1. An airplane flies with an airspeed (speed relative to the air) of 215
km/h. The wind is blowing at 45.0 km/h. Find the velocity of the
plane (relative to the ground) if the plane is pointed straight west and
the wind is:

A. blowing toward the
west (3 points)

B. blowing toward the east (3 points)

C. blowing toward the south (4 points)
(For part C be sure to calculate the angle as well.)

Answers

Answer 1

Answer:

Explanation:

A) 215 + 45.0 = 260 km/h West

B) 215 - 45.0 = 170 km/h West

C) √(215² + 45.0²) = 219.6588... 220 km/h

θ = arctan(45.0/215) = 11.8214... 11.8° S of W

Answer 2

The airplane's ground velocity is westward at 170 km/h regardless of wind direction. The south wind does not affect the plane's westward velocity in component C.

Vector addition will determine each scenario's plane velocity relative to the ground. We'll simplify by assuming a consistent horizontal breeze.

Let's divide the airplane's velocity and wind velocity into westward (negative) and eastward (positive) components.

Data: Plane airspeed = 215 km/h.

45 km/h wind.

A. West wind:

The airplane's westward (negative) ground velocity:

West aeroplane velocity = -215 km/h

-45 km/h west wind.

Find Vg when the wind blows west:

Vg (west) = Aeroplane velocity + Wind velocity (west)

Vg(west) = -215 km/h - (-45 km/h)

Vg (west)=-170 km/h

B. East wind: 2. Westward (negative) aeroplane velocity relative to the ground:

West aeroplane velocity = -215 km/h

Wind velocity (east) = 45 km/h (because the wind is blowing eastward)

Find Vg when the wind blows east:

Vg (west) = Aeroplane velocity (west) + Wind velocity (east).

Vg (west) = -215 + 45 km/h

Vg (west)=-170 km/h

C. Southerly wind: 3. Westward (negative) aeroplane velocity:

West aeroplane velocity = -215 km/h

Wind velocity (south) = 0 km/h (no effect since wind is not moving westward)

Find Vg when the wind blows south:

Vg (west) = Aeroplane velocity (west) + Wind velocity (south).

Vg (west) = -215 + 0 km/h.

Vg (west)=-215 km/h

All resultant velocities are westward and 170 km/h.

Part C's plane's westward velocity is unaffected by the south wind. The plane's ground velocity remains -215 km/h (westward). Since the heading is westward, no angle is needed.

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Related Questions

The gradual increase in activity over time is called

Answers

Answer:Progression is the way in which an individual should increase the load. It is a gradual increase either in frequency, intensity, or time or a combination of all three components.

Explanation:

the gradual increase in activity over time is called progression

I need help with this question!

Answers

Answer:

Explanation:

Conservation of momentum

In the x direction

2(15) + 2(-10) = 2(-5) + 2(vBfx)

vBfx = 10 m/s

In the y direction

2(30) + 2(5) = 2(20) + 2(vBfy)

vBfy = 15 m/s

vBf = 10i + 15j

KEi = ½(2)(15² + 30²) + ½(2)(-10² + 5²) = 1250 J

KEf =  ½(2)(-5² + 20²) + ½(2)(10² + 15²) = 750 J

KEf - KEi = 750 - 1250 = -500 J

please answer it I will mark it brainliest​

Answers

Explanation:

1) If you spot something you think might be hazardous in your workplace, report it to your employer and safety rep straight away. Your employer should then decide what harm the hazard could cause and take action to eliminate, prevent or reduce that harm. Read more about risk assessments .

2) Complex hazards are understood as various combinations of sources of hazards that lead to the accident occurrences. ... The term "natural-technological" applies to both human-induced intensification of natural risks and any accidents in the technosphere triggered by natural processes or phenomena.

3)Risk Evaluation : To determine who may be harmed. Risk Control : Taking preventive measures to control the impact of risk.

In general, to do an assessment, you should:

Identify hazards.

Determine the likelihood of harm, such as an injury or illness occurring, and its severity. ...

Identify actions necessary to eliminate the hazard, or control the risk using the hierarchy of risk control methods.

A hazard is something that can cause harm, e.g. electricity, chemicals, working up a ladder, noise, a keyboard, a bully at work, stress, etc. A risk is the chance, high or low, that any hazard will actually cause somebody harm.



Risk Identification : It is a process of characterizing hazards. Risk Analysis : A process of determining the impact of risk. Risk Evaluation : To determine who may be harmed.

