What statements accurately describe sunspots? Check all that apply.
Sunspots are storms on the Sun's surface.
Sunspots are marked by intense magnetic activity.
Sunspots produce solar flares and hot gassy ejections.
Sunspots can affect Earth's climate.
Sunspots are cool areas where the Sun is covered by clouds.

The pressure applied by the standing book on the table is 1,800 N/m².

option B is the correct answer.

The pressure applied by the standing book on the table is determined from the ratio of weight of the book and the area of the standing book.

Mathematically, the formula for the pressure of a material is given as;

P = F / A

where;

The weight of the book , F = mg

where;

F = 1.5 kg  x  9.8 m/s²

F = 14.7 N

The dimension of the book include;

The height of the book is not in contact with the surface of the table, so the area of the book in contact with the table becomes;

A = w x b

W = 0.21 m  x  0.04 m

W = 8.4 X 10⁻³ m²

P = F / A

P = ( 14.7 N ) / ( 8.4 X 10⁻³ m² )

P = 1,750 N/m² ≈ 1,800 N/m²

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When an object transported from earth to the moon, its mass remains same but weight becomes 1/6th that on earth..

In physics, mass is a quantitative measurement of inertia, a basic characteristic of all matter.

It essentially refers to a body of matter's resistance to changing its speed or location in response to the application of a force. The change caused by an applied force is smaller the more mass a body has.

The amount of a body's weight indicates how much gravity is pulling on it. Weight is calculated using the method w = mg. Since weight is a force, it has the same SI unit as a force, which is the Newton (N).

So, mass of an object is intrinsic property of the object - it remains same in earth and in moon. But weight is the amount of gravitational  attraction force. Moon has nearly 1/6th gravitational attraction field. So, the weight of the object on moon will be 1/6th that on earth.

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From pressure definition,  the area of the car tires that are in contact with the road is 0.079 m

Pressure can be defined as the ratio of force to area. It is measured in Pascal.

Given that a car weighs 15,000 N and its tires are I flared to a pressure of 190 kPa. How large is the area of the cars tires that are in contact with the road ?

From the definition of Pressure, P = F/A

Where

190000 = 15000/A

A = 15000/190000

A = 0.079 m

A car will have 4 tires. The area of one in contact with the road will be

A = 0.079/4

A = 0.02 m

Therefore, the area of the car tire that is in contact with the road is 0.02 m

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Answer:

you could put the iron filings on the peace of paper and hover a magnet over top of the paper and the iron filings would stand up, or even stick to the magnet

Explanation:

According to the magnitude of the peak that is produced when the centers of the pulses coincide, the four examples are ordered as follows: A, B, D, and C.

A transverse pulse is one in which the individual coils move in a direction that is antithetical to the direction in which you are moving the slinky. transverse and longitudinal waves the vibrational motion of the particles or layers of the medium in a longitudinal wave is parallel to the direction of the disturbance's propagation.

Destructive interference is when two waves superimpose on top of each other in the opposite phase, with the resultant wave's amplitude being equal to the individual waves' differences in amplitude and producing the least amount of light.

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Using properties of transverse wave, According to the magnitude of the peak that is produced when the centers of the pulses coincide, the four examples are ordered as follows: A, B, D, and C.

A transverse pulse is one in which the individual coils move in a direction that is antithetical to the direction in which you are moving the slinky. transverse and longitudinal waves the vibrational motion of the particles or layers of the medium in a longitudinal wave is parallel to the direction of the disturbance's propagation.

Destructive interference is when two waves superimpose on top of each other in the opposite phase, with the resultant wave's amplitude being equal to the individual waves' differences in amplitude and producing the least amount of light.

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Complete question -

Shown above right are four different pairs of transverse wave pulses that move toward each other. At some point in time, the pulses meet and interact (interfere) with each other. (Figure 1) Figure 1 of 1 ترتر А. B c D Part A Rank the four cases, from most to least, on the basis of the amplitude of the peak that results when the centers of the pulses coincide. Rank from most to least. To rank items as equivalent, overlap them. Reset Help B A с Least Most

(a) The mass of water is approximately 49.65 kg. (b) The final temperature of the water vapor will be 120°C. (c) The total enthalpy change is approximately 277,956 kJ. (d) Diagram shown below.

(a) To determine the mass of the water, we need to know its density at the given conditions. The density of water changes with temperature and pressure. At 40°C and 200 kPa, the density of water is approximately 993 kg/m³.

