Consider a boat heading due east at 15 miles/hour. The water's current is moving at 7.1 miles/hour at 45º south of east. Drag vectors for the boat and the current into the vector addition simulation.
What do Rx, Ry, , and |R| represent in terms of the force of the current, and what do they represent in terms of the forces moving the boat?

Answers

Answer 1

If a boat is going East at 15mph and there is a water current going southeast at 45° then the boat is being drifted southward.  So since the current is going at an angle then it has a x and y component.  So Rx refers to the x-component force of the current and Ry refers to the y-component of the current, and |R| refers to the magnitude of these forces.

Answer 2

Answer:

The boat velocity is 20.6 mph and  14 south of east

Explanation:

For this exercise it is best to write the speeds based on its components x and y

Boat    V1x = 15 mph

Water  V2 = 7.1 mph

           θ = - 45º

 Where the angle is measured from the x axis, the negative indicates that it is measured on an hourly basis

           V2x = V2 cos (-45) = 7.1 (0.707)

           V2y = V2 sin (-45) = 7.1 (-0.707)

           V2x = 5.0 mph

           V2y = -5.0 mph

The total system speed is the sum of each component

          Vx = V1x + V2x

          Vy = V1y + V2y

          Vx = 15 + 5.0

          Vy = 0 -5

          Vx = 20 mph

          Vy = -5.0 mph

We can give result in the form of magnitude and angle

          V² = Vx² + Vy²

          θ = tan-1 (Vy / Vx)

          V = √ [20² ++ (-5)²]

          θ = tan-1 (-5/20)

         

         V = 20.6 mph

         θ = -14º

 The boat velocity is 20.6 mph and  14 south of east


Related Questions

A 1.5m wire carries a 6 A current when a potential difference of 68 V is applied. What is the resistance of the wire?

Answers

Answer:

[tex]11.3 \Omega[/tex]

Explanation:

We can find the resistance of the wire by using Ohm's law:

[tex]V=RI[/tex]

where

V is the voltage applied

R is the resistance

I is the current

In this problem, we know I = 6 A and V = 68 V, so we can re-arrange the equation to find the resistance of the wire:

[tex]R=\frac{V}{I}=\frac{68 V}{6 A}=11.3 \Omega[/tex]

Which statement correctly describes sound waves? Sound waves do not travel through a medium. Sound waves are transverse waves. Sound waves are longitudinal waves. Sound waves are not considered mechanical waves.

Answers

Sound waves are longitudinal waves.

What is a mechanical wave?

A mechanical wave is a type of wave that do not require material medium for its propagation.

Examples of mechanical waves;Sound wavesWater wavesSpring wavesSeismic waves

A mechanical wave can be longitudinal or transverse depending on the direction of the propagation.

Longitudinal waves

Longitudinal waves travel parallel to their direction of propagation, example sound waves.

Thus,  Sound waves are longitudinal waves.

Learn more about sound waves here: https://brainly.com/question/1199084

Final answer:

Sound waves are longitudinal waves that require a medium to propagate. The disturbances in sound waves are periodic variations in pressure that are transmitted in fluids.

Explanation:

Sound waves are longitudinal waves that require a medium to propagate. Unlike light waves, which can travel through a vacuum, sound waves need a medium, such as air, water, or solids, to travel through. The disturbances in sound waves are periodic variations in pressure that are transmitted in fluids, like air and water.

many musical instruments use "boxes" as a part of an instrument. Some examples are acoustic guitars and pianos. From your experience in this lab on building your own musical instrument, write a brief essay on the purpose of these "boxes". Include description of a xylophone and what is uses of for the purpose of the box.

Answers

Answer :

Essay on Sound instruments/sound boxes:

Musical instruments also some times called "sound box",  because the box modifies the sound of an instrument. It also helps in transfer of sound to the surrounding air.

The box is the open chamber in the body of musical instrument

The sound box responds more strongly to vibrations at certain frequencies created by playing instrument. This phenomena is known as resonance.

The resonance impacts on the tone quality.

For example xylophone, which is a musical instrument with wooden bars. These wooden bars are arranged similar to the piano. under each bar of the xylophone there is a resonator tube, which amplifies the sound.