Physical Science A 2021-2022
Why does increasing the number of trials increase confidence in the results of the experiment?

Answers

Answer:

Increasing the number of trials reduces the impact of any one imprecise measurement. … To increase the number of attempts, you can find an average result for the experiment, as well as find and discrepancies as human error if you perform an experiment several times.

Explanation:

hope it helps :)

Answer:

It is because the increase in the number of trials reduces the impact of any one imprecise measurement. Using an average value for data points provides a better representation of the true value.

A student connects a 21.0 V battery to a capacitor of unknown capacitance. The result is that 52.8 µC of charge is stored on the capacitor. How much energy (in J) is stored in the capacitor?

Answers

Answer:

1.108 ×  [tex]10^{-3}[/tex]J

Explanation:

v=21.0v

Q=52.8×  [tex]10^{-6}[/tex]

E=?

V=E/Q

E=v ×Q

 =21 ×52.8 ×[tex]10^{-6}[/tex]

 =1108.8   ×[tex]10^{-6}[/tex]

E= 1.108 ×  [tex]10^{-3}[/tex]J

A car is moving north on a freeway. If a bug is flying south on the freeway, is the momentum of the bug positive or negative?
Neither

Positive


Negative


Can be both depending on the weather

Answers

Negative

Because the car is moving up and the bug is moving down. but it also depends on the weather so choice between one of those two I think is Negative but I may be wrong.

Which is better, forward bias or reverse bias, and why ?!

Answers

Answer:

reverse bias

Explanation:

bcz the potential barrier and impedes the flow of charge carriers. In contrast, a forward bias weakens the potential barrier, thus allowing current to flow more easily across the junction.

if the force acting on a body of mass 40 k.g is doubled by how much will the acceleration change​

Answers

Answer:

Explanation:

Ignoring friction, the acceleration will double

F = ma

2F = m(2a)

The magnetic field B at all points within the colored circle of the figure (Figure 1) has an initial magnitude of 0.780 T. (The circle could represent approximately the space inside a long, thin solenoid.) The magnetic field is directed into the plane of the diagram and is decreasing at the rate of 0.0300 T/s.
a) What is the magnitude of the induced electric field at any point on the circular conducting ring with radius r = 0.100 m ?
b) What is the direction of this field at any point on the circular conducting ring?
c) What is the current in the ring if its resistance is 4.00 Ω ?
d) What is the emf between points a and b on the ring?
e) If the ring is cut at some point and the ends are separated slightly, what will be the emf between the ends?

Answers

The magnitude of the induced electrical field is 0.0015V/m, the field is pointing towards the clockwise direction while the current in the ring will be 0.0002355A if the resistance is 4 ohms. The emf between point a and b is zero and the EMF across the point if they're slightly separated between the ends is 0.000942V

To solve this question, we would have to go about each one individually

Data:

[tex]r=10cm=0.1m\\[/tex]

a.

The magnitude of the induced electrical field at any point within the radius is

[tex]\int\limits^a_b {E} \, du=\frac{dU}{dt}=\pi \frac{dB}{dt}=\pi r^{2}\frac{dB}{dt}\\E*2\pi r=\pi r^{2}\frac{dB}{dt} \\E=\frac{r}{2}\frac{dB}{dt}=\frac{0.1}{2}*0.03=0.0015V/m[/tex]

b.

The field is pointing towards the clockwise direction.

c.

The current in the ring if we are given a resistance of 4ohms

[tex]I=\frac{emf}{R}=\frac{\pi r^{2}\frac{dB}{dt} }{R} =\frac{\pi (0.1)^2*0.03}{4} =0.0002355A[/tex]

d.

The emf between point a and b is zero

e.

The EMF across two points if they're separated by small distance across the ring is

we would use the formula to solve for the EMF

[tex]E=\pi r^{2}\frac{dB}{dt}=\pi (0.1)^2*0.03=0.000942V[/tex]

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The range of a Cannon ball fired horizontally from a laboratory table equals to 8/ 3 times the height of the table . what is the direction of the velocity of the projectile as it strikes the ground?

A,450 from the horizontal
B,600 from the horizontal
C,530 from the horizontal
D,370 from the horizontal​

Answers

Theprojectile launch allows to find the correct answer for the question about the direction of the ball when it hits the ground is:

       D)  θ = 37º measured clockwise from the x-axis.

projectile launch is an application of kinematics where there is no acceleration on the horizontal axis and the acceleration on the vertical axis is the gravity acceleration.

