It takes 525 J of work to compress a spring 25 cm. What is the force constant of the spring (in kN/m)?
Answer:
1.680kN/m
Explanation:
Work done by the spring is expressed as shown:
[tex]W = \frac{1}{2}ke^2[/tex] where:
k is the spring constant
e is the extension
Given
W = 525Joules
extension = 25cm = 0.25m
Substitute into the formula:
[tex]525 = \frac{1}{2}k(0.25)^{2} \\525 = \frac{0.0625k}{2}\\ 525 = 0.03125k\\k = \frac{525}{0.3125}\\k = 1680N/m\\k = 1.680kN/m[/tex]
Hence the force constant of the spring is 1.680kN/m
What (rather remarkable!) equation relates the speed of light to other fundamental electromagnetic constants?
Complete Question
The complete question is shown on the first uploaded image
Answer:
The equation is [tex]c = \frac{1}{\sqrt{ \mu_o * \epsilon_o} }[/tex]
The value of c is [tex]c = 2.998 *10^{8} \ m/s [/tex]
Explanation:
From the question we are told that
Generally the equation that relates the speed of light to other fundamental electromagnetic constants is
[tex]c = \frac{1}{\sqrt{ \mu_o * \epsilon_o} }[/tex]
Here c is the speed of light
[tex]\mu_o[/tex] is the permeability of free space with value
[tex]\mu_o = 4\pi *10^{-7} \ N/A^2[/tex]
and [tex]\epsilon_o[/tex] is the permittivity of free space with value
[tex]\epsilon_o = 8.85*10^{-12} \ C/V \cdot m[/tex]
So
[tex]c = \frac{1}{\sqrt{ 4\pi *10^{-7} * 8.85*10^{-12}} }[/tex]
=> [tex]c = 2.998 *10^{8} \ m/s [/tex]
Based on the information in the table, which elements are most likely in the same periods of the periodic table?
Answer:
Just to help, periods on the periodic table are those running horizontally from left to right
Answer:
The answer is A.Boron and carbon are likely together in one period because they have very close atomic numbers, while gallium and germanium are likely together in another period because they have very close atomic numbers.
Explanation:
just took test
Question C) needs to be answered, please help (physics)
(a) Differentiate the position vector to get the velocity vector:
r(t) = (3.00 m/s) t i - (4.00 m/s²) t² j + (2.00 m) k
v(t) = dr/dt = (3.00 m/s) i - (8.00 m/s²) t j
(b) The velocity at t = 2.00 s is
v (2.00 s) = (3.00 m/s) i - (16.0 m/s) j
(c) Compute the electron's position at t = 2.00 s:
r (2.00 s) = (6.00 m) i - (16.0 m) j + (2.00 m) k
The electron's distance from the origin at t = 2.00 is the magnitude of this vector:
||r (2.00 s)|| = √((6.00 m)² + (-16.0 m)² + (2.00 m)²) = 2 √74 m ≈ 17.2 m
(d) In the x-y plane, the velocity vector at t = 2.00 s makes an angle θ with the positive x-axis such that
tan(θ) = (-16.0 m/s) / (3.00 m/s) ==> θ ≈ -79.4º
or an angle of about 360º + θ ≈ 281º in the counter-clockwise direction.
Why does the brightness of a bulb not change noticeably when you use longer copper wires to connect it to the battery?
a. All the current is used up in the bulb, so the connecting wires don't matter.
b. Very little energy is dissipated in the thick connecting wires.
c. Electric field in the connecting wires is zero, so emf = E_bulb * L_bulb.
d. Current in the connecting wires is smaller than current in the bulb.
e. The electric field in connecting wires is very small, so emf almost = E_ bulb * L_bulb.
Answer:
Options B & E are correct
Explanation:
Looking at all the options, B & E are the correct ones.
Option B is correct because the thicker the wire per unit length, the lesser resistance it will posses and the lesser the energy that will be dissipated by the wire and in return more energy will be dissipated by the bulb.
Option E is also correct because the resistance of the copper wires is low enough to ensure that there's not much drop in voltage across the copper wires. Thus, there will not be any noticeable differences in the voltage across the bulb.
Option A is not correct because the current is not used up and thus the charge is conserved, and it will circulate just through the circuit.
Option C is not correct because although the Electric field along the wire is not zero, it is very small.
Option D is not correct because the wires and the light bulb are connected in series and as such, the current in both the wires and the light bulb will be identical.
