The atomic radii of a divalent cation and a monovalent anion are 0.19 nm and 0.126 nm, respectively. (a) Calculate the force of attraction between these two ions at their equilibrium interionic separation (i.e., when the ions just touch one another). Enter your answer for part (a) in accordance to the question statement N (b) What is the force of repulsion at this same separation distance

Answer:

the critical angle of the flint glass is 37.04⁰

Explanation:

to calculate the critical angle for total internal reflection.

given,

wavelength of the flint glass =  λ  = 580.0 nm

                                                       = 580 × 10⁻⁹ m        

critical angle  = sin^{-1}(\dfrac{\mu_a}{\mu_g})

at the wavelength of 580.0 nm the refractive index of the glass is 1.66

refractive index of air = 1                        

critical angle  = sin^{-1}(\dfrac{1}{1.66})

                      = 37.04⁰              

hence, the critical angle of the flint glass is 37.04⁰

Answer:

Thickness = 123.19 nm

Explanation:

Given that:

The refractive index of the glass = 1.50

The refractive index of thin layer of magnesium fluoride = 1.38

The wavelength of the light = 680 nm

The thickness can be calculated by using the formula shown below as:

[tex]Thickness=\frac {\lambda}{4\times n}[/tex]

Where, n is the refractive index of thin layer of magnesium fluoride = 1.38

[tex]{\lambda}[/tex] is the wavelength

So, thickness is:

[tex]Thickness=\frac {680\ nm}{4\times 1.38}[/tex]

Thickness = 123.19 nm

Answer:

[tex]D=V*\sqrt{\frac{2H}{g} } -\frac{H}{4}[/tex]

Explanation:

From the vertical movement, we know that initial speed is 0, and initial height is H, so:

[tex]Y_{f}=Y_{o}-g*\frac{t^{2}}{2}[/tex]

[tex]0=H-g*\frac{t^{2}}{2}[/tex]    solving for t:

[tex]t=\sqrt{\frac{2H}{g} }[/tex]

Now, from the horizontal movement, we know that initial speed is V and the acceleration is -g/4:

[tex]X_{f}=X_{o}+V*t+a*\frac{t^{2}}{2}[/tex]   Replacing values:

[tex]D=V*\sqrt{\frac{2H}{g} }-\frac{g}{4}*\frac{1}{2} *(\sqrt{\frac{2H}{g} })^{2}[/tex]

Simplifying:

[tex]D=V*\sqrt{\frac{2H}{g} } -\frac{H}{4}[/tex]

Answer:

Distance between Karl and Joe is 38.467 m

Solution:

Let us assume that you are at origin

Now, as per the question:

Joe's tent is 19 m away from yours in the direction [tex]19.0^{\circ}[/tex] north of east.

Now,

Using vector notation for Joe's location, we get:

[tex]\vec{r_{J}} = 19cos(19.0^{\circ})\hat{i} + 19sin(19.0^{\circ})\hat{j}[/tex]

[tex]\vec{r_{J}} = 17.96\hat{i} + 6.185\hat{j} m[/tex]

Now,

Karl's tent is 45 m away from yours and is in the direction [tex]39.0^{\circ}[/tex]south of east, i.e.,  [tex]- 39.0^{\circ}[/tex] from the positive x-axis:

Again,  using vector notation for Karl's location, we get:

[tex]\vec{r_{K}} = 45cos(-319.0^{\circ})\hat{i} + 45sin(- 39.0^{\circ})\hat{j}[/tex]

[tex]\vec{r_{K}} = 34.97\hat{i} - 28.32\hat{j} m[/tex]

Now,  obtain the vector difference between [tex]\vec{r_{J}}[/tex] and [tex]\vec{r_{K}}[/tex]:

[tex]\vec{r_{K}} - \vec{r_{J}} = 34.97\hat{i} - 28.32\hat{j} - (17.96\hat{i} + 6.185\hat{j}) m[/tex]

[tex]\vec{d} = \vec{r_{K}} - \vec{r_{J}} = 17.01\hat{i} - 34.51\hat{j} m[/tex]

Now, the distance between Karl and Joe, d:

|\vec{d}| = |17.01\hat{i} - 34.51\hat{j}|

[tex]d = \sqrt{(17.01)^{2} + (34.51)^{2}} m[/tex]

d = 38.469 m

The distance between Karl's and Joe's tent is:

The distance between Joe's tent and Karl's tent is approximately 36.84 m.

