.
What is Bill's average running speed?
Answer:
Hello!
Sorry you haven't put up an image of your question! Without it we can't answer your question!
Explanation:
Maybe put up another one and it'll be answered!
:D
A trumpet player on a moving railroad flatcar moves toward a second trumpet player standing alongside the track both play a 490 Hz note. The sound waves heard by a stationary observer between the two players have a beat frequency of 2.0 beats/s. What is the flatcar's speed? Take the speed of sound to be 343 m/s.
Answer:
The value is [tex]v_s = 1.394 \ m/s[/tex]
Explanation:
From the question we are told that
The frequency of the second player is [tex]f_2 = 490 \ Hz[/tex]
The beat frequency is [tex]f_b = 2.0 \ Hz[/tex]
The speed of sound is [tex]v_s = 343 \ m/s [/tex]
Generally the frequency of the note played by the first player is mathematically represented as
[tex]f_1 = f_2 + f_b[/tex]
=> [tex]f_1 = 490 + 2.0 [/tex]
=> [tex]f_1 = 492 Hz[/tex]
From the relation of Doppler Shift we have that
[tex]f_1 = \frac{ f_2 (v+ v_o )}{v-v_s }[/tex]
Here [tex] v_o\ is\ the\ velocity\ of\ the\ observer\ with\ value\ 0 \ m/s [/tex]
So
[tex]492 = \frac{ 490 (343+0 )}{343 -v_s }[/tex]
=> [tex]v_s = 1.394 \ m/s[/tex]
When the sun provides energy for photosynthesis, an interaction with the __________ takes place.
A 3.7-mm-diameter wire carries a 20 A current when the electric field is 8.2×10−2 V/m .
What is the wire's resistivity? (in Ωm)
Answer:
ρ = 4.4 10⁻⁸ Ω m
Explanation:
For this exercise let's start by finding the value of the resistance using Ohm's law
V = I R
R = V / I
The voltage is related to the electric field
V = E L
let's substitute
R = E L / I
R = 8.2 10⁻² / 20
R = 4.1 10⁻³ L
now we can use the resistance relation
R = ρ L / A
the area of a circular wire is
A = π r² = π d² / 4
ρ = R π d² / (4 L)
let's calculate
ρ = (4.1 10⁻³ L) π 0.0037² / (4 L) = (4.1 10⁻³) π 0.0037² / (4)
ρ = 4.4 10⁻⁸ Ω m
The resistivity of the wire is 4.4 10⁻⁸ Ω m.
We know that;
V = I R
R = V / I
The voltage the electric field can be connected using the formula;
V = E L
Hence;
R = E L /I
R = 8.2 10⁻² / 20
R = 4.1 10⁻³ L
For resistivity;
R = ρ L / A
the area can be obtained from;
A = π r² = π d² / 4
ρ = R π d² / (4 L)
ρ = (4.1 10⁻³ L) π 0.0037² / (4 L) = (4.1 10⁻³) π 0.0037² / (4)
ρ = 4.4 10⁻⁸ Ω m
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Two cylindrical resistors are made from the same material. The shorter one has a length LL and diameter DD . The longer one has a length 16L16L and diameter 4D4D . How do their resistances compare? The resistance of the longer resistor is four times the resistance of the shorter resistor. The resistance of the longer resistor is twice the resistance of the shorter resistor. The resistance of the longer resistor is the same as the resistance of the shorter resistor. The resistance of the longer resistor is half the resistance of the shorter resistor. The resistance of the longer resistor is a quarter of the resistance of the shorter resistor.
Answer:
The resistance of the longer resistor is a quarter of the resistance of the shorter resistor.Explanation:
If Two cylindrical resistors are made from the same material, then their resistivity will be the same. Formula for calculating resistivity of a material is expressed as;
[tex]\rho = \frac{RA}{L} \ where \ A = \frac{\pi d^2}{4}[/tex] where;
R is the resistance
A is the cross sectional area of the material
L is the length of the material
For the shorter cylinder:
Length = L
diameter = D
[tex]\rho = \dfrac{R_s(\frac{\pi D^2}{4})}{L} \\\\\rho = \dfrac{R_s{\pi D^2}}{4L}[/tex]
For the longer cylinder:
Length = 16L
diameter = 4D
[tex]\rho = \dfrac{R_l(\frac{\pi (4D)^2}{4})}{16L} \\\\\\\rho = \dfrac{R_l(\frac{\pi (16D^2)}{4})}{16L} \\\\\rho = \dfrac{R_l{16\pi D^2}}{16L}\\\\\rho = \dfrac{R_l{\pi D^2}}{L}[/tex]
Since their resistivity are the same then;
[tex]\dfrac{R_s{\pi D^2}}{4L} = \dfrac{R_l{\pi D^2}}{L} \\\\ \dfrac{R_s}{4} = {R_l} \\\\R_s = 4R_l\\\\R_l = \frac{R_s}{4}[/tex]
Hence the resistance of the longer resistor is a quarter of the shorter resistor.