Final answer:

The 'boxes' or 'bodies' of musical instruments like guitars and pianos serve as resonators, amplifying and enhancing the tone quality of the instrument. In the case of a xylophone, the boxes or tubes beneath the wooden bars act as amplifiers of the initial sound produced.

Explanation:

In the field of music, 'boxes' are components of many musical instruments, often called the instrument's 'resonating box' or 'body'. They contribute significantly to the amplification and tone quality of the instrument. For instance, in an acoustic guitar, the sound box amplifies the vibrations of the strings and projects the sound.

When playing a piano, the sound is created by hammers striking strings. These vibrations are then transmitted to the soundboard, basically a large 'box', which increases the surface area that vibrates to project the sound throughout the room.

In relation to a xylophone, the utility of the 'box' is a bit different. In a xylophone, the wooden bars produce sound which is amplified by resonator tubes or boxes below. Instead of acting as the primary sound source like in a guitar or piano, the tubes or boxes under a xylophone amplifies the initial sound created by the strike of the mallet.

Learn more about Boxes in Musical Instruments here:

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As you hold book at rest in your hand two forces are being exerted on the book identify the force

Answers

gravity dwn, your hand up and equal to grav

Final answer:

When you hold a book at rest, two forces act upon it: the downward gravitational force exerted by the Earth represented as -14ĵ N and the upward force exerted by your hand that equals 14 N, balancing out the weight of the book. These forces cancel each other out, keeping the book at rest as per Newton's Second Law.

Explanation:

When you hold a book at rest in your hand, two primary forces are acting upon it. These are the gravitational force, which is the book's weight pulling it downwards, and the force exerted by your hand, pushing upwards against the book. The downward gravitational force is caused by the earth's mass attracting the book's mass. This force is represented as -14ĵ N in physics. The force exerted by your hand counters this gravitational pull, allowing the book to remain at rest in your hand.

By Newton's second law, since the book is at rest, the net force acting on the book is zero. Meaning, the gravitational force and the force exerted by your hand cancel each other out. This is represented as: FPH + FEH = 0, where FPH is the force exerted by your hand and FEH is the force exerted by the Earth. With proper calculation, the force exerted by your hand on the book equals 14 N in the upward direction, counteracting the book's weight.

Learn more about Forces on a Resting Book here:

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During the operation of a laser, as photons interacts with atoms inside the laser the photons _________.

Answers

Final answer:

Photons in a laser induce stimulated emission in atoms, creating additional photons of the same frequency and phase, which results in coherent and monochromatic light.

Explanation:

During the operation of a laser, as photons interact with atoms inside the laser, the photons stimulate those atoms to emit additional photons through a process known as stimulated emission. The original and the newly created photons have the same frequency and phase, leading to coherent and monochromatic light. The photons are initially absorbed by the electrons in the atoms, which are then elevated to higher energy levels. Most of these electrons drop back to their ground state immediately, while some remain in a metastable state. A population inversion is created when more atoms are in the excited state than in the ground state. When a photon with the correct energy interacts with an electron in the metastable state, it triggers the emission of a second photon with the same energy, contributing to the coherence of the laser light. The laser light amplification is thus derived from the energy input, such as from a flash tube or an electrical discharge, leading to a cascade or chain reaction of photon production.

A force is a push or a pull applied to a body or an object. Consider the workout you just participated in. What forces were present on your body as you practiced your running stride or your jump shot? Were the forces helpful or hurtful? Explain using the ideas of external forces, internal forces, tension, and compression.

Answers

Gravity and friction

Answer: One external force was the ground pushing into my foot as I ran. This is a compression force. Gravity is also an external force that is working to pull my body down toward the ground. Since I’m trying to move my body forward, gravity is probably hurting my performance. Since I lean slightly forward when I run, there is a small tension force in my lower back. There is also tension in my shoulders as my arms swing by my sides.

Explanation: EDMENTUM

A 50-foot flagpole is at the entrance of a building that is 300 feet tall. If the length of the flagpole's shadow is 30 feet at a certain time of day, how long is the building's shadow at that time? A. 75 feet B. 150 feet C. 180 feet D. 200 feet

Answers

First you will want to sketch out both of the situations. It should be two sketches, one for the flagpole and one for the building.

