Look at the attached photo:

Look At The Attached Photo:

Answers

Answer 1

Answer:

C) Mass of the ball

Explanation:

Independent variable is the variable the researcher changes.

Since Martin is testing the mass of the ball, he'll be using different balls and that is the only thing he changes.

The distance traveled by the ball is the dependent variable since it depends on the mass of the ball.

The height and length of the ramp are the constant variables since that's the only ones that remain the same throughout this experiment.


Related Questions

A very long straight current-carrying wire produces a magnetic field of 20 mT at a distance d from the wire. To measure a field of 5 mT due to this wire, you would have to go to a distance from the wire of A very long straight current-carrying wire produces a magnetic field of 20 mT at a distance d from the wire. To measure a field of 5 mT due to this wire, you would have to go to a distance from the wire of:_____.
a. 4d.
b. 16d.
c. 2d.
d. 8d.

Answers

Answer:

A. 4d

Explanation:

Let's begin with the formula for the magnetic field produced by a long wire.

[tex]B = \frac{\mu_0I}{2\pi d}[/tex]

So [tex]d=\frac{\mu_0 I}{2\pi B }[/tex]

at point d_{1} is

[tex]d_{1}=\frac{\mu_{0} i}{2 \pi B_{1}} \\ \frac{d_{1}}{d}=\frac{\frac{\mu_{0} i}{2 \pi B_{1}}}{\frac{\mu_{0} i}{2 \pi B}} \\ d_{2}=d\left(\frac{B}{B}\right) \\ =d\left(\frac{20 \mathrm{mT}}{5 \mathrm{mT}}\right) \\ =4 d[/tex]

Hence, option  (A) is correct answer

The indices of refraction for her contact lens, cornea, and the fluid behind her cornea are 1.6, 1.4, and 1.3, respectively. Light is incident from air onto her contact lens at an angle of 30 ∘∘ from the normal of the surface. At what angle is the light traveling in the fluid behind her cornea?

Answers

Answer:

[tex]23^{\circ}[/tex]

Explanation:

n = Refractive index of air = 1

[tex]n_1[/tex] = Refractive index of contact lens = 1.6

[tex]n_2[/tex] = Refractive index of cornea = 1.4

[tex]n_3[/tex] = Refractive index of fluid = 1.3

From Snell's law

[tex]n\sin30^{\circ}=n_1\sin\theta\\\Rightarrow \theta=\sin^{-1}\dfrac{1\sin30^{\circ}}{1.6}\\\Rightarrow \theta=18.21^{\circ}[/tex]

[tex]n_1\sin\theta=n_2\sin\theta_1\\\Rightarrow \theta_{1}=\sin^{-1}\dfrac{1.6\times \sin18.21^{\circ}}{1.4}\\\Rightarrow \theta_1=20.92^{\circ}[/tex]

[tex]n_2\sin\theta_1=n_3\sin\theta_3\\\Rightarrow \theta_3=\sin^{-1}\dfrac{1.4\sin20.92^{\circ}}{1.3}\\\Rightarrow \theta_3=22.62^{\circ}\approx 23^{\circ}[/tex]

The angle is the light traveling in the fluid behind her cornea is [tex]23^{\circ}[/tex].

The angle is the light traveling in the fluid will be 23⁰. Light is traveling in a particular direction with an angle.

What is snell law?

"The ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant, for the light of a given color and for a given set of media,

The given data in the problem is;

n is the refractive index of air = 1

n₁ is the refractive index of contact lens = 1.6

n₂ is the refractive index of cornea = 1.4

n is the refractive index of fluid = 1.3

According to Snell's law. The formula for Snell's law is

[tex]\rm n sin30^0 = n_1 sin\theta \\\\ \theta = sin^{- 1}(\frac{1sin30^0}{1.6} )\\\\ \theta = 18.21 ^0[/tex]

For contact lenses;

[tex]\rm n_1sin\theta = n_2 sin\theta_1 \\\\ \theta_1 = sin^{-1}\frac{1.6 \times sin 18.21^0}{1.4} \\\\ \theta_1 =20.92 ^0[/tex]

For fluid;

[tex]n_2 sin\theta_1 = n_2 sin \theta_3\\\\ \theta_3 = sin^{-1}\frac{1.4 sin 20.92^0}{1.3} \\\\ \theta_3 = 22.62 ^ 0 =23^0[/tex]

Hence the angle is the light traveling in the fluid will be 23⁰.