If a person is standing up in a moving bus that stops suddenly,the person can easily fall forwards.Has a force acted to push the person forwards? Use Newton's law of motion to explain what is happening.
Answer:
i think gravity created a force strong enough to push him down.but it also depends how slow or fast the bus is going
Explanation:
gravity as we all now it is pretty strong so if the bus is going fast then the person standing up will fall but if the bus is going pretty slow then it'll just nudge him.
An airplane with a hot-wire anemometer mounted on its wing tip is to fly through the turbulent boundary layer of the atmosphere at a speed of 50 m/sec. The velocity fluctuations in the atmosphere are of order 0.5 m/sec, the length scale of the large eddies is about 100 m. The hot-wire anemometer is to be designed so that it will register the motion of the smallest eddies.What is the highest frequency the anemometer will encounter
Answer:
0.55 hz
Explanation:
Given that the plane fly through the turbulent boundary layer of the atmosphere at a speed of 50 m/sec. And the velocity fluctuations in the atmosphere are of order 0.5 m/sec, the length scale of the large eddies is about 100 m.
The maximum speed attained will be
Maximum speed = 50 + 0.5 = 5.5 m/s
The Length = 100m
Speed = FL
Where F = frequency
Substitute speed and distance length into the formula
55 = 100F
F = 55/100
F = 0.55 Hz
Therefore, the highest frequency the anemometer will encounter will be 0.55 Hz
Answer:
40079 Hz
Explanation:
1. The first step is to calculate energy dissipation ∈=u^3/l and here u is fluctuating velocity
∈=(0.5^3)/100 = 0.00125 m^2/s^3
2. Find out the length scale of the small eddies
η=(viscosity/∈)^1/4
η=(1.470e-5/0.00125)^1/4 = 0.00126 m
3. The frequency associated with these small-scale eddies will be the greatest frequency the anemometer will encounter, thus:
u_max=f_max * η
u_max = u + u' = 50+0.5=50.5 m/s
f_max = u_max/η = 50.5/0.00126 = 40079 Hz
This is the heighest frequency the anemometer will encounter.
A magnetic field is created by ____.
A moving electric charges
b elctromagnetic pulses
c a strong current
d a change in the current of a wire
A basketball leaves a player's hands at a height of 2.20 m above the floor. The basket is 2.70 m above the floor. The player likes to shoot the ball at a 36.0 ∘ angle. Of the shot is made from a horizontal distance of 9.10 m and must be accurate to ±0.23m (horizontally), what is the range of initial speeds allowed to make the basket
Answer:
The range of initial speeds allowed to make the basket is: [tex]9.954\,\frac{m}{s}\leq v \leq 10.185\,\frac{m}{s}[/tex].
Explanation:
We must notice that basketball depicts a parabolic motion, which consists of combining a constant speed motion in x-direction and free fall motion in the y-direction. The motion is described by the following kinematic formulas:
x-Direction
[tex]x = x_{o} + v_{o}\cdot t \cdot \cos \alpha[/tex]
y-Direction
[tex]y = y_{o} + v_{o}\cdot t\cdot \sin \alpha +\frac{1}{2}\cdot g\cdot t^{2}[/tex]
Where:
[tex]x_{o}[/tex], [tex]y_{o}[/tex] - Initial position of the basketball, measured in meters.
[tex]x[/tex], [tex]y[/tex] - Final position of the basketball, measured in meters.
[tex]v_{o}[/tex] - Initial speed of the basketball, measured in meters per second.
[tex]t[/tex] - Time, measured in seconds.