Since we have 50 L of water, we need to convert it to cubic meters:

50 L = 0.05 m³

Now we can calculate the mass of water:

Mass = Density * Volume

Mass = 993 kg/m³ * 0.05 m³

Mass ≈ 49.65 kg

Therefore, the mass of water is approximately 49.65 kg.

(b) To find the final temperature, we need to consider the phase change from liquid to vapor. At constant pressure, the temperature will remain constant until all the liquid water has vaporized. This temperature is called the saturation temperature.

We can determine the saturation temperature at 200 kPa using a steam table or other relevant data sources. Let's assume that the saturation temperature is 120°C.

Therefore, the final temperature of the water vapor will be 120°C.

(c) To calculate the total enthalpy change, we need to consider the energy required to heat the water from its initial temperature to the saturation temperature, as well as the energy required for the phase change from liquid to vapor.

The enthalpy change during heating can be calculated using the formula:

ΔH1 = Mass * Specific Heat Capacity * ΔT1

Where:

Mass = 49.65 kg (from part a)

Specific Heat Capacity = specific heat capacity of water at constant pressure = 4.18 kJ/(kg·°C)

ΔT1 = final temperature - initial temperature = 120°C - 40°C = 80°C

ΔH1 = 49.65 kg * 4.18 kJ/(kg·°C) * 80°C

ΔH1 ≈ 165,938 kJ

The enthalpy change during phase change can be calculated using the formula:

ΔH2 = Mass * Latent Heat of Vaporization

Where:

Mass = 49.65 kg (from part a)

Latent Heat of Vaporization = energy required to vaporize 1 kg of water = 2257 kJ/kg

ΔH2 = 49.65 kg * 2257 kJ/kg

ΔH2 ≈ 112,018 kJ

The total enthalpy change is the sum of ΔH1 and ΔH2:

Total Enthalpy Change = ΔH1 + ΔH2

Total Enthalpy Change ≈ 165,938 kJ + 112,018 kJ

Total Enthalpy Change ≈ 277,956 kJ

Therefore, the total enthalpy change is approximately 277,956 kJ.

(d) The process can be shown on a T-v (temperature-volume) diagram with respect to saturation lines. In this case, the process starts at the initial temperature and pressure (40°C, 200 kPa), and moves along the constant pressure line until reaching the saturation temperature (120°C). Then, the process follows the saturation line until the entire liquid is vaporized.

Here is a simplified representation of the process on a T-v diagram:

           |

Saturation |                       |

 Line   |                             |

           |                             |

           |                             |

           |                             |

           |                             |

           |                             |

           |                             |

 Initial |-----------------------------| Final

           |                             |

           |                             |

           |                             |

           |                             |

           |                             |

           |                            

This diagram is a rough representation and does not accurately reflect specific volume values or scale. It simply illustrates the general process from initial conditions to the final state along the constant pressure and saturation lines.

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According to Coulomb's law, the force between the given charges is 0.225N which is explained below.

Force on two identical charges q separate by a distance of r is given by:

F = kq²/r²

where k is Coulomb's constant

q is the charge

r is the separation between the charges

Given that q = 1×10⁻⁵C,

and r = 2m

So, the force between the given charges will be:

F = (9×10⁹)(1×10⁻⁵)²/2²

F = 0.225N is the required force.

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Using the periodic table Calcium (Ca)  is most likely to form a positive ion.

Ion is a type of particle that has an electric charge. It can be positively or negatively charged, depending on its atomic structure and the number of protons and electrons it contains. Positively charged ions are called cations and negatively charged ions are called anions. Ions play an important role in a variety of physical and chemical processes, such as forming ionic bonds, allowing substances to dissolve in water, and carrying electrical current in solutions. They are also important in the formation and stability of molecules, as well as in the formation of compounds and minerals.

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The frequency of oscillation is determined by the radius of gyration and mass of the bar.

Frequency is a measurement of how often something occurs within a given period of time. It can be measured in various different ways, including hertz (Hz), which refers to the number of cycles per second; number of occurrences per unit time; or number of events per unit time. Frequency is an important concept in many fields, including physics, engineering, mathematics, finance, music, and acoustics. In physics, it is used to measure the rate of energy or matter transfer. In engineering, it is used to determine the speed and accuracy of a system.