To learn more about snell's law refer to the link;

https://brainly.com/question/10112549

Plz help

What factors determine
how the speed of the marbles changes in a
collision?

Answers

Answer:

Force,friction,inertia and momentum

Explanation:

The speed that the marble is moving at can be determined by the amount of force used when pushed or pulled and what kind of surface it's on.Momentum is also a factor because of the mass of the marbles.

Frequency more than 20,000 HZ​

Answers

Answer:

dvhd

Explanation:

xhxjjdvcbxhjddvifidid

Answer:

The units of frequency are called hertz (Hz). Humans with normal hearing can hear sounds between 20 Hz and 20,000 Hz. Frequencies above 20,000 Hz are known as ultrasound

Protons, neutrons, electrons, and a nucleus are

Answers

It would be Atoms, they’re all made up of these tiny particles

What is characteristic of a good insulator?
A. Electrons are usually not moving at all.
B. Electrons are free to move around.
C. Electrons are semi-free to move around.
D. Electrons are tightly bound to the nuclei.

Answers

Answer:

D. Electrons are tightly bound to the nuclei.

Explanation:

In an insulator, the electrons of the outer most shell are bound with a very high electrostatic forces coming from the nucleus of each atom so electrons cannot flow around all atoms making up the material as in a conductor.

The characteristic of a good insulator is Electrons are tightly bound to the nuclei. (option d)

In a good insulator, electrons are tightly bound to the nuclei of their atoms. This means that they are not free to move around within the material, unlike conductors where electrons are relatively loosely bound and can move freely. Due to this strong binding, electrons in insulating materials cannot carry an electric charge or energy easily from one atom to another.

When an electric field is applied to an insulator, the electrons may experience a small displacement within their respective atoms, but they generally do not move from one atom to another or flow through the material like they would in a conductor. As a result, insulators prevent the flow of electric current and are used to isolate or protect conductive elements from accidental contact.

So, the correct answer is D. Electrons are tightly bound to the nuclei.

To know more about insulator here

https://brainly.com/question/2619275

#SPJ6

PLEASE HELP! I'LL GIVE BRAINLEST​

Answers

Answer:

In this conversation the Neil astronaut is right

What is Force ?.............​

Answers

Answer:

Push or pull of an object is considered a force. Push and pull come from the objects interacting with one another. Terms like stretch and squeeze can also be used to denote force.

In Physics, force is defined as:

The push or pull on an object with mass that causes it to change its velocity.

Force is an external agent capable of changing the state of rest or motion of a particular body. It has a magnitude and a direction. The direction towards which the force is applied is known as the direction of the force and the application of force is the point where force is applied.

Question 8: Unspooling Thread (100 points) A 110 g spool of thread with a 4.2 cm radius is held up by a peg through its center and allowed to freely rotate. Assume the thread is ideal (i.e., it does not stretch or slip, and its mass is negligibly small). A 160 g needle is tied to the loose end of the thread. The needle is dropped, and it accelerates to the floor as the thread unwinds. Find the tension in the thread and the magnitude of the acceleration of the needle as it falls.