[tex]\alpha[/tex] - Tilt angle, measured in sexagesimal degrees.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
If we know that [tex]x_{o} = 0\,m[/tex], [tex]y_{o} = 2.20\,m[/tex], [tex]\alpha = 36^{\circ}[/tex], [tex]g = -9.807\,\frac{m}{s^{2}}[/tex], [tex]x = (9.10\pm0.23)\,m[/tex] and [tex]y = 2.70\,m[/tex], the system of equation is reduce to this:
[tex](9.10\pm 0.23)\,m = 0\,m + v_{o}\cdot t \cdot \cos 36^{\circ}[/tex]
[tex]9.10\pm 0.23 = 0.809\cdot v_{o}\cdot t[/tex] (Ec. 1)
[tex]2.70\,m = 2.20\,m + v_{o}\cdot t \cdot \sin 36^{\circ} + \frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right) \cdot t^{2}[/tex]
[tex]0.50 = 0.588\cdot v_{o}\cdot t-4.904\cdot t^{2}[/tex] (Ec. 2)
At first we clear [tex]v_{o}\cdot t[/tex] in (Ec. 1):
[tex]v_{o}\cdot t = \frac{9.10\pm 0.23}{0.809}[/tex]
[tex]v_{o}\cdot t = 11.248\pm 0.284[/tex]
(Ec. 1) in (Ec. 2):
[tex]0.5 = 0.588\cdot (11.248\pm 0.284)-4.904\cdot t^{2}[/tex]
Now we clear the time in the resulting expression:
[tex]4.904\cdot t^{2} = 0.588\cdot (11.248\pm 0.284)-0.5[/tex]
[tex]t = \sqrt{\frac{0.588\cdot (11.248\pm 0.284)-0.5}{4.904} }[/tex]
There are two solutions:
[tex]t_{1} = \sqrt{\frac{0.588\cdot (11.248- 0.284)-0.5}{4.904} }[/tex]
[tex]t_{1} \approx 1.101\,s[/tex]
[tex]t_{2} = \sqrt{\frac{0.588\cdot (11.248+ 0.284)-0.5}{4.904} }[/tex]
[tex]t_{2}\approx 1.131\,s[/tex]
The initial velocity is cleared within (Ec. 2):
[tex]v_{o}=\frac{0.50+4.904\cdot t^{2}}{0.588\cdot t}[/tex]
The bounds of the range of initial speed is determined hereafter:
[tex]t_{1} \approx 1.101\,s[/tex]
[tex]v_{o} = \frac{0.50+4.904\cdot (1.101)^{2}}{0.588\cdot (1.101)}[/tex]
[tex]v_{o} = 9.954\,\frac{m}{s}[/tex]
[tex]t_{2}\approx 1.131\,s[/tex]
[tex]v_{o} = \frac{0.50+4.904\cdot (1.131)^{2}}{0.588\cdot (1.131)}[/tex]
[tex]v_{o} = 10.185\,\frac{m}{s}[/tex]
The range of initial speeds allowed to make the basket is: [tex]9.954\,\frac{m}{s}\leq v \leq 10.185\,\frac{m}{s}[/tex].
A hockey puck initially travelling to the right at 34 m/s. It moves for 7 before
coming to a stop. How far did it move in 7 seconds?
You can use kinematic equations
Answer:
[tex]x=119m[/tex]
Explanation:
Hello,
In this case, since the hockey puck was moving at 34 m/s and suddenly stopped (final velocity is zero) in 7 seconds, we can first compute the acceleration via:
[tex]a=\frac{v_f-v_o}{t}=\frac{0m/s-34m/s}{7s}\\ \\a=-4.86m/s^2[/tex]
In such a way, we can compute the displacement via:
[tex]x=\frac{v_f^2-v_o^2}{2a}\\ \\x=\frac{0^2-(34m/s)^2}{2*-4.86m/s^2}\\ \\x=119m[/tex]
Best regards.
A bucket is being lowered by a very light rope with a constant downward velocity. The tension in the rope must be
Answer:
The tension in the light rope must be equal to the weight of the bucket
Explanation:
Given that,
Constant velocity of bucket and direction of bucket in downward
We need to find the tension in the rope
Using given data,
When a bucket moves downward with a constant velocity then the net force does not applied on the bucket.