The brightness of a bulb that not change noticeably when you use longer copper wires to connect it to the battery is :
b. Very little energy is dissipated in the thick connecting wires.
e. The electric field in connecting wires is very small, so emf almost = E_ bulb * L_bulb.
"Energy"The brightness of a bulb that not change noticeably when you use longer copper wires to connect it to the battery is very little energy is dissipated in the thick connecting wires and the electric field in connecting wires is very small, so emf almost = E_ bulb * L_bulb.
The thicker the wire per unit length, the lesser resistance it'll posses and the lesser the vitality that will be scattered by the wire and in return more vitality will be disseminated by the bulb.
The resistance of the copper wires is low sufficient to guarantee that there's not much drop in voltage over the copper wires. Hence, there will not be any noticeable contrasts within the voltage over the bulb.
Thus, the correct answer is B and E.
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A plane travelling at 100 m/s accelerates at 5 m/s² for a distance of 125 m. What is the final velocity of the plane?
Analyzing the question:
We are given:
initial velocity (u) = 100 m/s
final velocity (v) = v m/s
distance (s) = 125 m
acceleration (a) = 5 m/s²
Solving for Final Velocity (v):
from the third equation of motion:
v² - u² = 2as
v² - (100)² = 2(5)(125)
v² - 10000 = 1250
v² = 1250 + 10000
v² = 11250
v = 106.06 m/s
You release a ball from rest at the top of a ramp. 6 s later it is moving at 4.0
m/s. What is the acceleration? (in meters per second squared) *
Your answer
[tex]a = \frac{vf - vi}{t} [/tex]
here initial velocity vi=0 as ball release from rest
the final velocity is vf=4.0
time is t=6
so putting all these values in above equation
[tex]a = \frac{ 4.0- 0}{6} [/tex]
[tex]a = 0.6667m \s {}^{2} [/tex]
Question 1-1: In each case, lifting or pushing, why must you exert a force to keep the object moving at a constant velocity?
Answer:
We must apply a force to keep the object moving at a constant velocity due to gravitational force or weight (in case of lifting), and due to frictional force (in case of pushing).
Explanation:
LIFTING:
When an object is lifted, we first need to overcome the force exerted on it by the field of gravity. Due to this force, which is also called the weight of object, we must apply a force on the object to keep it moving at constant speed, otherwise the gravity force will cause the object to slow down and eventually fall back on ground.
PUSHING:
When pushing an object the person must apply the force to first overcome the frictional force. The frictional force acts in opposite direction of motion. Thus, to move the object at constant speed we must apply force to it.
Hence, we must apply a force to keep the object moving at a constant velocity due to gravitational force or weight (in case of lifting), and due to frictional force (in case of pushing).
If you weigh 660 N on the earth, what would be your weight on the surface of a neutron star that has the same mass as our sun and a diameter of 20.0 km? Take the mass of the sun to be 1.99×10^30, the gravitational constant to be G = 6.67×10^−11Nm^2/kg^2, and the acceleration due to gravity at the earth's surface to be g = 9.810 m/s^2.p
Answer:
8.93*10^13 N.
Explanation:
Assuming that in this case, the weight is just the the force exerted on you by the mass of the star, due to gravity, we can apply the Universal Law of Gravitation:[tex]F_{g}= \frac{G*m_{1}*m_{s}}{r_{s}^{2} }[/tex]
where, m1 = mass of the man = 660 N / 9.81 m/s^2 = 67.3 kg, ms = mass of the star = 1.99*10^30 kg, G= Universal Constant of Gravitation, and rs= radius of the star = 10.0 km. = 10^4 m.Replacing by the values, we get:[tex]F_{g}= \frac{6.67e-11Nm^2/kg^2*1.99e30 kg*67.3 kg}{10e4m^2} = 8.93e13 N[/tex]
Fg = 8.93*10^13 N.7. A 1,500-N force is applied to a 1,000-kg car. What is the car's acceleration?
Answer:
1.5m/s^2
Explanation:
Answer:
1.5 m/s2. accerelation =force ÷mass
help me get the answer in Physical Science.
Answer:
lithium
Explanation:
I took physical science 2 years ago and passed with an A
While making some observations at the top of the 66 m tall Astronomy tower, Ron
accidently knocks a 0.5 kg stone over the edge. How long will a student at the bottom
have to get out of the way before being hit?