Answers

Answer:

   a = 7.29 m / s²,      T = 0.40 N

Explanation:

To solve this exercise we must apply Newton's second law to each body

The needle

              W -T = m a

              mg - T = ma

The spool, which we will approach by a cylinder

             Σ τ = I α

             T R = I α

the moment of inertia of a cylinder with an axis through its center is

             I = ½ M R²

angular and linear variables are related

            a = α R

            α = a / R

we substitute

           T R = (½ M R²) a / R

            T = ½ M a

we write our system of equations together

              mg - T = m a

                      T = ½ M a

we solve

              m g = (m + ½ M) a

              a = [tex]\frac{m}{m + \frac{1}{2} M} \ g[/tex]

let's calculate

              a = [tex]\frac{0.160}{0.160 + \frac{1}{2} 0.110} \ 9.8[/tex]

              a = 7.29 m / s²

now we can look for the tension

              T = ½ M a

              T = ½ 0.110 7.29

               T = 0.40 N

Which statement is true about how early humans met their needs?

Answers

Answer:

they were hunter gatherers

Explanation:

What do meteorologists call the lines
that join places with the same
temperature?
A. isobars
B. isotherms
C. anisobars
D. anisotherms

Answers

The answer is B.isotherms

A 0.545-kg ball is hung vertically from a spring. The spring stretches by 3.56 cm from its natural length when the ball is hanging at equilibrium. A child comes along and pulls the ball down an additional 5cm, then lets go. How long (in seconds) will it take the ball to swing up and down exactly 4 times, making 4 complete oscillations before again hitting its lowest position

Answers

Answer:

t = 9.52 s

Explanation:

This is an oscillatory motion exercise, in which the angular velocity is

         w = [tex]\sqrt{ \frac{k}{m} }[/tex]

Let's use hooke's law to find the spring constant, let's write the equilibrium equation

        F_e - W = 0

        F_e = W

        k x = m g

        k = [tex]\frac{m g}{x}[/tex]

        k = 0.545 9.8 /0.0356

        k = 150 N / m

now the angular velocity is related to the period

          W = 2π / T

we substitute

          4π² T² = k /m

          T = 4pi² [tex]\sqrt{ \frac{m}{k} }[/tex]

we substitute

           T = 4 pi² [tex]\sqrt{ \frac{0.545}{150} }[/tex]

            T = 2.38 s

therefore for the spring to oscillate 4 complete periods the time is

            t = 4 T

            t = 4 2.38

            t = 9.52 s

What is the mass of a mallard duck whose speed is 9.1 m/s and whose momentum has a magnitude of 12 kg⋅m/s?

Answers

Answer:

m = 1.31 kg

Explanation:

Given that,

The speed of duck, v = 9.1 m/s

The magnitude of momentum, p = 12 kg-m/s

We need to find the mass of the duck. We know that the momentum of an object is given by :

p = mv

Where

m is the mass of the duck

[tex]m=\dfrac{p}{v}\\\\m=\dfrac{12\ kg-m/s}{9.1\ m/s}\\\\m=1.31 kg[/tex]

So, the mass of the duck is equal to 1.31 kg.

PROJECT: VIRTUAL LABS — CIRCUITS

Answers

Stop with that link need to stop

Answer:

Table A

Measuring Current as a Function of Voltage with a 20 Ω Resistor

Voltage

(V)

Current: Calculated

(A)

Current: Experimental

(A)

1 0.05 0.05

5 0.25 0.25

10 0.50 0.50

20 1.00 1.00

50 2.50 2.50

Table B

Measuring Current as a Function of Resistance at 25 V

Resistance

(Ω)

Current: Calculated

(A)

Current: Experimental

(A)

10 2.50 2.50

20 1.25 1.25

100 0.25 0.25

200 0.12 0.12

Table C

Measuring Current in a Parallel Circuit

Resistor Set

(Ω)

Total  

Resistance

(Ω)

Calculated

Current

(A)

Observed  

Current

(A)

Observed Current  

through Each Resistor

(A)

20, 20, 20 6.67 3.75 3.74 1.25, 1.25, 1.25

20, 20, 200 9.52 2.63 2.62 1.25, 1.25, 0.12

Voltage needed to raise current to 3.75 A (20, 20, 200 resistor set):