To find the distance between Joe's tent and Karl's tent, we can use the concept of vector addition. We first need to break down the given distances and angles into their respective components:

Next, we can use the components to find the displacement from Joe's tent to Karl's tent:

Using the components, we can calculate the displacement from Joe's tent to Karl's tent:

Finally, we can use the Pythagorean theorem to find the magnitude of the displacement:

Magnitude = sqrt((-33.46 m)^2 + (16.49 m)^2) = 36.84 m

Therefore, the distance between Joe's tent and Karl's tent is approximately 36.84 m.

Answer:

16.17 m/s

Explanation:

h = 3.2 m

u = 18.1 m/s

Angle of projection, θ = 49°

Let H be the maximum height reached by the ball.

The formula for the maximum height is given by

[tex]H=\frac{u^{2}Sin^{2}\theta }{2g}[/tex]

[tex]H=\frac{18.1^{2}\times Sin^{2}49 }{2\times 9.8}=9.52 m[/tex]

The vertical distance fall down by the ball, h'  H - h = 9.52 - 3.2 = 6.32 m

Let v be the velocity of ball with which it strikes the ground.

Use third equation of motion for vertical direction

[tex]v_{y}^{2}=u_{y}^{2}+2gh'[/tex]

here, uy = 0

So,

[tex]v_{y}^{2}=2\times 9.8 \times 6.32[/tex]

vy = 11.13 m/s

vx = u Cos 49 = 18.1 x 0.656 = 11.87 m/s

The resultant velocity is given by

[tex]v=\sqrt{v_{x}^{2}+v_{y}^{2}}[/tex]

[tex]v=\sqrt{11.87^{2}+11.13^{2}}[/tex]

v = 16.27 m/s

Answer:

plane speed: 525mph, jetstream speed=75mph, in explanation it is solved with a linear equations system

Explanation:

First lets name each speed

vs:=speed of the jetstream

vp:=speed of the plane

Now when in the jetstream direction the speeds are added and on the opposite direction are subtracted, then we get these equations, that are linear.

1800 mi=(vp+vs)*3h

1800 mi=(vp-vs)*4h

which is a linear equation system equivalent to:

600 mph=vp+vs (1)

450 mph=vp-vs  (2)

Now from (2) vp= 450mph+vs (3), replacing this in (1) we get:

600mph=(450mph+vs)+vs=450mph+2*vs, then 2*vs=150mph or vs=*75mph, this is the jetstream speed, replacing this in (3) we get the plane speed too vp=450 mph +75mph = 525 mph

Answer:

a)  1.28 *10^5 N/C

b)2.05 *10^{-14} N

c) 4.83 *10^{-17} J

Explanation:

Given Data:

Distance between the plates, d = 4.67 mm

[tex]= (4.67) *10^{-3} m[/tex]

[/tex]= 4.67 *10^{-3} m[/tex]

Potential difference, V = 600 V

Solution:

(a) The  magnitude of the electric field between the plates is,

    [tex]E = \frac{V}{d}[/tex]  

[tex]= \frac{600 V}{4.67 *10^{-3}} m[/tex]

  [tex]= 1.28 *10^5 V/m or 1.28 *10^5N/C[/tex]  

(b) Force on electron btwn the plates is,

   F = q E

 [tex]= (1.6 *10^{-19} C) (1.28 *10^5N/C[/tex]

 [tex]= 2.05 *10^{-14} N[/tex]  

(c) Work done on the electron is

   W = F * s

 [tex]= (2.05 *10^{-14} N) * (5.31 *10^{-3} m - 2.95 *10^{-3} m)[/tex]

 [tex]= 4.83 *10^{-17} J[/tex]

Answer:

a) t = 3.01s

b) 15th floor

Explanation:

First we need to know the distance the elevator has descended before the bolt fell.