So, The weight of the bucket will be equal to the tension in the light rope
In mathematically,
[tex]T=mg[/tex]
Where, T = tension
m = mass of bucket
g = acceleration due to gravity
Hence, The tension in the light rope must be equal to the weight of the bucket.
True or false: points that lie on the same plane are Collinear
QUICK
a sound wave has a frequency of 85 Hz and a wavelength of 4.2 m. what is the wave speed of the sound wave?
A. 357 seconds
B. 89.2 m/s
C. 357 m/s
D. 20 m/s
Answer:C
Explanation:
I did the test
a sound wave has a frequency of 85 Hz and a wavelength of 4.2 m. Then the wave speed of the sound wave is 20.23 m/s. Hence option D is correct.
What is wave ?Wave is is a disturbance in a medium that carries energy as well as momentum . wave is characterized by amplitude, wavelength and phase. Amplitude is the greatest distance that the particles are vibrating. especially a sound or radio wave, moves up and down. Amplitude is a measure of loudness of a sound wave. More amplitude means more loud is the sound wave.
Wavelength is the distance between two points on the wave which are in same phase. Phase is the position of a wave at a point at time t on a waveform. There are two types of the wave longitudinal wave and transverse wave.
The wave speed is given by,
c = νλ
where λ is wavelength, ν is frequency.
Given,
frequency ν = 85 Hz
wavelength λ = 4.2 m
The speed of the wave,
c = 85 × 4.2 = 20.23 m/s
Hence option D is correct.
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Mathew has a filtration kit, which consists of a funnel, a flask, and filter papers. Which of these mixtures can he separate using filtration?
Answer:
C. Muddy Water
PLZ PLZ PLZ I VOTE BRAINLY PLZPLZPLZ what is the theory of an atom created by a Dr. VAN C.
Answer: Bohr developed the Bohr model of the atom, in which he proposed that energy levels of electrons are discrete and that the electrons revolve in stable orbits around the atomic nucleus but can jump from one energy level (or orbit) to another. ... During the 1930s Bohr helped refugees from Nazism.
A 0.091-in-diameter electrical wire at 90°F is covered by 0.02-in-thick plastic insulation (k = 0.075 Btu/h·ft·°F). The wire is exposed to a medium at 50°F, with a combined convection and radiation heat transfer coefficient of 2.5 Btu/h·ft2·°F. Calculate the critical radius (rcr) of the plastic insulation (in inches).
Answer:
The critical radius of the plastic insulation is 0.72 inches.
Explanation:
Given that,
Diameter = 0.091 in
Thickness = 0.02 in
Initial temperature = 90°F
Final temperature = 50°F
Heat transfer coefficient = 2.5 Btu/h.ft²°F
Material conductivity = 0.075 Btu/h.ft °F
We need to calculate the critical radius of the plastic insulation
Using formula of critical radius
[tex]r_{cr}=\dfrac{2K}{h}[/tex]
Where, k = Material conductivity
h = Heat transfer coefficient
Put the value into the formula
[tex]r_{cr}=\dfrac{2\times0.075}{2.5}[/tex]
[tex]r_{cr}=0.06\ ft[/tex]
[tex]r_{cr}=0.72\ inches[/tex]
Hence, The critical radius of the plastic insulation is 0.72 inches.
Using the equation for Impact, can you explain the following:
Why are car steering rods designed to collapse?
Why are highway guard rails designed to crumple up on impact?
Why are traffic saftey barrels filled with water or sand?
Explanation:
Equation for Impact
FΔt = ΔP,
F = force
Δt = Impact of time
ΔP = Change in momentum
Car steering is engineered to fail in order to maximize the time of contact and hence reduce the initial impact and mitigate the damage incurred.
Road guard railing crumple on contact to maximize impact time and hence reduce impact intensity and mitigate damage.
Road safety containers are loaded with liquid or sand as they improve the period of impact.
A column of soldiers, marching at 100 steps per minute, keep in step with the beat of a drummer at the head of the column. It is observed that the soldiers in the rear end of the column are striding forward with the left foot when the drummer is advancing with the right. What is the approximate length of the column? (Take the speed of sound to be 343 m/s.)