Calculated: 35.7

Observed: 36

Table D

Calculating Power of Circuit ComponentsTeacher Guide (continued)

Observed Total Current

(A)

Current through Each Bulb

(A)

Power Usage per Bulb

(W)

2.00 0.67 6.7

Explanation:

got this from the teachers guide

Determiner l'interfrange i sur le plan d'observation π distant de L de D:
Determine the interfringe i on the observation plane π distant from L from D:
1) pour D=f
if D=f
2)pour D=2f
if D=2f

Answers

Answer:

can you explain in Hindi language

because i learn hindi

The work done is a vector quantity and SI base unit is J

Answers

Answer:

Is this your question? Also I think work done is a scalar quantity.

Explanation:

In the process of fluorescence, a molecule in its ground state will absorb a photon with a certain energy Eex, called the excitation energy, and then emit a photon with energy Eem, the emission energy. Obviously, the molecule cannot emit more energy than was absorbed.
a) You wrote in your notebook that you excited a fluorescent bead at λ = 640 nm and found that it fluoresced (emitted) at λ = 520 nm. Or was it the other way around? Do a calculation and see if those figures should be switched.
b) You shine a laser on your bead at the excitation wavelength. The laser has a power of 1 mW. How many photons are emitted by the laser in one second?
c) You wish to build a microscope that allows you to excite the bead at its excitation wavelength and detect the fluorescence at the emission wavelength. This requires a pair of filters, one that allows only the excitation wavelength to pass through, and the other that allows only the emission wavelength to pass through. What color will these filters appear (i.e. what color light will they transmit?) You may have to look at an electromagnetic spectrum to figure this out…

Answers

Answer:

a) the excitation energy is E₂ λ = 520 nm

the emission energy is E₁,  λ= 640 nm

b)   #_photons = 2.6 10¹⁸ photons,  

c) he excitation wavelength   λ = 520 nm is green, therefore the filter is also green

the emission wavelength is lam = 640 nm is orange

Explanation:

a) the energy of a photo is given by the planck relation

           E = h f

the speed of light is

          c = λ f

          f = c /λ

we substitute

          E = hc /λ

let's calculate the energy for the two photons

λ = 640 nm = 640 10⁻⁰ m

            E₁ = 6.63 10⁻³⁴ 3 10⁸/640 10⁻⁹

            E₁ = 3.1 10⁻¹⁹ J

λ = 520 nm = 520 10⁻⁹ m

             E₂ = 6.63 10⁻³⁴ 3 10⁸/520 10⁻⁹

            E₂ = 3.825 10⁻¹⁹  J

therefore the excitation energy is E₂ λ = 520 nm

the emission energy is E₁,  λ= 640 nm

b) For this part let's use a direct proportion rule (rule of three). If a photon (lam = 520 nm) has an energy of 3.825 10⁻¹⁹ J, how many photons have an energy of E = 1 10-3 J. Remember that the power is the energy per unit of time

        #_photons = 1 10⁻³ J (1 photon / 3.825 10⁻¹⁹ J)

        #_photons = 2.6 10¹⁸ photons

c) the excitation wavelength   λ = 520 nm is green, therefore the filter is also green

the emission wavelength is lam = 640 nm is orange

Hydrocarbons are by-products of which of the following:
a. Fossil oils, mines and fossil fuels
b. Burning fires, water treatment plants and fossil oils
C. Combustion in fossil, lighting and exhaust fumes
d. Petrol, sea waves and combustion in vehicles
e. Burning fires and combustion in factories and vehicles​

Answers

Answer:

A

Explanation:

Key word part: hydro - water
I believe the answer is A
Please give me Brainliest

A 3kg horizontal disk of radius 0.2m rotates about its center with an angular velocity of 50rad/s. The edge of the horizontal disk is placed in contact with a wall, and the disk comes to rest after 10s. Which of the following situations associated with linear impulse is analogous to the angular impulse that is described?

a. A 3kg block is initially at rest. An applied force of 3N is applied to the block, but the block does not move.
b. A 3kg block is initially at rest. A net force of 3N is applied to the block until it has a speed of 10m/s.
c. A 3kg block is initially traveling at 10m/s. An applied force of 3N is applied to the block in the direction of its velocity vector for 10s.
d. A 3kg block is initially traveling at 10m/s. The block encounters a 3N frictional force until the block eventually stops.