[tex]\Delta Y_{e} = -V_{e}*t = -5.7 * 4.95 = -28.215m[/tex]

Now we can calculate the time that passed before both elevator and bolt had the same position:

[tex]Y_{b}=Y_{e}[/tex]

[tex]Y_{ob}+V_{ob}*t-g*\frac{t^{2}}{2} = Y_{oe} - V_{e}*t[/tex]

[tex]0+0-5*t^{2} = -28.215 - 5.7*t[/tex]   Solving for t:

t1 = -1.87s    t2 = 3.01s

In order to know how the amount of floors, we need the distance the bolt has fallen:

[tex]Y_{b}=-g*\frac{t^{2}}{2}=-45.3m[/tex]  Since every floor is 3m:

Floors = Yb / 3 = 15 floors.

Answer:

Explanation:

Resistance in series is given by the sum of all the resistor in series

value of Total Resistance is given by

[tex]R_{th}=R_1+R_2+R_3+R_4+..............R_n[/tex]

Where [tex]R_{th}[/tex] is the total resistance

[tex]R_1,R_2[/tex] are the resistance in series

Current in series remains same while potential drop is different for different resistor

The value of net resistor is always greater than the value of individual resistor.

If a there is a defect in a single resistor then it affects the whole circuit in series.

Answer:

The second system must be set in motion [tex]t=0.70s[/tex] seconds later

Explanation:

The oscillation time, T, for a mass, m, attached to spring with Hooke's constant, k, is:

[tex]T=2\pi\sqrt(\frac{m}{k} )[/tex]

One oscillation takes T secondes, and that is equivalent to a 2π phase. Then, a difference phase of π/2=2π/4, is equivalent to a time t=T/4.

If the phase difference π/2 of the second system relative to the first oscillator. The second system must be set in motion [tex]t=\frac{\pi}{2}\sqrt(\frac{m}{k})=\frac{\pi}{2}\sqrt(\frac{0.55}{2.8}= 0.70s)[/tex] seconds later

Answer:

width of fringes are increased

Explanation:

The width of central maxima is given by the following expression

Width = 2 x Dλ / d

Due to increased width ,  position of a fringe  moves away from the centre.

Answer:67.45 m

Explanation:

Given

at t=2.5 s it passes the top of a tall building and after 1.5 s it reaches maximum  height

let u is the initial velocity of stone

v=u+at

0=u-gt

[tex]u=9.81\times 4=39.24 m/s[/tex]

Let us take h be the height of building

[tex]h=ut+\frac{-1}{2}gt^2[/tex]

[tex]h=39.24\times 2.5-\frac{1}{2}\times 9.81\times 2.5^2[/tex]

h=67.45 m

Answer and Explanation:

The gravitational acceleration 'g' depend directly on the mass of the object or body and inversely on the distance or radius squared:

[tex]g = \frac{Gm_{o}}{R^{2}}[/tex]

where

[tex]m_{o}[/tex] = mass of the object

G = Gravitational constt

Thus the plastic ball is lighter and have low mass as compared to the steel and golf balls.

This is the reason that a plastic ball have a low value of acceleration as compared to that of steel and golf balls with higher values of acceleration.

Final answer:

The acceleration due to gravity is a constant (g) for all objects in the same gravitational field, and any observed difference in falling rates is likely due to air resistance, not gravitational pull. Experiments have confirmed that objects fall at the same rate regardless of their mass or composition, assuming no air resistance.

Explanation:

The acceleration due to gravity should not vary with the material of the object if we ignore effects such as air resistance. This is because, according to Newton's second law, the force acting on an object is the product of its mass and the acceleration (F = ma). Here, the force due to gravity is mg, where m is the mass and g is the acceleration due to gravity. Since a = F/m, the mass cancels out, leaving a = g, which is constant for all objects in the same gravitational field.

The question assumes that the plastic ball has a lower acceleration due to gravity, which contradicts known physics principles. All objects, regardless of their mass or composition, feel the same acceleration due to gravity near the Earth's surface, assuming no other forces, like air resistance, play a significant role. Historical experiments by scientists such as Galileo and Eötvös have confirmed the equality of gravitational acceleration (g) for different substances within exceptionally high precision.