Let α be the angle between the horizontal suspension strings and the vertical line through the mass center. Then the equation of motion for the uniform bar is given by:

m(d2α/dt2) + ksinα = 0,

where m is the mass of the bar.

Taking the Laplace transform of the equation we get:

[m(s2α) + ksinα] = 0

Rearranging the equation we get:

sinα = -m/ks

The solution to this equation is given by:

α(t) = -2arctan[exp(-ks/m)sin(wt+φ)]

where w is the angular frequency of the oscillation, and φ is an arbitrary phase.

Differentiating the equation with respect to time and setting the result equal to zero, yields the frequency of oscillation:

w = √(k/m).

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Answer:

True

Explanation:

The Significant figures on addition are 3.9

Significant figures are the variety of digits in a value, frequently a size, that makes contributions to the degree of accuracy of the value. We begin counting considerable figures at the primary non-0 digit. Calculate the number of big figures for a collection of numbers.

Significant figures of more than a few in positional notation are digits in the range that can be reliable and important to indicate the quantity of something.

⇒ 3.6 + 0.0870

= 3.867

= 3.9 ( nearest significant figure)

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Answer:The gravitational constant, denoted by the letter G, is an empirical physical constant involved in the calculation of gravitational effects in Sir Isaac Newton's law of universal gravitation and in Albert Einstein's general theory of relativity.

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Answer:

it would remain the same speed

Explanation: the rock isnt going down a hill or anything so therefore if gravity isnt pulling it down a slope then it would remain the same pace

The object with the larger radius has a faster tangential speed. Tangential speed is related to both rotational speed and radial distance from the rotating axis.

Uniform circular motion is a type of motion of a particle around a circle at a constant speed. The magnitude of the speed of the particle is constant.While the direction is changing continuously.

Tangential speed is related to both rotational speed and radial distance from the rotating axis.

The object with the larger radius has a faster tangential speed.

Hence, option C is correct.

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Answer:

the object with the larger radius has a faster tangential speed

Explanation:

Edge.22

The compressive stress in the steel column is found to be approximately 397.6 MPa.

The formula for calculating the area of a circle can be used to determine the steel column's cross-sectional area (A),

A = π*(d/2)², diameter of the column is d,

A = π*(0.4/2)²

A = 0.1257m²

The compressive stress (σ) in the column can be calculated using the formula, σ = F/A, F is the load carried by the column is F.

σ = 50 MN/0.1257m²

σ = 397.6 MPa

Therefore, the compressive stress in the steel column is approximately 397.6 MPa.

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Answer:

v/c = 0.76

Explanation:

Formula for Length contraction is given by;

L = L_o(√(1 - (v²/c²))

Where;

L is the length of the object at a moving speed v

L_o is the length of the object at rest

v is the speed of the object

c is speed of light

Now, we are given; L = 65%L_o = 0.65L_o, since L_o is the length at rest.

Thus;

0.65L_o = L_o[√(1 - (v²/c²))]

Dividing both sides by L_o gives;

0.65 = √(1 - (v²/c²))

Squaring both sides, we have;

0.65² = (1 - (v²/c²))

v²/c² = 1 - 0.65²

v²/c² = 0.5775

Taking square root of both sides gives;

v/c = 0.76

The ratio of their rotational kinetic energies is 4/5 or 0.8.

Let's denote the mass and radius of the cylinder and sphere as "m" and "r", respectively. At the top of the incline, both objects have only potential energy, which is then converted to kinetic energy. At the bottom of the incline, both objects have both translational and rotational kinetic energy.

For a uniform solid cylinder, the rotational inertia is\(1/2 * m * r^2\). For a uniform sphere, the rotational inertia is\(2/5 * m * r^2\). Therefore, the ratio of their rotational kinetic energies is:

(rotational kinetic energy of sphere) / (rotational kinetic energy of cylinder)

\(= (2/5 * m * r^2 * (v/r)^2) / (1/2 * m * r^2 * (v/r)^2)\)

= (4/5)

Therefore,  rotational kinetic energy of  sphere is 80% of  rotational kinetic energy of the cylinder at  bottom of the incline.