Answers

Answer:

D

Explanation:

From the information given:

The angular speed for the block [tex]\omega = 50 \ rad/s[/tex]

Disk radius (r) = 0.2 m

The block Initial velocity is:

[tex]v = r \omega \\ \\ v = (0.2 \times 50) \\ \\ v= 10 \ m/s[/tex]

Change in the block's angular speed is:

[tex]\Delta _{\omega} = \omega - 0 \\ \\ = 50 \ rad/s[/tex]

However, on the disk, moment of inertIa is:

[tex]I= mr^2 \\ \\ I = (3 \times 0.2^2) \\ \\ I = 0.12 \ kgm^2[/tex]

The time t = 10s

Frictional torques by the wall on the disk is:

[tex]T = I \times (\dfrac{\Delta_{\omega}}{t}) \\ \\ = 0.12 \times (\dfrac{50}{10}) \\ \\ =0.6 \ N.m[/tex]

Finally, the frictional force is calculated as:

[tex]F = \dfrac{T}r{}[/tex]

[tex]F= \dfrac{0.6}{0.2} \\ \\ F = 3N[/tex]

Medical devices implanted inside the body are often powered using transcutaneous energy transfer (TET), a type of wireless charging using a pair of closely spaced coils. And emf is generated around a coil inside the body by varying the current through a nearby coil outside the body, producing a changing magnetic flux. Calculate the average induced emf, of each 10-turn coil has a radius of 1.50 cm and the current in the external coil varies from its maximum value of 10.0 A to zero in 6.25 x10-6s.

Answers

Answer:

[tex]0.475\ \text{V}[/tex]

Explanation:

n = Number of turns = 10

r = Radius = 1.5 cm

I = Current = 10 A

t = Time = [tex]6.25\times 10^{-6}\ \text{s}[/tex]

[tex]\mu_0[/tex] = Vacuum permeability = [tex]4\pi\times 10^{-7}\ \text{H/m}[/tex]

Magnetic field is given by

[tex]B=\dfrac{\mu_0I}{2r}\\\Rightarrow B=\dfrac{4\pi 10^{-7}\times 10}{2\times 1.5\times 10^{-2}}\\\Rightarrow B=0.00042\ \text{T}[/tex]

EMF is given by

[tex]\varepsilon=\dfrac{nBA}{t}\\\Rightarrow \varepsilon=\dfrac{10\times 0.00042\times \pi (1.5\times 10^{-2})^2}{6.25\times 10^{-6}}\\\Rightarrow \varepsilon=0.475\ \text{V}[/tex]

The average induced emf is [tex]0.475\ \text{V}[/tex].

7) If someone behaves against your company's code of ethics, what should you do?
A) Ignore it and mind your own business.
B) Suggest that they talk to the human resources department.
C) Talk to them about the situation.
D) Report the problem to your supervisor

Answers

Answer:

D) Report the problem to your supervisor.

Explanation:

This is probably the most efficient way get them to stop or to get them to follow the rules <3

D. Because you should report it on how the person is acting

Acellus
A motion sensor emits sound, and
detects an echo 0.0115 s after. A
short time later, it again emits a
sound, and hears an echo after
0.0183 s. How far has the
reflecting object moved?
Help Resources
(Speed of sound = 343 m/s)
(Unit = m)

Answers

Answer:

1.17m

Explanation:

The formula to find distance is d=vt/2

This problem is asking for how far the reflecting object has moved so you need to find the distance from the motion sensor at both times.