If you observe differing acceleration rates, this can be often attributed to air resistance, not a difference in gravitational pull. Objects with a larger surface area or less aerodynamic shape, like the plastic ball, may experience greater air resistance and thus appear to fall slower, even though their acceleration due to gravity is the same as denser objects like the steel or golf ball.

Answer: Yes, he is exceeding the speed limit

Explanation:

Hi!

This is problem about unit conversion

1 mile = 1,609.344 m

Then the speed limit v is:

v = 75 mi/h = 120,700.8 m/h

1 hour = 60 min = 60*60 s = 3,600 s

v = (120,700.8/3,600) m/s = 33.52 m/s

38 m/s is higher than the speed limit v.

Answer:

1.708*10^6 tons.

Explanation:

Number of Aluminum cans used by 1 person in 1 year = 365/2=182.5 say it as 183 cans per year.

Total number of people in US= 280,000,000

Total number of cans used by americans.

[tex] = 5.12×10^10 cans[/tex]

Weight of 1 can =1/15 pounds

Weight of all cans used in 1 year

[tex]= \frac{5.12*10^10}{15} =3.41*10^9pounds.[/tex]

we know that

1ton=2000pounds.

[tex]\frac{3.41*10^9 pounds}{2000} = 1.708*10^6 tons.[/tex]

Answer:

[tex]t=2.8h[/tex]

[tex]t=10080s[/tex]

[tex]t=168 min[/tex]

Explanation:

From this exercise we have velocity and distance. Using the following formula, we can calculate time:

[tex]v=\frac{d}{t}[/tex]

Solving for t

[tex]t=\frac{d}{v}=\frac{26.22mi}{9.39mi/h} =2.8h[/tex]

[tex]t=2.8h*\frac{3600s}{1h} =10080s[/tex]

[tex]t=2.8h*\frac{60min}{1h} =168min[/tex]

Answer:

(C) The frequency decrease and intensity decrease

Explanation:

The Doppler effect describes the change in frequency or wavelength of a wave in relation to an observer who is moving relative to the wave source, or the wave source is moving relative to the observer, or both.

if the observer and the source move away from each other as is the case for this problem, the wavelength heard by the observer is bigger.

The frequency is the inverse from the wavelength, so the frequency heard will increase.

The sound intensity depends inversely on the area in which the sound propagates. When the buzzer is close, the area is from a small sphere, but as the buzzer moves further away, the wave area will be from a larger sphere and therefore the intensity will decrease.

Explanation:

Given that,

Initial sped of the runner, u = 0

Final speed of the runner, v = 30 ft/s

Acceleration of the runner, [tex]a=8\ ft/s^2[/tex]

Let t is the time taken by the runner. It can be calculated using first equation of motion as :

[tex]t=\dfrac{v-u}{a}[/tex]

[tex]t=\dfrac{30-0}{8}[/tex]

t = 3.75 seconds

Let s is the distance covered by the runner. Using the second equation of motion as :

[tex]s=ut+\dfrac{1}{2}at^2[/tex]

[tex]s=\dfrac{1}{2}\times 8\times (3.75)^2[/tex]

s = 56.25 feet

Hence, this is the required solution.

Final answer:

To reach a maximum speed of 30 f/s from rest with an acceleration of 8 f/s², it takes 3.75 seconds. During this time, the runner travels a distance of 56.25 feet.

Explanation:

The question involves calculating the time it takes for a runner to reach a maximum speed of 30 feet per second (f/s) from rest with an acceleration of 8 feet per second squared (f/s²), and then finding the distance traveled during this time. This can be solved using the basic kinematics equations.

Calculating Time to Reach Max Speed

To find the time, we use the equation v = at, where v is the final velocity (30 f/s), a is the acceleration (8 f/s²), and t is the time. Rearranging the equation to solve for t, we get t = v/a. Plugging in the values, t = 30 f/s / 8 f/s² = 3.75 seconds.

Calculating Distance Traveled

To find the distance traveled, we use the equation d = 0.5 * a * t², where d is the distance, a is the acceleration, and t is the time. Substituting the given values, d = 0.5 * 8 f/s² * (3.75 s)² = 56.25 feet.