Analysing the question:
Since the stone was dropped, there was no initial velocity applied on it and hence it's initial velocity of the stone is 0 m/s
We are given:
height of the tower (h) = 66 m
mass of the stone (m) = 0.5 kg
initial velocity of the stone (u) = 0 m/s
time taken by the stone to reach the ground (t) = t seconds
acceleration due to gravity = 10 m/s²
** Neglecting air resistance**
Finding the time taken by the stone to reach the ground:
from the second equation of motion
h = ut + 1/2at²
replacing the variables
66 = (0)(t) + 1/2 (10)(t)²
66 = 5t²
t² = 13.2
t = 3.6 seconds
I initially wanted to subtract the height of the student from the height of the tower since the time i calculated is the time taken by the stone to reach the ground and that means that the stone has already hit the student before 3.6 seconds
but since we were NOT given the height of a student, the person who posed this question wants the time taken by the stone to reach the ground and that is what we solved
The emf of the battery is 1.5 V. In Nichrome there are 9 × 1028 mobile electrons per m3, and the mobility of mobile electrons is 7 × 10−5 (m/s)/(N/C). Each thick wire has length 29 cm = 0.29 m and cross-sectional area 9 × 10−8 m2. The thin wire has length 6 cm = 0.06 m and cross-sectional area 1.3 × 10−8 m2. (The total length of the three wires is 64 cm.) In the steady state, calculate the number of electrons entering the thin wire every second. Do not make any approximations, and do not use Ohm's law or series-resistance equations.
Answer:
The number of electrons entering the thin wire every second is 1.75 x 10⁻³ mobile electrons / second
Explanation:
Given;
emf of the battery, V = 1.5 V
electron density, = 9 × 10²⁸ mobile electrons per m³
mobility of electron, u = 7 × 10⁻⁵ (m/s)/(N/C)
length of thin wire, L = 6 cm = 0.06 m
cross sectional area of the thin wire, A = 1.3 x 10⁻⁸ m²
The magnitude of the electric field in the thin wire is given by;
E = V/L
E = (1.5) / (0.06)
E = 25 N/C
the number of electrons entering the thin wire every second is given by;
[tex]e/s = mobility \ x \ Electric \ field\\\\number \ of \ electrons \ per \ second =\frac{7*10^{-5} (m/s)}{N/C} *25 (N/C)\\\\number \ of \ electrons \ per \ second = 1.75*10^{-3} \ m/s[/tex]
Therefore, the number of electrons entering the thin wire every second is 1.75 x 10⁻³ mobile electrons / second
The number of electrons entering the thin wire every second is 1.75 x 10⁻³ mobile electrons / second
Calculation of the number of electrons:Since
emf of the battery, V = 1.5 V
electron density, = 9 × 10²⁸ mobile electrons per m³
mobility of electron, u = 7 × 10⁻⁵ (m/s)/(N/C)
length of thin wire, L = 6 cm = 0.06 m
cross sectional area of the thin wire, A = 1.3 x 10⁻⁸ m²
So here the magnitude should be
E = V/L
E = (1.5) / (0.06)
E = 25 N/C
Now the number of electrons should be
= 7 × 10⁻⁵ *25
= 1.75 x 10⁻³ mobile
hence, The number of electrons entering the thin wire every second is 1.75 x 10⁻³ mobile electrons / second
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A soccer player kicking a ball; the ball soaring through the air and landing on the ground
the peripheral nervous system is responsible for both sending and receiving signals to and from the brain
Answer:
its true trust me
Explanation:
Answer: true
Explanation: edge
An FM radio station, 20 miles away, broadcast at a 93.4 MHz frequency(a) What is the wavelength of the radio wave associated with this signal ?(b) How long does it take for the signal to reach your radio from the station ?
Answer:
(a) Wavelength = 3.21 m (b) Time = [tex]1.07\times 10^{-4}\ s[/tex]
Explanation:
Given that,
The frequency of FM radio station, f = 93.4 MHz
(a) We need to find the wavelength of the radio wave associated with this signal. The relation between wavelength and frequency is given by :
[tex]c=f\lambda\\\\\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{93.4\times 10^6}\\\\\lambda=3.21\ m[/tex]
(b) It is given that, an FM radio station, 20 miles away. Let t is time taken for signal to reach your radio from the station. So,
[tex]t=\dfrac{d}{c}\\\\t=\dfrac{20\times 1609.34}{3\times 10^8}\\\\t=1.07\times 10^{-4}\ s[/tex]
Hence, this is the required solution.
21. Prediction: If you were to measure the current at points A, B and C, how do you think the values would compare? Why? 22. Prediction: If you were to measure the potential differences across these bulbs (what the voltmeter measures) how do you think the values will compare to each other and to the potential difference across the battery pack or the power supply? Why?