(343)(0.115) / (2) = 1.97

(343)(0.0183) /(2) =3.14

After that, all you have to do is find the difference so

3.14 - 1.97

= 1.17

A 25.0kg girl pushes a 50.0kg boy so that he accelerates at 4.00m/s2. What is the force of the boy on the girl? A. 200N B. 100N C. 12.5 D. 400N

Answers

Answer:

a

Explanation:

so the answer is 200N

and I hope it is correct

A textbook of mass 2.10 kg rests on a frictionless, horizontal surface. A cord attached to the book passes over a pulley whose diameter is 0.100 m , to a hanging book with mass 3.10 kg . The system is released from rest, and the books are observed to move a distance 1.29 m over a time interval of 0.850 s . Part A What is the tension in the part of the cord attached to the textbook

Answers

Answer:

the tension in the part of the cord attached to the textbook is 7.4989 N

Explanation:

Given the data in the question;

As illustrated in the image below;

first we determine the value of the acceleration,

along vertical direction; we use the second equation of motion;

y = ut + [tex]\frac{1}{2}[/tex]a[tex]_y[/tex]t²

we substitute;

0 m/s for u, 1.29 m for y, 0.850 s for t,

1.29 = 0×0.850  + [tex]\frac{1}{2}[/tex]×a[tex]_y[/tex]×(0.850)²

1.29 = 0.36125a[tex]_y[/tex]

a[tex]_y[/tex] = 1.29 / 0.36125

a[tex]_y[/tex] = 3.5709 m/s²

Now when the text book is moving with acceleration , the dynamic equation will be;

T₁ = m₁a[tex]_y[/tex]

where m₁ is the mass of the text book ( 2.10 kg )

a[tex]_y[/tex] is the vertical acceleration ( 3.5709 m/s² )

so we substitute

T₁ = 2.10 × 3.5709

T₁ = 7.4989 N

Therefore, the tension in the part of the cord attached to the textbook is 7.4989 N

A certain type of laser emits light that has a frequency of 4.9 x 1014 Hz. The light, however, occurs as a series of short pulses, each lasting for a time of 2.9 x 10-11 s. The light enters a pool of water. The frequency of the light remains the same, but the speed of light slows down to 2.3 x 108 m/s. In the water, how many wavelengths are in one pulse

Answers

Answer:

N = 1.42 × 10⁴ cycles

Explanation:

Given that:

frequency f = 4.9 × 10¹⁴ Hz

Time = 2.9 × 10⁻¹¹ s

Speed = 2.3 × 10⁸ m/s

Recall that:

wavelength [tex]\lambda = \dfrac{c}{f} \\ \\[/tex]

Horizontal distance [tex]\Delta x = ct[/tex]

Number of wavelengths [tex](N) = \dfrac{\Delta x}{\lambda}[/tex]

[tex]N = \dfrac{ct}{c/f} \\ \\ N= ft[/tex]

N = (4.9 × 10¹⁴ cycles/s) (2.9 × 10⁻¹¹ s)

N = 14210

N = 1.42 × 10⁴ cycles

using the human species as an example, explain what is meant by variation of traits

Answers

a variation of traits could be a variation of personality traits? everybody is different? what I do know Is that variation can be inherited or environmental which means that you get it from your ancestors through DNA or you get it from the environment around you

Calculate the acceleration of a car if the force on the car is 450 Newtons and the mass is 1300 kilograms.

Answers

[tex] \Large {\underline { \sf {Required \; Solution :}}}[/tex]

We have

Force, F = 450 NMass of the car, m = 1300 kg

We have been asked to calculate the acceleration of the car.