Answer:
The value is [tex]D = 205.8 \ m [/tex]
Explanation:
The time taken for the column to take a step mathematically represented as
100 steps => 1 minutes => 60 seconds
1 step => t
=> [tex]t = 0.6 \ s [/tex]
Generally the length of the column is mathematically represented as
[tex]D = v * t[/tex]
substituting 343 m/s for v we have
[tex]D = 343 * 0.6 [/tex]
=> [tex]D = 205.8 \ m [/tex]
What does this picture show?
A. Good accuracy, poor precision
B. Poor accuracy, good precision
C. Good accuracy, good precision
D. Poor accuracy, poor precision
A watermelon is dropped from the edge of the roof of a build- ing and falls to the ground. You are standing on the sidewalk and see the watermelon falling when it is 30.0 m above the ground. Then 1.50 s after you first spot it, the watermelon lands at your feet. What is the height of the building
Answer:
The hight of the building is 38.16 m
Explanation:
These two pieces of information given, first, the watermelon is 30 m above the ground and after 1.50 s the watermelon has been spotted. Now we are required to find the height of the building.
Use the below formula to find the height of buildings.
S = ut + ½ gt^2
30 =1.5u + (1/2) × 9.8 (1.5)^2
u = 12.65 m/sec
v^2 – u^2 = 2gs
(12.65)^2 = 2×9.8 s’
S’ = 8.16 m
h = s + s’
h = 30 + 8.16 = 38.16 m
The hight of the building is 38.16 m.
The height of the building is 38.16 m.
Given data:
The height above the ground is, h = 30.0 m.
The time interval after observation of first spot is, t = 1.50 s.
We need to find the height of building. And since two pieces of information given, first, the watermelon is 30 m above the ground and after 1.50 s the watermelon has been spotted. So, using the second kinematic equation of motion as,
[tex]h = ut + \dfrac{1}{2}gt^{2}[/tex]
Here, u is the initial speed. Solving as,
[tex]30 = (u \times 1.50) + \dfrac{1}{2} \times 9.8 \times (1.50)^{2}\\\\u =12.65 \;\rm m/s[/tex]
Now landing distance (s') is calculated using the third kinematic equation of motion as,
[tex]v^{2} =u^{2}+2(-g)s\\\\0^{2} =12.65^{2}+2(-9.8)s\\\\s = 8.16 \;\rm m[/tex]
Then the height of building is given as,
H = h + s
H = 30 m + 8.16 m
H = 38.16 m
Thus, we can conclude that the height of the building is 38.16 m.
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Personal behavior is the only factor that determines if a person becomes ill.
Please select the best answer from the choices provided.
ОТ
OF
Answer:
FALSE
Explanation:
half life of a given sample of radium is 22 years the sample will reduce to 25% of its original value after
Answer:
44 years
Explanation:
Use half life equation:
A = A₀ (½)^(t / T)
where A is the final amount,
A₀ is the initial amount,
t is time,
and T is the half life.
0.25 A₀ = A₀ (½)^(t / 22)
0.25 = (½)^(t / 22)
t / 22 = 2
t = 44
44 sec is half life of a given sample of radium is 22 years the sample will reduce to 25% of its original value.
What is half-life?The period of time it takes for one-half of a radioactive isotope to decay is known as the half-life. A given radioactive isotope's half-life is constant; it is unaffected by external factors and independent of the isotope's starting concentration.
Use half life equation:
A = A₀ (½)^(t / T)
where A is the final amount,
A₀ is the initial amount,
t is time,
and T is the half life.
0.25 A₀ = A₀ (½)^(t / 22)
0.25 = (½)^(t / 22)
t / 22 = 2
t = 44 sec
44 sec is half life of a given sample of radium is 22 years the sample will reduce to 25% of its original value.
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On a still water, a speedboat decreases its speed uniformly from 30 m/s to 20 m/s. How long does it take the boat to travel a distance of 200m?
a. -8 s
b. 8 s
c. -200 s
d. 200 s
Answer:
t=8
Explanation:
u have solution I give solution also
don't mark plzz follow y
what is the mass of an object that is experiencing a net force of 200N and an acceleration of 500m/s2
Answer:
F=200N
a=500m/s2
Mass=?
Explanation:
F=ma
200=m*500
200/500=m
Mass=0.4kg
A corvette starts from rest and travels 69.0 meters in 50 s. What is its acceleration?