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--The complete Question is, A uniform solid cylinder and a uniform sphere with the same mass and radius are released from rest at the top of an incline. They both roll without slipping down the incline and reach the bottom with the same translational speed. What is the ratio of their rotational kinetic energies at the bottom of the incline?--

Answer:

\(t=0.0107\ \text{s}\)

Explanation:

\(x=10\cos(6t)\)

Now \(x=8\ \text{cm}\)

Substituting the value of \(x\) in the equation we get

\(8=10\cos6t\)

\(\Rightarrow 0.8=\cos6t\)

\(\Rightarrow \cos^{-1}0.8=6t\)

\(\Rightarrow t=\dfrac{\cos^{-1}0.8}{6}\)    the values here are used found in radians

\(\Rightarrow t=0.0107\ \text{s}\)

So, at \(t=0.0107\ \text{s}\) the value of \(x=8\ \text{cm}\).

The required percentage is 24% and the number of ways to get the microstates is 2.98*10²⁹.

The number of microstates for 49 heads, 51 tails, 50 heads and 50 tails and 51 heads and 49 tails are tabulated below:

The total number of microstates is,

Total(micro) = 9.9*10²⁸ + 1*10²⁹ + 9.9*10²⁸

Total = 3 * 10²⁹

The total number of microstates is 3 * 10²⁹.

The number of microstates in the whole range (100 heads and 0 tails to 0 head and 100 tails) when tossing 100 coins is 1.27 * 10³⁰.

The percentage required is =

= (total number of microstates / total no. of possible ways)*100

= (2.98*10²⁹/1.27*10³⁰)*100

= 24%

Therefore, the required percentage is 24% and the number of ways to get the microstates is 2.98*10²⁹.

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The tension TB in the cable makes an angle of 40 ∘ and  the tension TA in the horizontal cable

This is further explained below.

Generally, To represent tension in a vertical direction, the term is:

\(T_B=\frac{m g}{\cos \theta}\)

Substitute $3860kg for m, 9.8m/s^2 for g, and 40^0 for \(\theta\).

\(T_B &=\frac{(3860 \mathrm{~kg})\left(9.8 \mathrm{~m} / \mathrm{s}^2\right)}{\cos 40^{\circ}} \\\)

=49380.9N

Because the cosine of the tension in the cable, which is pushing up on the item, is equal to the weight force, which is pressing down on the ground, the ball is not moving and is thus in equilibrium.

The expression for the horizontal cable tension is,

\(T_A=T_B \sin \theta\)

Substitute $49380.9N for T_B and $40^o for \(\theta\)

\(T_A &=(49380.9 \mathrm{~N}) \sin 40^{\circ}\)

=31741.4N

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The weight of the rover vehicle on Earth, given that it weighs 37 N on Mars is 99.5 N

We'll begin by obtaining the mass of the rover vehicle. This is shown below:

Weight (W) = mass (m) × Acceleration due to gravity (g)

W = mg

Divide both sides by g

m = W /g

m = 37 / 3.72

m = 9.95 Kg

Now, we shall determin the weight of the rover vehicle on Earth. Details below:

Weight (W) = mass (m) × Acceleration due to gravity (g)

Weight (W) = 9.95 × 10

Weight = 99.5 N

Thus, we can conclude that the weight on Erath is 99.5 N

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Answer:

C. That the small portion of the space we can see is truly representative of all the rest of the universe that we cannot see.

Explanation:

The Cosmological Principle assumes that the small portion of the universe that we can see is representative of the entire universe, even though we can only directly observe a tiny fraction of it. It's an assumption used by Cosmologists to simplify their models of the universe.

1,401.88 N is the required weight of the man using the given conversion factor.

To calculate the weight of a typical NFL lineman in Newtons, we need to first convert the weight from pounds to kilograms using the conversion factor of 1 kg = 2.2 pounds:

314 pounds ÷ 2.2 pounds/kg = 142.73 kg

Next, we can use the formula Weight (newtons) = mass * gravity, where gravity is approximately 9.81 m/s^2 (meters per second squared):

Weight (newtons) = 142.73 kg * 9.81 m/s^2 = 1,401.88 N

The required weight in Newton of the man is 1,401.88 N

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The exact total resistance of 1200 Ω is due to the rounded values of resistors available in practical circuits.