To solve this, you will want to create a proportion.

Flagpole height/flagpole shadow=building height/building shadow

Therefore, it should look like this:
50/30= 300/x

Solve for x:
50x=9,000
X= 180 feet

YOUR ANSWER IS C.

Answer:

c on plato

Explanation:

At an auto race, a member of the pit crew stands beside the track. A car approaches him at 100 m/s and emits a sound at frequency 1100Hz. The air is still and the speed of sound is 340mls. What frequency will the pit crew member hear? A.1560Hz
B.1420Hz
C.1640Hz
D.850Hz

Answers

As per the formula of Doppler's effect we know that

[tex]f = f_o\frac{v}{v - v_s}[/tex]

now we know that

v = 340 m/s

[tex]v_s = 100 m/s[/tex]

[tex]f_o = 1100 Hz[/tex]

now we will have

[tex]f = 1100(\frac{340}{340 - 100})[/tex]

[tex]f = 1560 Hz[/tex]

so the frequency heard by the crew will be approx 1560 Hz

Please help me out! Thanks! (:
A physical quantity X depends on mass, distance, and time. Use dimensional analysis to decide which of these is a possible expression for X, where a is acceleration, v is velocity, and F is force.

A. X=1/2at²
B. X=at
C. X=Ft
D. X=vt

Answers

X=Ft is a possible equation for X because by using SI units its final term became Kg metre per second

The correct expression for a physical quantity depending on mass, distance, and time is C. X = Ft, as it has the dimensions MLT⁻¹ and includes all three: mass (M), distance (L), and time (T).

In order to determine which expression for a physical quantity X that depends on mass m, distance s, and time t, is possible using dimensional analysis, we examine the provided options. Using the dimensions given [m] = M, [s] = L, [v] = LT⁻¹, [a] = LT⁻², and [t] = T:

X = 1/2at² has dimensions of L( T⁻² )( T² ) = L which does not involve mass.

X = at has dimensions of L( T⁻² )( T ) = LT⁻¹ which is velocity, not involving mass.

X = Ft has dimensions of ( M LT⁻² )( T ) = MLT⁻¹ which represents impulse or momentum, and it involves mass, length, and time.

X = vt has dimensions of L( T⁻¹ )( T ) = L which is distance, but it does not involve mass.

The correct option that includes dimensions of mass, distance, and time is C. X = Ft which represents impulse or momentum.

WILL GIVE BRAINLIEST PLEASE

1. An amount of heat equal to 10.5 x 105 J is supplied to 5 kg of water to raise its temperature from 25°C to 75°C. What is the specific heat capacity of water?

4.2 J/kg.°C

4200 J/kg.°C

1000 J/kg.°C

4200 kcal/kg.°C

2. What is the heat required in kilocalories to convert 2 kg of ice at 0°C completely into steam at 100°C?

80 Calories

1440 Calories

4186 Calories

540 Calories

3. How many kg of ice needs to be added to 1.31 kg of water at 64.9°C to cool the water to 14.9°C when the Latent heat of ice = 80 kcal/kg?

0.69

6.9

690

6900

4. 540 g of ice at 0°C is mixed with 540 g of water at 80°C. What is the final temperature of the mixture?

0°C

40°C

80°C

less than 0°C

5. If ∆Q is the heat supplied to a gas so that the internal energy of the system increases by ∆U, then the work done ∆W by the gas is equal to:

∆Q – ∆U

∆Q + ∆U

∆Q x ∆U

∆U/∆Q

6. The internal energy of the gas in a gasoline engine cylinder decreases by 210 J. If 60 J of work is done by the gas, the energy transferred as heat is:

270 J

150 J

-150 J

-270 J

7. A gaseous refrigerant undergoes compression over 160 J of work done on it. If the internal energy of the gas increases by 130 J, the amount of heat transfer is:

290 J

-290 J

30 J

-30 J

Answers

the amount of heat trader is

The amount of kinetic energy an object has can And is related to the objects And.