The horizontal distance traveled until it reaches the ground is

          x = v₀ₓ t

The height of the bullet is

          y = y₀ + [tex]v_o_y[/tex]  t - ½ g t²

The initial vertical velocity is zero and when it reaches the ground its height is zero.

         0 = y₀ - ½ g t²

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

We substitute

          [tex]x = v_{ox} \ \sqrt{\frac{2y_o}{g} }[/tex]  

They indicate that the distance is:

          x = [tex]\frac{8}{3} \ y_o[/tex]  

We substitute

          [tex]\frac{8}{3} y_o = v_{ox} \sqrt{\frac{2y_o}{g} }[/tex]  

          [tex]v_{ox} = \frac{8}{3} \sqrt{\frac{y_o g}{2} }[/tex]  

The vertical speed when reaching the ground is:

         [tex]v_y = v_{oy} - g t \\v_y = - g t \\v_y = - g \sqrt{\frac{2y_o}{g} }[/tex]

Let's use trigonometry for the direction at the point of hitting the ground.

         tan θ = [tex]\frac{v_y}{v_x}[/tex]  

          θ = tan⁻¹ ( [tex]- \frac{3}{8} \ \frac{\sqrt{2y_o/g}}{\sqrt{y_o g /2} }[/tex] )

          θ = tan⁻¹ ( [tex]tan ^{-1} (- \frac{3}{8} \sqrt{4} )= tan^{-1 } \ (- \frac{3}{4} )[/tex]

          θ = -36.9º

The negative sign indicates that the angle is clockwise from the x-axis.

In conclusion, using the projectile launch we can find the correct answer for the question about the direction of the bullet when it hits the ground is:

      D) θ = 37º measured clockwise from the x-axis.

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A teacher took two latex balloons and blew them up with helium gas to the same size. She took one and labeled it Balloon A and placed it in a -15o C freezer. The second one she labeled BALLOON B, and she took it outside and tied it to the railing in the sun on a 30o C day. After a half hour, she had the students measure the circumference of each balloon. Which TWO outcomes do you predict the students will find and why?

Answers

Answer:

n

Explanation:

TWO outcomes can be predicted the students will find:

The size of balloon A becomes smaller.The size of balloon B becomes larger.

What is the relation between temperature and volume of the gases?

When a constant mass of gas is cooled, its volume falls, and when the temperature is raised, its volume grows. The volume of the gas rises by 1/273 of its initial volume at 0 °C for every degree of temperature rise.

In layman's words, the volume of a fixed mass of gas is exactly proportional to temperature at constant pressure.

The teacher took two latex balloons and blew them up with helium gas to the same size. As she placed Balloon A in a -15° C freezer, its temperature decreases and that's why, the size of balloon A becomes smaller. Again she  placed Balloon B in the sun on 30° C day, its temperature increases and that's why, the size of balloon B becomes larger.

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The intensity of a sound wave increases as the wave spreads out from the source of the sound. True or false

Answers

Answer:False

Explanation:

the intensity of the sound wave decreases with increasing distance from the source.

13) What property of matter would be measured with this piece of equipment?
A) The mass of an apple
hing
)
By the temperature of a room
The volume of water in a glass.
D) The length of a piece of string.

Answers

Answer:

a

Explanation:

i took the test 100%

A metal bar has a volume of 32 cm3. The mass of the bar is 256 g. What is the density of the metal? A. 290 g/cm3 6 B. 8,200 g/cm C. 8.0 g/cm3 O D. 220 g/cm​

Answers

The density of the metal is ρ = 8.0g/cm³.

Why is density important?

The measure of material how densely it is packed together is called density. As the mass per unit volume, it has that definition. Symbol for density: D or Formula for Density: When is the density, m is the object's mass, and V is its volume, the equation is: = m/V.

Because it enables us to predict which compounds will float and which will sink in a liquid, density is a crucial notion. As long as an object's density is lower than the liquid's density, it will often float.

Equation :

To the given equation we have :

mass of the bar = 256g

volume of metal bar = 32cm³

So according to the formula of density

ρ = m/V

So, putting values

ρ = 256g /32cm³

ρ = 8.0g/cm³

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how is the atomic mass determined?

Answers

Answer:

Atomic mass is defined as the number of protons and neutrons in an atom, where each proton and neutron has a mass of approximately 1 amu (1.0073 and 1.0087, respectively). The electrons within an atom are so miniscule compared to protons and neutrons that their mass is negligible.