Answer:

[tex]a=4m/s^{2}[/tex]

Explanation:

From the concept of average acceleration we know that

[tex]a=\frac{v_{2}-v_{1} }{t_{2}-t_{1}  }[/tex]

From the exercise we know that

[tex]v_{2}=30m/s\\v_{1}=10m/s\\t_{2}=5s\\t_{1}=0s[/tex]

So, the average acceleration of the car is:

[tex]a=\frac{30m/s-10m/s}{5s}=4m/s^{2}[/tex]

Answer:[tex]3.874 m/s^2[/tex]

Explanation:

Given

Car speed decreases at a constant rate from 64 mi/h to 30 mi/h

in 3 sec

[tex]60mi/h \approx 26.8224m/s[/tex]

[tex]34mi/h \approx 15.1994 m/s[/tex]

we know acceleration is given by [tex]=\frac{velocity}{Time}[/tex]

[tex]a=\frac{15.1994-26.8224}{3}[/tex]

[tex]a=-3.874 m/s^2[/tex]

negative indicates that it is stopping the car

Distance traveled

[tex]v^2-u^2=2as[/tex]

[tex]\left ( 15.1994\right )^2-\left ( 26.8224\right )^2=2\left ( -3.874\right )s[/tex]

[tex]s=\frac{488.419}{2\times 3.874}[/tex]

s=63.038 m

Answer:920.31 J

Explanation:

Given

Volume of water (V)[tex]=14.1 cm^3 [/tex]

mass(m)[tex]=\rho \times V=1000\times 14.1\times 10^{-6}=14.1 gm[/tex]

Temperature [tex]=8.4^{\circ} C[/tex]

Final Temperature [tex]=-7.2 ^{\circ}C[/tex]

specific heat of water(c)[tex]=4.184 J/g-^{\circ}C[/tex]

Therefore heat required to removed is

[tex]Q=mc(\Delta T)[/tex]

[tex]Q=14.1\times 4.184\times (8.4-(-7.2))[/tex]

[tex]Q=920.31 J[/tex]

Answer:

1.  1.568 x 10^6 N m^2 / C

2. -  1.568 x 10^6 N m^2 / C

Explanation:

E = 4.9 x 10^4 N/C

Side of square, a = 8 m

Area, A = side x side = 8 x 8 = 64 m^2

Angle between lane of sheet and electric field = 30°

Angle between the normal of plane of sheet and electric field,

θ = 90°- 30° = 60°

The formula for the electric flux is given by

[tex]\phi = E A Cos\theta[/tex]

(1) [tex]\phi = E A Cos\theta[/tex]

By substituting the values, we get

Ф = 4.9 x 10^4 x 64 x Cos 60 = 1.568 x 10^6 N m^2 / C

(2) [tex]\phi = E A Cos\theta[/tex]

By substituting the values, we get

Ф = - 4.9 x 10^4 x 64 x Cos 60 = - 1.568 x 10^6 N m^2 / C

Answer:

306500 N/C

Explanation:

The magnitude of an electric field around a single charge is calculated with this equation:

[tex]E(r) = \frac{1}{4 \pi *\epsilon 0} \frac{q}{r^2}[/tex]

With ε0 = 8.85*10^-12 C^2/(N*m^2)

Then:

[tex]E(0.89) = \frac{1}{4 \pi *8.85*10^-12} \frac{27*10^-6}{0.89^2}[/tex]

E(0.89) = 306500 N/C

Answer: The average speed is 27,24 mph (exactly 1008/37 mph)

Explanation:

This is solved using a three rule: We know the speeds and the distances, what we can obtain from it is the time used. It is done like this:

1h--->18mi

X ---->20 mi, then X=20mi*1h/18mi= 10/9 h=1,111 h

1h--->56mi

X ---->20 mi, then X=20mi*1h/56mi= 5/14 h=0,35714 h

Then the average speed is calculated by taking into account that it was traveled 40mi and the time used was 185/126 h=1,468 h and since speed is distance over time we get the answer. Average speed= 40mi/(185/126 h)=1008/37 mph=27,24 mph.

Answer:

The snail travel at the end of 5 s with a velocity of 12 m/s and the distance of the snail is 22.5 m.