Answers

he kinetic energy of an object is the energy it has because of its motion. In Newtonian (classical) mechanics, which describes macroscopic objects moving at a small fraction of the speed of light, the kinetic energy ( E) of a massive body in motion can be calculated as half its mass ( m) times the square of its velocity.

What force besides gravity would act on something that had been thrown in the air

Answers

Air resistance could be acting on the object

A 0.500-kilogram cart traveling to the right on a horizontal, frictionless surface at 2.20 meters per second collides head on with a 0.800-kilogram cart moving to the left at 1.10 meters per second. What is the magnitude ofthe total momentum of the two-cart system after the collision?

Answers

Consider the motion towards right as positive and motion towards left as negative.

m₁ = mass of the cart moving to right = 0.500 kg

v₁ = initial velocity before collision of the cart moving towards right = 2.2 m/s

m₂ = mass of cart moving to left = 0.800 kg

v₂ = initial velocity before collision of the cart moving towards left = - 1.1 m/s

initial momentum of the system of carts before the collision is given as

P₁ = m₁ v₁ + m₂ v₂

P₁ = (0.500) (2.2) + (0.800) (- 1.1)

P₁ = 0.22 kgm/s

P₂ = momentum of system of carts after collision

As per conservation of momentum,

Momentum of system of carts after collision = Momentum of system of carts before collision

P₂ = P₁

P₂ = 0.22 kgm/s

Final answer:

The magnitude of the total momentum of the two-cart system after the collision is 0.220 kg*m/s in the opposite direction of the initial velocity.

Explanation:

The total momentum of the two-cart system after the collision can be determined by applying the law of conservation of momentum. According to this law, the total momentum before the collision is equal to the total momentum after the collision.

Before the collision, the momentum of the first cart is calculated by multiplying its mass (0.500 kg) by its velocity (2.20 m/s) and the momentum of the second cart is calculated by multiplying its mass (0.800 kg) by its velocity (-1.10 m/s) since it is moving in the opposite direction. Therefore, the total momentum before the collision is 0.500 kg * 2.20 m/s + 0.800 kg * (-1.10 m/s) = 1.100 kg*m/s - 0.880 kg*m/s = 0.220 kg*m/s.

After the collision, the carts stick together and move as one. Since they have the same velocity, the total momentum after the collision is the sum of the masses multiplied by the common velocity. The total mass of the two carts is 0.500 kg + 0.800 kg = 1.300 kg. Therefore, the total momentum after the collision is 1.300 kg * V, where V is the common velocity. Since the carts move in opposite directions, the common velocity is negative. Setting the total momentum before the collision equal to the total momentum after the collision, we have 0.220 kg*m/s = 1.300 kg * V. Solving for V, we find V = 0.220 kg*m/s / 1.300 kg = -0.169 m/s.

In the diagram below, a student compresses the spring in a pop-up toy a distance of 0.020 m. If 0.068 J of energy are stored in the toy, what is the toy’s spring constant? a) 120 N/m b) 170 N/m c) 225 N/m d) 340 N/m

Answers

Answer: option d) 340 N/m

Explanation:

Energy stored in the spring of the toy,

[tex] E = \frac{1}{2}kx^2 = 0.068 J [/tex]  ( given)

Where, k is the spring constant and x is the length of the string after compression.

x = 0.020 m

Then spring constant, [tex]k = \frac{2E}{x^2}[/tex]

[tex]k = \frac{2\times 0.068 J}{(0.020 m)^2} = 340 N/m[/tex]

Thus, the spring constant of toy is 340 N/m. Correct option is d.

11. A 3.8 kg object is lifted 12 meters. Approximately how much work is performed during the lifting?

Answers

Hope this answers your question!! Ask any help at anytime

a missile is moving 1350 m/s at a 25.0 deg angle. it needs to hit a target 23,500 m away in 55.0 deg direction in 10.20 s. what is the magnitude of its final velocity?