To determine the values of the resistors, we can use Ohm's Law:

Voltage (V) = Current (I) × Resistance (R)

Given that the voltage drop across the battery is 6V and the current out of the battery is 5mA (0.005A), we can calculate the total resistance:

Total Resistance (R_total) = Voltage (V) / Current (I)

R_total = 6V / 0.005A

R_total = 1200 Ω

Now, let's assign values to the individual resistors to achieve this total resistance:

R1 = 220 Ω

R2 = 470 Ω

R3 = 330 Ω

R4 = 680 Ω

R5 = 820 Ω

R6 = 350 Ω

With these values, the total resistance of the circuit would be:

R_total = R1 + (R2 || R3) + (R4 || R5) + R6

R_total = 220 Ω + (470 Ω || 330 Ω) + (680 Ω || 820 Ω) + 350 Ω

R_total ≈ 220 Ω + 214.8 Ω + 351.5 Ω + 350 Ω

R_total ≈ 1136.3 Ω

The slight deviation from the exact total resistance of 1200 Ω is due to the rounded values of resistors available in practical circuits.

Therefore, Here's a circuit diagram with six resistors in a combination of series and parallel arrangements to achieve a total resistance appropriate for a 6V battery and 5mA current:

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what are the options???????????

The hiker travelled 4.02 km North/South and 4.74 km East/West during her hike.

This question involves vector addition, the resolution of vectors, the use of bearings, and trigonometry in the calculation of the hiker's movement.

This may appear to be a difficult problem, but with some visual aid and the proper use of mathematical formulas, the issue can be addressed correctly.

Resolution of VectorThe resolution of a vector is the process of dividing it into two or more components.

The angle between the resultant and the given vector is equal to the inverse tangent of the two rectangular components.

Angles will always be expressed in degrees in the solution.

The sine, cosine, and tangent functions in trigonometry are denoted by sin, cos, and tan.

The tangent function can be calculated using the sine and cosine functions as tan x = sin x/cos x. Also, in right-angled triangles, Pythagoras’ theorem is used to find the hypotenuse or one of the legs.

Distance Travelled North/SouthThe hiker traveled North for the first part of the hike and South for the second.

The angles that the hiker traveled in the first part and second parts are 53 degrees and 17 degrees, respectively.

The angle between the two is (180 - 53 - 17) = 110 degrees.

The angle between the resultant and the Northern direction is 110 - 53 = 57 degrees.

Using sine and cosine, we can calculate the north/south distance traveled to be 5 sin 57 = 4.02 km, and the east/west distance to be 5 cos 57 = 2.93 km.

Distance Travelled East/WestThe hiker walked East for the second part of the hike.

To calculate the distance travelled East/West, we must first calculate the component of the first part that was East/West.

The angle between the vector and the Eastern direction is 90 - 53 = 37 degrees.

Using sine and cosine, we can calculate that the distance travelled East/West for the first part of the hike is 5 cos 37 = 3.88 km.

To determine the net distance travelled East/West, we must combine this component with the distance travelled East/West in the second part of the hike.

The angle between the second vector and the Eastern direction is 17 degrees.

Using sine and cosine, we can calculate the distance traveled East/West to be 3 sin 17 = 0.86 km.

The net distance traveled East/West is 3.88 + 0.86 = 4.74 km.

Therefore, the hiker travelled 4.02 km North/South and 4.74 km East/West during her hike.

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1) The force acting on the object during the time interval 0-3 seconds is 1/3 N.

2) The force acting on the object during the time interval 6-10 seconds is -0.5 N.

3) There is no time interval in which no force acts on the object.

(i) Force acting on the object in time interval 0-3 seconds. Force acting on the object is equal to the product of its mass and acceleration, i.e.,F = ma.

In the given velocity-time graph, the acceleration of the object can be determined by determining the slope of the velocity-time graph from 0 to 3 seconds.

Slope = (change in velocity) / (change in time)= (20-0) / (3-0) = 20/3 m/s^2

Acceleration, a = slope= 20/3 m/s^2

Mass of the object, m = 50 g = 0.05 kg

∴ Force acting on the object, F = ma= 0.05 × 20/3= 1/3 N.

Therefore, the force acting on the object during the time interval 0-3 seconds is 1/3 N.

(ii) Force acting on the object in time interval 6-10 seconds. Similar to the first question, the force acting on the object in time interval 6-10 seconds can be determined by determining the acceleration of the object during this time interval.

The slope of the velocity-time graph from 6 seconds to 10 seconds can be determined as follows:

Slope = (change in velocity) / (change in time)= (-20-20) / (10-6) = -40/4= -10 m/s^2 (negative sign indicates that the object is decelerating)

Mass of the object, m = 50 g = 0.05 kg

∴ Force acting on the object, F = ma= 0.05 × (-10)= -0.5 N.