Explanation:

explain how wind erosion changes land forms

Answers

Answer:

the wind carries abrasive materials

Explanation:

such as sand and salt over time theses small particles slowly strip way at the land form sculpting it by eroding the softer layers first


The potential at point P is the work required to bring a one-coulomb test charge from far
away to the point P?

True or false ?

Answers

true so true Its so true

A 0.050 kg metal bolt is heated to an unknown initial temperature. It is then dropped into calorimeter containing 0.15 kg of water with an initial temperature of 21.0°C. The bolt and the water then reach a final temperature of 25.0°C. If the metal has a specific heat capacity of 899 J/kg•°C, find the initial temperature of the metal.

Answers

The initial temperature of the metal bolt is 80.8 °C

We'll begin by calculating the heat absorbed by the water.

Mass of water (M) = 0.15 Kg Initial temperature (T₁) = 21 °CFinal temperature (T₂) = 25 °CChange in temperature (ΔT)  = T₂ – T₁ = 25 – 21 = 4 °CSpecific heat capacity of water (C) = 4184 J/KgºCHeat absorbed (Q) =?

Q = MCΔT

Q = 0.15 × 4184 × 4

Q = 2510.4 J

Finally, we shall determine the initial temperature of the metal bolt.

Heat absorbed by water = 2510.4 JHeat released by metal (Q) = –2510.4 JMass of metal (M) = 0.050 Kg Final temperature (T₂) = 25 °CSpecific heat capacity of metal (C) = 899 J/Kg°CInitial temperature (T₁) =?

Q = MC(T₂ – T₁)

–2510.4 = 0.050 × 899 (25 – T₁)

–2510.4 = 44.95 (25 – T₁)

Clear bracket

–2510.4 = 1123.75 – 44.95T₁

Collect like terms

–2510.4 – 1123.75 = –44.95T₁

–3634.15 = –44.95T₁

Divide both side by –44.95

T₁ = –3634.15 / –44.95

T₁ = 80.8 °C

Thus, the initial temperature of the metal is 80.8 °C.

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when do we say a curvilinear motion is accelerated?​

Answers

Answer:

There always is an acceleration in a curvilinear motion, as the velocity vector changes, so always.

What is the average momentum of Cart 1 during the entire time shown before
kg-m
the collision, in units .?
$
Note that the collision appears to take place somewhere between 0.5 s and 0.6
s (perhaps at 0.6 s exactly, but we cannot be sure), so we can safely say that
data up to and including 0.5 s is "before the collision."

Answers

The average momentum of Cart 1 during the entire time shown before the collision is 0.24 kgm/s.

The given parameters:

Mass of the cart 1 = 500 g = 0.5 kg

The average velocity of the cart 1 before collision is calculated as follows;

[tex]v = \frac{\Delta x}{\Delta t} \\\\v = \frac{0.55 - 0.31}{0.5 - 0} \\\\v = 0.48 \ m/s\\\\[/tex]

The average momentum of Cart 1 during the entire time shown before the collision is calculated as follows;

[tex]P = mv\\\\P = 0.5 \times 0.48\\\\P = 0.24 \ kgm/s[/tex]

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Q1) Assertion : a vector can be resolved into maximum two components
Reason : in projectile motion the velocity is resolved into vertical and horizontal components

a) Both A and R are correct; R is the correct explanation of A
b) Both A and R are correct; R is not correct explanation of A
c) A is correct; R is not correct
d) A is not correct; R is correct

Q2) Assertion : speed is constant in uniform circular motion
Reason : acceleration is constant in uniform circular motion

a) Both A and R are correct; R is the correct explanation of A
b) Both A and R are correct; R is not correct explanation of A
c) A is correct; R is not correct
d) A is not correct; R is correct

Answers

Answer:

Explanation:

Q1) d) A is not correct; R is correct (sort of) but is not the reason for A.

A vector may be resolved into an infinite number of components.

Breaking a projectile motion into the vertical and horizontal components is the EASIEST way to solve it because acceleration can be assumed to be constant in both directions. Zero for horizontal and g(ravity) in the vertical. However it is not the ONLY way to do it. Other ways are much more complex.

Q2) a) Both A and R are correct; R is the correct explanation of A  (sort of)

A is correct. Speed is constant in uniform circular motion.

R is correct if you call CENTRIPETAL acceleration constant. It is only constant in magnitude. The direction is always changing as the vector always points toward the center of rotation.

How should an artistic statement be written?

ОА.

from your own perspective

OB.

in a non-factual, persuasive manner

O C.

as a lengthy and overly extensive essay

OD. by the artist alone with no outside critique

Answers

An artistic statement should be written from your own perspective.