Answers

In this question we have given

velocity of missile=1350m/s

angle at which missile is moving=25degree

distance between missile and targets=23500m

angle between target and missile=55degree

time=10.2s

To find the final velocity of missile we will first find the acceleration required

Let x be the horizontal component of distance

x - vertical component of distance

t-time

ax- horizontal component of acceleration

ay-Vertical component of acceleration

Vx-horizontal component of velocity

Vy-Vertical component of velocity


horizontally: x = Vx*t + ½*ax*t²  

23500m * cos55.0º = 1350m/s * cos25.0º * 10.20s + ½ * ax * (10.20s)²  

ax = 19.2 m/s²  

V'x = Vx + ax*t = 1350m/s * cos25.0º + 19.2m/s² * 10.20s = 1419 m/s  

similarly vertically:

y = Vy*t + ½*ay*t² 

23500m * sin55.0º = 1350m/s * sin25.0º * 10.20s + ½ * ay * (10.20s)²  

ay = 258 m/s²  


V'y = Vy + ay*t = 1350m/s * sin25.0º + 258m/s² * 10.20s = 3204 m/s  

Therefore

V = √(V'x² + V'y²) = 3504 m/s  

therefore magnitude of final velocity of missile=3504m/s

The picture above shows a football player kicking a football. This is known as two dimensional motion. In which direction does the football move?
A) vertical only (y)
B) horizontal only (x)
C) horizontal and vertical (x, y)
D) horizontal, vertical, and side to side (x, y, and z)

The diagram shows the motion of a tennis ball that has just been hit with a racket (air resistance is neglected). Which of these is true of the horizontal and vertical components of the ball’s velocity?
A) Both the horizontal and the vertical components are constant.
B) Both the horizontal and the vertical components are accelerated.
C) The horizontal component is constant but the vertical component is accelerated.
D) The horizontal component is accelerated but the vertical component is constant.

Answers

1. C) horizontal and vertical (x, y)

The picture shows the motion of a projectile, which consists of two separate motions along two different directions:

- horizontal (x): along this direction, the football has a uniform motion, with constant horizontal speed [tex]v_0 cos \theta[/tex],where [tex]v_0[/tex] is the magnitude of the initial velocity of the ball and [tex]\theta[/tex] the angle at which it has been thrown

- vertical (y): along this direction, the football has an accelerated motion, with initial vertical velocity [tex]v_0 sin \theta[/tex] upward and constant acceleration [tex]g=9.8 m/s^2[/tex] downward (acceleration due to gravity)


2. C) The horizontal component is constant but the vertical component is accelerated.

As described in the previous part of the exercise:

- along the horizontal direction there are no forces exerted on the ball, so it is a uniform motion, therefore the acceleration is zero and the horizontal component of the velocity is constant

- along the vertical direction there is one force acting on the ball (the force of gravity, which pushes downward), so there is an acceleration (downward) equals to [tex]g=9.8 m/s^2[/tex] and therefore the vertical component of the velocity is not constant.

If we relied solely on the nonrenewable resources found in the U.S., which one would we run out of first at current usage levels?


A. natural gas

B. oil

C. uranium

D. coal

Answers

If we are depending on nonrenewable resources only then we will run out first at that resource which we are using at large level

that resource which is used at all levels at very fast level but we can't get it at that rate from natural sources

So out of all given choices we know that at most fastest rate we are using the OIL which we extract from the earth crust.

The rate with which we are extracting oil is very fast and if we use it at same rate then we will definitely run out of it

so correct answer would be

B. oil

Answer:

B: Oil

Explanation:

If consumption of natural resources continues at the 1994 rate, then without importing resources from other countries, the U.S. has approximately enough oil to last for 23 years, enough natural gas to last for 68 years, enough uranium to last for 364 years, and enough coal to last for 7,007 years.

Reference: Energy Information Administration's State Energy Data Report 1994

( study island )

Does displacement = Δx?

Answers

Typically no. Displacement can be in multiple directions as a vector. of something is traveling only along x, then it would be true though this is usually not the case.

Displacement is the distance and direction from the start point to the end point.

If the motion starts and stops on the x-axis, then the displacement is equal to delta-x toward either the left or the right. But that would be rare and unusual.

According to Newton, there were two things needed for an object to fall around or orbit the earth. Label the diagram below with these two things

Answers

Centripetal force and centrifugal force


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