Answer:
hello your question is incomplete attached below is the complete question
21) The current at points B and C would be the same ( identical bulbs) while the current at Point A will be greater than the currents at point B and C. i.e. twice the current at either point B or point C
22) The potential difference across the bulbs will be the same and this is because the bulbs are connected in parallel to the the power source ( battery)
hence the voltage in the battery will be equal to the voltage across each bulb
Explanation:
The current at points B and C would be the same ( identical bulbs) while the current at Point A will be greater than the currents at point B and C. i.e. twice the current at either point B or point C
The potential difference across the bulbs will be the same and this is because the bulbs are connected in parallel to the the power source ( battery)
hence the voltage in the battery will be equal to the voltage across each bulb
A lamp of mass m hangs from a spring scale which is attached to the ceiling of an elevator. When the elevator is stopped at the fortieth floor, the scale reads mg. What does it read as the elevator slows down to stop at the ground floor?
a. more than mg
b. mg
c. less than mg
d. zero
e. can't tell
Answer:
The correct answer is (a)
Explanation:
A spring scale measures the weight of an object not the mass because according to hooke's law the extension of a spring is directly proportional to the load or force attached/applied to it. The force of gravity acting on the mass of any substance as it goes up actually reduces and increases as it comes down.
If F = ma, as a increases, F will also increase and vice versa
Where F = force
m = mass
a = acceleration (due to gravity in this case)
From the above explanation, it can be deduced that the scale will read more than mg as it gets to the ground because of an increase in the force of gravity (which also increases a) as it approaches the ground.
a jogger travels at 4 m/s for 100 s what is the distance covered
400m
Explanation:
given,
v= 4m/s
t= 100s
d= ?
since, v = d / t
therefore, d = v * t (velocity multiplied by time)
=> d = 4 * 100
= 400m.
For both resonance curves and Fourier spectra, amplitude is plotted vs frequency, but these two types of plots are not the same. Describe how they are different.
Answer:
he peaks are the natural frequencies that coincide with the excitation frequencies and in the second case they are the natural frequencies that make up the wave.
Explanation:
In a resonance experiment, the amplitude of the system is plotted as a function of the frequency, finding maximums for the values where some natural frequency of the system coincides with the excitation frequency.
In a Fourier transform spectrum, the amplitude of the frequencies present is the signal, whereby each peak corresponds to a natural frequency of the system.
From this explanation we can see that in the first case the peaks are the natural frequencies that coincide with the excitation frequencies and in the second case they are the natural frequencies that make up the wave.
A small compass is held horizontally, the center of its needle has a distance of 0.270 m directly north
of a long wire that is perpendicular to the Earth's surface. When there is no current in the wire, the
compass needle points due north, which is the direction of the horizontal component of the Earth's
magnetic field at that location. This component is parallel to the Earth's surface. When the current in
the wire is 26.3 A, the needle points 22.9∘ east of north.
(a) Does the current in the wire flow toward or away from the Earth's surface? ( 2 marks)
(b) What is the magnitude of the horizontal component of the Earth's magnetic field at the location of
the compass? (3 marks)
Answer:
Explanation:
The needle is showing north south direction . when current starts flowing in the wire which is held vertical to the ground , it deflects towards east .
a )
Therefore a magnetic field towards east has been created . It is possible only if current flows towards the surface in the vertical wire .
b )
magnetic field created at the magnetic needle B = 10⁻⁷ x 2I / d where I is current and d is distance .
B = 10⁻⁷ x 2 x 26.3 / .27
= 194.81 x 10⁻⁷ T
angle of deflection of solenoid = 22.9°
Tan 22.9 = B /H
.422 = 194.81 x 10⁻⁷ / H
H = 461.63 x 10⁻⁷ T
= .46 x 10⁻⁴ T .