Explanation:

Given that, the initial velocity of the snail is,

[tex]u=2m/s[/tex]

And the acceleration of the snail is,

[tex]a=1m/s^{2}[/tex]

And the time taken by the snail is,

[tex]t=5 sec[/tex]

Now according to first equation of motion,

[tex]v=u+at[/tex]

Here, u is the initial velocity, t is the time, v is the final velocity and a is the acceleration.

Now substitute all the variables

[tex]v=2m/s+ 1 \times 5 sec\\v=7m/s[/tex]

Therefore, the snail travel at the end of 5 s with a velocity of 7 m/s.

Now according to third equation of motion.

[tex]v^{2}- u^{2}=2as\\ s=\frac{v^{2}- u^{2}}{2a} \\[/tex]

Here, u is the initial velocity, a is the acceleration, s is the displacement, v is the final velocity.

Substitute all the variables in above equation.

[tex]s=\dfrac{7^{2}- 2^{2}}{2(1)}\\s=\dfrac{45}{2}\\ s=22.5m[/tex]

Therefore the distance of the snail is 22.5 m.

Answer: Kilogram- newton is wrong unit for Torque

Idem pound-inch is also wrong unit for Torque

Explanation: As it is well known the torque is defined as :

Τorque= F x R

so its UNITS are Newton*meter (SI)

or in the Imperial System is often use  Pound-foot

Answer:

19.642 cm

Explanation:

f₁ = Focal length of first lens = 11 cm

f₂ = Focal length of second lens = -25 cm

Combined focal length formula

[tex]\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}\\\Rightarrow \frac{1}{f}=\frac{1}{11}+\frac{1}{-25}\\\Rightarrow \frac{1}{f}=\frac{14}{275}\\\Rightarrow f=\frac{275}{14}\\\Rightarrow f=19.642\ cm[/tex]

Combined focal length is 19.642 cm

The focal length of the combination of lenses is approximately 19.64 cm.

To find the focal length of the combination of lenses in this scenario, we need to use the lensmaker's formula for thin lenses in combination.

Given:

Focal length of lens 1, [tex]\( f_1 = +11 \)[/tex] cm

Focal length of lens 2, [tex]\( f_2 = -25 \)[/tex] cm

The formula for the focal length of two thin lenses in contact is given by:

[tex]\[ \frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2} \][/tex]

[tex]\[ \frac{1}{f} = \frac{1}{+11 \text{ cm}} + \frac{1}{-25 \text{ cm}} \][/tex]

[tex]\[ \frac{1}{f} = \frac{1}{11} - \frac{1}{25} \][/tex]

To subtract the fractions, find a common denominator, which is 275:

[tex]\[ \frac{1}{f} = \frac{25}{275} - \frac{11}{275} \][/tex]

[tex]\[ \frac{1}{f} = \frac{14}{275} \][/tex]

Now, invert both sides to solve for (f):

[tex]\[ f = \frac{275}{14} \][/tex]

[tex]\[ f \approx 19.64 \text{ cm} \][/tex]

Answer:

a)11.6m

b)45.55s

Explanation:

A body that moves with constant acceleration means that it moves in "a uniformly accelerated movement", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.

When performing a mathematical demonstration, it is found that the equations that define this movement are as follows.

Vf=Vo+a.t  (1)\\\\

{Vf^{2}-Vo^2}/{2.a} =X(2)\\\\

X=Xo+ VoT+0.5at^{2}    (3)\\

X=(Vf+Vo)T/2 (4)

Where

Vf = final speed

Vo = Initial speed

T = time

A = acceleration

X = displacement

In conclusion to solve any problem related to a body that moves with constant acceleration we use the 3 above equations and use algebra to solve

a)

for this problem

Vo=0

Vf=319m/min=5.3m/s

a=1.2m/s^2

we can use the ecuation number 1 to calculate the time

t=(Vf-Vo)/a

t=(5.3-0)/1.2=4.4s

then we use the ecuation number 3 to calculate the distance

X=0.5at^2

X=0.5x1.2x4.4^2=11.6m

b)second part

We know that when the elevator starts to accelerate and decelerate, it takes a distance of 11.6m and a time of 4.4s, which means that if the distance is subtracted 2 times this distance (once for acceleration and once for deceleration)

we will have the distance traveled in with constant speed.