What is an artistic statement ?

An artist's statement, also known as an artist statement, is a written summary of their creative output. The brief essay serves as an explanation of and a defense of their own work for the spectator. As a result, it is didactic, descriptive, or reflective in nature and seeks to inform, connect with an artistic environment, and reveal the inspiration behind the work.

The writing created by the artist aims to contextualize, justify, extend, and/or explain their body of work. It situates—or tries to situate—the piece in relation to art history, art theory, the contemporary art scene, and the times. The statement also demonstrates that the artist is aware of their aims, their practice, its place within the parameters of art, and the debate that surrounds it.

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Two steel guitar strings have the same length. String A has a diameter of 0.489 mm and is under 410 N of tension. String B has a diameter of 1.27 mm and is under a tension of 809 N. Calculate the ratio of the wave speeds, vA/vB, in these two strings.

Answers

Answer:

Explanation:

vA / vB = √(TA/(m/L)) / √(TB/(m/L))

The lengths are the same, so the L divides out to 1

The material is identical so the mass will be directly proportional to the cross sectional area of the string

vA / vB = √(TA/(πdA²/4)) / √(TB/(πdB²/4))

π/4 is common so divides out to 1

vA / vB = √(TA/dA²) / √(TB/dB²)

vA / vB = √(410/0.489²) / √(809/1.27²)

vA / vB = 41.407 / 23.396

vA / vB = 1.8488961...

vA / vB = 1.85

7. Cellular respiration that uses oxygen is called ​

Answers

Answer:

Cellular respiration that uses oxygen is called ​Cellular respiration

Explanation:

Without oxygen, it's call fermentation.

is xenon a pure substance​

Answers

[tex]\large\huge\green{\sf{Yes}}[/tex]

a
A person throws a ball up into the air, and the ball falls back towar
would the kinetic energy be the lowest? (1 point)
at a point before the ball hits the ground
when the ball leaves the person's hand
o when the ball is at its highest point
at a point when the ball is still rising

Answers

Answer:

when the ball is at its highest point

Explanation:

Provided the ball returns to where it was thrown. The velocity, and therefore kinetic energy, will be momentarily zero at the highest point of the throw.

A block of mass m=10 kg at rest slides down a rough incline plane of θ=30° and length l=5 m. The coefficient of kinetic friction between the block and the incline is μ_k=0.1. At the bottom of the plane the block continues to slide on a frictionless surface and hits a spring with spring constant k=100 N/m.

Answers

The speed of the block at point B = 6.36 m/s

Given data :

Mass ( m ) = 10 kg

Angle of inclination ( θ ) = 30°

Length of incline = 5 m

Determine the speed of the block at a point B on the incline

First step : Calculate the work done by frictional force

given that there is no friction on the horizontal plane. Vinclined = Vhorizontal

∴ Work done by frictional force ( Wf ) = F * L cos 180

                                                              = - ( μ_k * N ) L

                                                              = - (0.1 * 84.86   ) * 5  = - 42.4 Joule

where N = mg cos 30°

              = 10 * 9.8 * 0.866 = 84.86

Next step : Calculate the speed of the block at point B

applying work energy theorem

Wf = ΔK.E + ΔP.E

     = ( 1/2 mv² - 0 ) + ( 0 - mgh )

∴ - 42.48  = 1/2 mv² - mgh

1/2 mv² = mgh - 42.48

            = 10 * 9.8 * 5 sin 30° - 42.4

       v²  = 40.52

V ( speed of the block at point B ) = 6.36 m/s

Hence we can conclude that the speed of the block at point B = 6.36 m/s

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Attached below is the complete question related to your question

If a force of 20kn acts on a circular rod of a diameter 10mm, calculate the stress of the rod

Answers

σ = F/A = F/ (πD²/4) = 20000 / (π0.10²/4) = 2.55 MN/m²

What are the characteristics of high energy wave?

A. Low frequencies and short wavelengths.

B. High frequencies and long wavelengths.

C. Low frequencies and long wavelengths.

D. High frequencies and short wavelengths

Answers

Answer:

D. High frequency and short wavelengths.

Explanation:

If a wave is high in energy it will have a higher frequency.

High frequency = short wavelengths

how long is the photons journey from the milky way to earth

Answers

well i’d love to answer your question but its missing some information. firstly the earth is within the milky way galaxy so something cannot travel “from” but if you said “through” the answer would be 299,792,458 meters a second
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