[tex]\qquad \implies\boxed{\red{\sf{ F = ma }}}\\[/tex]

F denotes Forcem denotes massa denotes acceleration

[tex] \quad \twoheadrightarrow\sf { 450 = 1300a} \\ [/tex]

[tex] \quad \twoheadrightarrow\sf {\cancel{ \dfrac{450}{1300}} = 1300a} \\ [/tex]

[tex]\quad\twoheadrightarrow\boxed{\red{\sf{0.346 \; ms^{-2} = a }}}\\[/tex]

Therefore, acceleration of the car is 0.346 m/.

If a virtual image is formed 10.0 cm along the principle axis from a convex mirror of focal length-15.0 cm, how far is the object from the mirror

Answers

Answer:

U=30cm

Explanation:

All you have to do is to put

Mirror formula , 1/f=1/u + 1/v

You should be careful in sign convention .

Virtual image is negative

we take focal length of convex lens negative even if its not given and so on...

A metal can containing condensed mushroom soup has mass 215 g, height 10.8 cm, and diameter 6.38 cm. It is placed at rest on its side at the top of a 3.00 m long incline that is at 25.0 degrees to the horizontal, and it is then released to roll straight down. Assuming mechanical energy conservation, calculate the moment of inertia of the can if it takes 1.50 s to reach the bottom of the incline. Which pieces of data, if any, are unnecessary for calculating the solution

Answers

Answer:

I = 1.093 x 10⁻⁴ kg.m²

Here, all the other data, namely, the height of the can, length of the inclined plane, angle of inclination, time to reach the bottom, are unnecessary.

Explanation:

The can which is filled with the soup can be modelled as a solid cylinder. The moment of inertia of this solid cylinder about its axis of rotation can be given by the following formula:

[tex]I = \frac{1}{2}mr^2[/tex]

where,

I = moment of inertia of can = ?

m = mass of can with soup = 215 g = 0.215 kg

r = radius of can = diameter/2 = 6.38 cm/2 = 3.19 cm = 0.0319 m

Therefore,

[tex]I = \frac{1}{2}(0.215\ kg)(0.0319\ m)^2 \\[/tex]

I = 1.093 x 10⁻⁴ kg.m²

Here, all the other data, namely, the height of the can, length of the inclined plane, angle of inclination, time to reach the bottom, are unnecessary.

An aircraft has to fly between two cities, one of which is 600.0 km north of the other. The pilot starts from the southern city and encounters a steady 100.0 km/h wind that blows from the northeast. The plane has a cruising speed of 236.0 km/h in still air. In what direction (relative to east) must the pilot head her plane

Answers

Answer:

72.57° North of east

Explanation:

From the given information:

We can compute the velocity plane that is related to the ground in air in the North direction as;

[tex]v^{\to} _{PG} = v \\ \\ v^{\to} _{PG,x} = 0 \\ \\ v^{\to} _{PG,y} = v[/tex]

However, the velocity of the wind-related to the ground from the NorthEast  direction is;

[tex]v^{\to}_{wG}=100 \ km/h \\ \\ \text{from North East} \\ \\ v_{wG,x} = (-100 \ km/h ) cos 45 = -70.7 km/h \\ \\ v_{wG,y} = (-100 \ km/h ) sin 45 = -70.7 km/h[/tex]

Now,

Since the plane is moving with a 236 km/h speed in the Northeast direction;

Then;

[tex]v^{\to} _{pw} = 236 \ km/h \\ \\ v^{\to} _{pw.x} = (236 m/s) cos \theta \\ \\ v^{\to} _{pw,y} = (236\ m/s) sin \theta \\ \\ v_{pG,x} = v_{pw,x} + v_{w G,x} \\ \\ \implies 0 = (236 \ km/h) sin \theta -( 70.7 \ km/h) \\ \\ \implies cos \theta = \dfrac{70.7 \ km/h}{236 \ km/h} \\ \\ \theta = cos^{-1} (0.2996) \\ \\ \mathbf{\theta = 72.57}[/tex]

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