With this information we will find the time, and then we will add it with the time it takes for the elevator to accelerate and decelerate

X=218-11.6x2=194.8m

X=VT

T=X/v

t=194.8/5.3=36.75s

Total time=36.75+2x4.4=45.55s

The problem requires vector operations in two dimensions. Displacement is broken down into x and y-components which follow the east-west and north-south directions respectively. Total displacement being zero means the sum of the x and y components of displacement will also be zero. The fourth displacement is determined by negating the total x and y components of the first three displacements. The magnitude and direction are then obtained using Pythagorean theorem and arctan function respectively.

To solve this problem, we need to deal with the changes in displacement in terms of vector operations. Given the different directions, we need to break down the vectors into their x and y components, where x represents east-west direction, and y north-south direction. Since our spelunker starts and ends in the same place, the sum of the displacements in each dimension will also be zero.

For displacement 1, moving 180m west, the x component would be -180, and y component would be 0. For displacement 2, moving 230m in a direction 45° east of south, the x would be -230×sin(45) and y would be -230×cos(45). For displacement 3, moving 280m at 30° east of north, the x would be 280×cos(30) and y would be 280×sin(30).

To determine the fourth displacement, we sum up the x and y components for displacement 1,2 and 3 and then negate them to get the x and y component of the fourth displacement. We then use the Pythagorean theorem to calculate the magnitude of the 4th displacement which is square root of (sum x² + sum y²). The direction can be obtained by calculating the arctan of the total y component / total x component.

https://brainly.com/question/20047824

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Answer:

a) 231 yards

b) 55.2 yards

c) 0 yards to 260 yards

d) Infinite values

Explanation:

This situation can be described as a horizontal line that begins at point [tex]P_{1}=0 yards[/tex] (Meredith's house) and ends at point [tex]P_{2}=260 yards[/tex] (Bus stop). Where [tex]x[/tex] is the varying distance from her house, which can be calculated in the following way:

x=Final Position - Initial Position

or

[tex]x=x_{f} - x_{i}[/tex]

a) For the first case Meredith is at position [tex]x_{i}=29 y[/tex] and the bus stop at position [tex]x_{f}=260 y[/tex]. So the distance Meredith is from the bus stop is:

[tex]x=260 y - 29 y=231 y[/tex]

b) For the second case the initial position is [tex]x_{i}=204.8 y[/tex] and the final position [tex]x_{f}=260 y[/tex]. Hence:

[tex]x=260 y - 204.8  y=55.2  y[/tex]

c) If we take Meredith's initial position at her house  [tex]x_{i}=0 y[/tex] and her final position at the bus stop  [tex]x_{f}=260 y[/tex], the value of  [tex]x[/tex] varies from 0 yards to 260 yards.

d) As Meredith walks from her house to the bus stop, the variable [tex]x[/tex] assumes infinite values, since there are infinite position numbers from [tex]x=0 yards[/tex] to [tex]x=260 yards[/tex]

The answers to the possible distance covered by Meredith at the various distances from her house are;

A) distance = 231 yards

A) distance = 231 yardsB) distance = 55.2 yards

A) distance = 231 yardsB) distance = 55.2 yardsC) x will vary from 0 m to 260 m i.e 0 ≤ x ≤ 260

A) We are told that meredith walks from her house to a bus stop that is 260 yards away.

After walking, she is now 29 yards from her house. This means that she has walked a total of 29 yards from her house.

Distance left to reach bus stop = 260 - 29 = 231 yards

B) We are told that Meredith is now 204.8 yards from her house. This means that she has walked a total of 204.8 yards from here house. Thus;

Distance left to reach bus stop = 260 - 204.8 = 55.2 yards.

C) This question is basically asking for all the possible values that Meredith could have walked from her house to the bus stop.

Since she starts from her house at 0m, then it means that if the bus stop is 260 m away, then if x is the possible distance, we can say that x will vary from 0 m to 260 m i.e 0 ≤ x ≤ 260

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