A) The current in the wire flows towards the Earth's surface
B) The magnitude of the horizontal component of the Earth's magnetic field is : 0.46 x 10⁻⁴ T
A) The compass needle held horizontally points in a North-south direction of the earth and also deflects eastwards when current is allowed to flow through it. The deflection of the needle indicates the presence/generation of a magnetic field on the earth surface. which is facilitated by the flow of the current in the wire towards the Earth's surface
B) Determine The magnitude of the horizontal component of the Earth's magnetic field
B ( magnetic field ) = 10⁻⁷ * 2I / d ---- ( 1 )
where : l = 26.3 A, d = 0.27 m
Back to equation ( 1 )
B = 10⁻⁷ * 2 * 26.3 / 0.27
= 194.81 * 10⁻⁷ T
Final step : Calculate the magnitude of horizontal component ( H )
Tan ∅ = B / H ---- ( 2 )
where : ∅ ( angle of deflection ) = 22.9°
∴ H = B / Tan ( 22.9° )
= ( 194.81 * 10⁻⁷ ) / 0.422
= 0.46 x 10⁻⁴ T
Hence we can conclude that The current in the wire flows towards the Earth's surface and The magnitude of the horizontal component of the Earth's magnetic field is : 0.46 x 10⁻⁴ T
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what is the force produced on a body of 30kg mass when a body moving with the velocity of 26km/hr is acceleted to gain the velocity of 54 km/hr in 4 sec
Answer:
F = 58.35 [N]
Explanation:
To solve this problem we must use Newton's second law, which tells us that force is equal to the product of mass by acceleration. But first we must use the following equation of kinematics.
We have to convert speeds from kilometers per hour to meters per second
[tex]\frac{26km}{hr}*\frac{1000m}{1km}*\frac{1hr}{3600s}=\frac{7.22m}{s} \\\frac{54km}{hr}*\frac{1000m}{1km}*\frac{1hr}{3600s}=15\frac{m}{s}[/tex]
[tex]v_{f}=v_{o}+(a*t) \\[/tex]
where:
Vf = final velocity = 15 [m/s]
Vi = initial velocity = 7.22 [m/s]
a = acceleration [m/s^2]
t = time = 4 [s]
Note: the positive sign of the above equation is because the car increases its speed
15 = 7.22 + (a*4)
a = 1.945 [m/s^2]
Now we can use the Newton's second law:
F = m*a
F = 30*1.945
F = 58.35 [N]
You are working out on a rowing machine. Each time you pull the rowing bar (which simulates the oars) toward you, it moves a distance of 1.1 m in a time of 1.8 s. The readout on the display indicates that the average power you are producing is 90 W. What is the magnitude of the force that you exert on the handle?
Answer:
147.27N
Explanation:
Power = workdone/time
Power = Force*distance/time
Given
Power = 90Watts
Distance = 1.1m
Time = 1.8secs
Force = ?
Substitute the given parameters into the formula:
[tex]90 = \frac{1.1d}{1.8}\\cross \ multiply\\ 90 \times 1.8 = 1.1F\\162 = 1.1F\\1.1F = 162\\F = \frac{162}{1.1} \\F = 147.27N[/tex]
Hence the magnitude of the force that you exert on the handle is 147.27N
If it takes you 5 minutes to dry your hair using a 1200-W hairdryer plugged into a 120-V power outlet, how many Coulombs of charge pass through your hair dryer
Answer:
The charge pass through your hair dryer is 3000 C.
Explanation:
Given that,
Power = 1200 W
Voltage = 120 V
Flow time = 5 min
We need to calculate the current
Using formula of power
[tex]P=VI[/tex]
[tex]I=\dfrac{P}{V}[/tex]
Put the value into the formula
[tex]I=\dfrac{1200}{120}[/tex]
[tex]I=10\ A[/tex]
We need to calculate the charge pass through your hair dryer
Using formula of current
[tex]I=\dfrac{Q}{t}[/tex]
[tex]Q=It[/tex]
Put the value into the formula
[tex]Q=10\times5\times60[/tex]
[tex]Q=3000\ C[/tex]
Hence, The charge pass through your hair dryer is 3000 C.
A coin rests on a record 0.15 m from its center. The record turns on a turntable that rotates at variable speed. The coefficient of static friction between the coin and the record is 0.30.
Required:
What is the maximum coin speed at which it does not slip?
Answer:
0.66m/sExplanation:
We are expected to solve for the velocity with no slip condition
we know that the expression that relate coefficient of friction and velocity is given as
μs = v^2/rg
Given
coefficient of friction μs = 0.3
radius r= 0.15
assume g=9.81m/s^2
substituting into the expression we have
0.3= v^2/0.15*9.81
v^2=0.3*0.15*9.81
v^2=0.44145
v=√0.44145
v=0.66
therefore the velocity is 0.66m/s
How much voltage (in terms of the power source voltage bV) will the capacitor have when it has started at zero volts potential difference, it is connected to the power supply and resistor and onehalf the characteristic time has passed (i.e. t= T(tau)/2)?