Answer:
0.0552 m/s²
Explanation:
Given:
Δx = 69.0 m
v₀ = 0 m/s
t = 50 s
Find: a
Δx = v₀ t + ½ at²
69.0 m = (0 m/s) (50 s) + ½ a (50 s)²
a = 0.0552 m/s²
If 13 is added to a number, the result is 43 less than twice the number. Find the number,
Explanation:
13+x=43>2x
13+x=43>2x+43>2x
13+x=86>4x
x-4x=86-13
3x=73
x=73/3
x=24.333
x=24.4
what would the answer be ?
Answer:
im going between 2 of them b and c but i would have choose b
Given three vectors A = 24i + 33j, B = 55i - 12j and C = 2i + 43j (a) Find the magnitude of each vector. (b) Write an expression for the vector difference A - C. (c) Find the magnitude and direction of the vector difference A-B. (d) In a vector diagram show vector A + B, and A - B, and also show that your diagram agrees qualitatively with your answer.
Answer:
(a) , . and .
(b)[tex]\vec A - \vec C=22 \hat i -10 \hat j[/tex].
(c)[tex]|\vec A - \vec B|=63.13[/tex] and the direction [tex]\theta =[/tex] 124.56°.
Explanation:
Given that,
,
and
[tex]\vec {C}=2 \hat i +43 \hat j[/tex]
(a) The magnitude of a vector is the square root of the sum of the square of all the components of the vector, i.e. for a ,.
So, the magnitude of the is
[tex]|\vec A|=\sqrt {24^2+ 33^2}[/tex]
[tex]\Rightarrow |\vec A|=\sqrt {1665}[/tex]
.
The magnitude of the is
[tex]|\vec B|=\sqrt {55^2+ (-12)^2}[/tex]
[tex]\Rightarrow |\vec B|=\sqrt {3169}[/tex]
.
And, the magnitude of the is
[tex]|\vec C|=\sqrt {2^2+ 43^2}[/tex]
[tex]\Rightarrow |\vec C|=\sqrt {1853}[/tex]
.
(b) The difference between the two vectors is the difference between the corresponding components of the vectors. So, the required expression of is
[tex]\vec A - \vec C=(24 \hat i +33 \hat j) - (2 \hat i +43 \hat j)[/tex]
[tex]\Rightarrow \vec A - \vec C=24 \hat i +33 \hat j - 2 \hat i -43 \hat j[/tex]
[tex]\Rightarrow \vec A - \vec C=22 \hat i -10 \hat j[/tex]
(c) The expression of is
[tex]\vec A - \vec N=(24 \hat i +33 \hat j) - (55 \hat i -12 \hat j)[/tex]
[tex]\Rightarrow \vec A - \vec B=24 \hat i +33 \hat j - 55\hat i +12 \hat j[/tex]
[tex]\Rightarrow \vec A - \vec B=-31 \hat i +45 \hat j\;\cdots (i)[/tex]
The magnitude of is
[tex]|\vec A - \vec B|=\sqrt {(-31)^2+55^2}[/tex]
[tex]\Rightarrow |\vec A - \vec B|=\sqrt {3986}[/tex]
[tex]\Rightarrow |\vec A - \vec B|=63.13[/tex]
Now, if a vector [tex]\vec V= -\alpha \hat i +\beta \hat j[/tex] in 3rd quadrant having direction [tex]\theta[/tex] with respect to [tex]\hat i[/tex] direction, than
in the anti-clockwise direction.
Here, from equation (i), for the vector [tex]\vec A - \vec C[/tex], [tex]\alpha=31[/tex] and [tex]\beta=45[/tex].
[tex]\Rightarrow \theta = \pi-\tan ^{-1}\left(\frac {45}{31}\right)[/tex]
180°-55.44° [as \pi radian= 180°]
124.56° in the anti-clockwise direction.
(d) Vector diagrams for [tex]\vec A +\vec B[/tex] and [tex]\vec A - \vec B[/tex] has been shown
in the figure(b) and figure(c) recpectively.
Vector [tex]\vec A - \vec B[/tex] is in 3rd quadrant as calculated in part (c).
While Vector [tex]\vec A +\vec B=(24 \hat i +33 \hat j)+(55 \hat i -12 \hat j)[/tex]
[tex]\Rightarrow \vec A +\vec B=79 \hat i +21 \hat j[/tex], which is in 1st quadrant as both the components are position has been shown in figure(b).