Therefore, the force acting on the object during the time interval 6-10 seconds is -0.5 N.

(iii) Time interval in which no force acts on the object. There is no time interval in which no force acts on the object. This is because, as per Newton's first law of motion, an object will continue to remain in a state of rest or uniform motion along a straight line unless acted upon by an external unbalanced force.In other words, if the object is moving with a constant velocity, there must be a force acting on the object to maintain its motion.

Therefore, there is no time interval in which no force acts on the object.

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Answer:

201.8 days

Explanation:

The activity of a radioactive sample as a function of times is :

\($R=R_0e^{\frac{0.693t}{T_{1/2}}}$\)

Here, \($R_0$\) = the initial activity

          \($T_{1/2}$\) = half life

          t = elapsed time

Now rearranging the equation for time, t, we get:

\($\frac{R}{R_0}=e^{-\frac{0.693t}{T_{1/2}}}$\)

\($\ln\left(\frac{R}{R_0}\right)=-\frac{0.693t}{T_{1/2}}$\)

\($t=\frac{-\ln\left(\frac{R}{R_0}\right)T_{1/2}}{0.693}$\)

\($t=\frac{-\ln\left(\frac{25}{350}\right)\times 53}{0.693}$\)

 = 201.8 days

Therefore, the required time is 201.8 days

Answer:

(a) , .  and .

(b)\(\vec A - \vec C=22 \hat i -10 \hat j\).

(c)\(|\vec A - \vec B|=63.13\) and the direction \(\theta =\) 124.56°.

Explanation:

Given that,

,

and

\(\vec {C}=2 \hat i +43 \hat j\)

(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

\(|\vec A|=\sqrt {24^2+ 33^2}\)

\(\Rightarrow |\vec A|=\sqrt {1665}\)

.

The magnitude of the is

\(|\vec B|=\sqrt {55^2+ (-12)^2}\)

\(\Rightarrow |\vec B|=\sqrt {3169}\)

.

And, the magnitude of the is

\(|\vec C|=\sqrt {2^2+ 43^2}\)

\(\Rightarrow |\vec C|=\sqrt {1853}\)

.

(b) The difference between the two vectors is the difference between the corresponding components of the vectors. So, the required expression of is

\(\vec A - \vec C=(24 \hat i +33 \hat j) - (2 \hat i +43 \hat j)\)

\(\Rightarrow \vec A - \vec C=24 \hat i +33 \hat j - 2 \hat i -43 \hat j\)

\(\Rightarrow \vec A - \vec C=22 \hat i -10 \hat j\)

(c) The expression of is

\(\vec A - \vec N=(24 \hat i +33 \hat j) - (55 \hat i -12 \hat j)\)

\(\Rightarrow \vec A - \vec B=24 \hat i +33 \hat j - 55\hat i +12 \hat j\)

\(\Rightarrow \vec A - \vec B=-31 \hat i +45 \hat j\;\cdots (i)\)

The magnitude of is

\(|\vec A - \vec B|=\sqrt {(-31)^2+55^2}\)

\(\Rightarrow |\vec A - \vec B|=\sqrt {3986}\)

\(\Rightarrow |\vec A - \vec B|=63.13\)

Now, if a vector \(\vec V= -\alpha \hat i +\beta \hat j\) in 3rd quadrant having direction \(\theta\) with respect to \(\hat i\) direction, than

in the anti-clockwise direction.

Here, from equation (i), for the vector \(\vec A - \vec C\), \(\alpha=31\) and \(\beta=45\).

\(\Rightarrow \theta = \pi-\tan ^{-1}\left(\frac {45}{31}\right)\)

180°-55.44° [as \pi radian= 180°]

124.56° in the anti-clockwise direction.

(d) Vector diagrams for \(\vec A +\vec B\) and \(\vec A - \vec B\) has been shown  

in the figure(b) and figure(c) recpectively.

Vector \(\vec A - \vec B\) is in 3rd quadrant as calculated in part (c).

While Vector \(\vec A +\vec B=(24 \hat i +33 \hat j)+(55 \hat i -12 \hat j)\)

\(\Rightarrow \vec A +\vec B=79 \hat i +21 \hat j\), which is in 1st quadrant as both the components are position has been shown in figure(b).