Answer:
The voltage is [tex]V = 0.993V_b[/tex]
Explanation:
From the question we are told that
The time that has passed is [tex]t = \frac{\tau}{2}[/tex]
Here [tex]\tau[/tex] is know as the time constant
The voltage of the power source is [tex]V_b[/tex]
Generally the voltage equation for charging a capacitor is mathematically represented as
[tex]V = V_b [1 - e^{- \frac{t}{\tau} }][/tex]
=> [tex]V = V_b [1 - e^{- \frac{\frac{\tau}{2}}{\tau} }][/tex]
=> [tex]V = V_b [1 - e^{- \frac{\tau}{2\tau} }][/tex]
=> [tex]V = V_b [1 - e^{- \frac{1}{2} }][/tex]
=> [tex]V = 0.993V_b[/tex]
If the shoe has less mass, it will experience _______________ (more, less, the same) friction as it would with more mass.
What are two ways that an object can have kinetic energy?
Answer:
The object has to have mass and speed
Explanation:
You can increase both speed and mass to increase the kinetic energy, hope this answers your question.
Happy Halloween!
When a parachute opens, the air exerts a large drag force on it. This upward force is initially greater than the weight of the sky diver and, thus, slows him down. Suppose the weight of the sky diver is 915 N and the drag force has a magnitude of 1061 N. The mass of the sky diver is 93.4 kg. Take upward to be the positive direction. What is his acceleration, including sign
Explanation:
According to newton's second law of motion.
[tex]\sum Fx = ma\\\\\sum Fx = 1061 - 915\\\\\sum Fx = 146N[/tex]
m is the mas of the sky diver = 93.4kg
a is the acceleration of the skydiver
From the formula above;
[tex]a = \frac{\sum Fx}{m}\\ \\a = \frac{146}{93.4}\\\\a = 1.563m/s^2[/tex]
Hence the acceleration of the sky diver is 1.563m/s²
. Suppose you walk 18.0 m straight west and then 25.0 m straight north. How far are you from your starting point? What is your displacement vector? What is the direction of your displacement? Assume the +x-axis is to the east.
Answer:
Explanation:
The total distance is how far you walk from the starting point.
Distance through west = 18.0m
Distance through north = 25.0m
Total distance covered = 18.0+25.0m
Total distance covered = 43.0m
This means that I am 43.0m from the starting point
Displacement is the distance covered in a specified direction. The displacement will be gotten using the Pythagoras theorem as shown:
[tex]d^2 = 25^2 + 18^2\\d^2 = 625+324\\d^2 = 949\\d = \sqrt{949}\\ d = 30.81m[/tex]
The direction of your displacement is 30.81m
Direction is gotten according to the formula;
[tex]\theta = tan ^{-1}{\frac{y}{x} }\\\theta = tan ^{-1}{\frac{25}{-18} }\\\theta = tan ^{-1}-1.3889}\\\theta = -60.27^0\\\theta = 180-60.27\\\theta = 119.7^0[/tex]
Note that the direction to the west is negative, that is why the x is -18.0m
The distance from the starting point is 43 m, the displacement vector is 30.81 m and the direction of the displacement is 119.7 degrees.
Given-
Distance travel through the west is 18 m.
Distance travel through the north is 25 m.
Distance from starting point-
To know the total distance, add both the covered distance. Thus total distance x is,
[tex]x=18+25[/tex]
[tex]x=43[/tex]
Hence, the distance from the starting point is 43 m.
The displacement vector-
Displacement is calculated as the shortest distance between starting and final point. This shortest distance can be calculated using the Pythagoras theorem which states that in a right-angled triangle, the square of the hypotenuse [tex]d[/tex] is equal to the sum of the squares of the other two sides. Therefore,
[tex]d^2=18^2+25^2[/tex]
[tex]d^2=324+625[/tex]
[tex]d^2=949[/tex]
[tex]d=\sqrt{949}[/tex]
[tex]d=30.81[/tex]
The displacement vector is 30.81 m.
The Direction of displacement-The direction of displacement [tex]\theta[/tex] with these two sides can be calculated with the formula,
[tex]\theta=tan^{-1}\dfrac{25}{-18}[/tex]
Here due to the west direction(opposite side), the sign is taken negatively.
[tex]\theta=tan^{-1}(-1.389)[/tex]
[tex]\theta=-60.27^o[/tex]
For the other quarter,
[tex]\theta=180-60.27=119.7^o[/tex]
Hence, the distance from the starting point is 43 m, the displacement vector is 30.81 m and the direction of the displacement is 119.7 degrees.
For more about the displacement, follow the link below-
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