A boat that travels 3.00 m/s relative to the water is crossing a river that is 1.00 km wide. The destination on the far side of the river is 0.500 km downstream from the starting point. (a) If the river current is 2.00 m/s, in what direction should the boat be pointed in order to reach the destination? (b) How much time will the trip take?

Answers

Answer 1

Answer:

Answer:

a) 10.29° upstream

b) t=338.7s

Explanation:

If the river is 1km wide and the destination point is 0.5km away downstream, then the angle and distance the the boat has to travel is:

$\alpha =atan((0.5)/(1))=26.56°$

$D=√(1^2+0.5^2)=1.118km$

The realitve velocity of the boat respect to the water is:

$V_(B/W)=[3*cos\beta ,3*sin\beta ]$ where β is the angle it has to be pointed at.

From the relative mvement equations:

$V_(B/W)=V_B-V_W$ where $V_B=[V*cos\alpha ,-V*sin\alpha ]$

From this equation we get one equation per the x-axis and another for the y-axis. If we square each of them and add them together, we will get 2 equations:

$(3*cos\beta )^2+(3*sin\beta )^2=(V*cos\alpha )^2+(-V*sin\alpha +2)^2$

$V^2-4*V*sin\alpha -5=0$ Solving for V:

V = 3.3m/s and V=-1.514m/s Replacing this value into one of our previous x or y-axis equations:

$\beta =acos((V*cos\alpha )/(3) ) = 10.29°$

The amount of time:

$t = D/V = (1118m)/(3.3m/s) =338.7s$

Related Questions

Calculate the linear momentum per photon,energy per photon, and the energy per mole of photons for radiation of wavelength; (a) 600 nm (red), (b) 550 nm (yellow), (c) 400 nm (violet), (d) 200 nm (ultraviolet), (e) 150 pm (X-ray), and (f ) 1.0 cm (microwave).
A small charged sphere is attached to a thread and placed in an electric field. The other end of the thread is anchored so that when placed in the field the sphere is in a static situation (all the forces on the sphere cancel). If the thread is horizontal, find the magnitude and direction of the electric field. The sphere has a mass of 0.018 kg and contains a charge of + 6.80 x 103 C. The tension in the thread is 6.57 x 10-2 N. Show your work and/or explain your reasoning. (20 pts)
A metal block of mass 3 kg is falling downward and has velocity of 0.44 m/s when it is 0.8 m above the floor. It strikes the top of a relaxed vertical spring of length 0.4 m. The spring constant is 2000 N/m. After striking the spring, the block rebounds. What is the maximum height above the floor that the block reaches after the impact
Explain the difference between kingdom and a species
A weather balloon is designed to expand to a maximum radius of 16.2 m when in flight at its working altitude, where the air pressure is 0.0282 atm and the temperature is â65âC. If the balloon is filled at 0.873 atm and 21âC, what is its radius at lift-off?

What If? Fluoride ions (which have the same charge as an electron) are initially moving with the same speed as the electrons from part (a) through a different uniform electric field. The ions come to a stop in the same distance d. Let the mass of an ion be M and the mass of an electron be m. Find the ratio of the magnitude of electric field the ions travel through to the magnitude of the electric field found in part (a). (Use the following as necessary: d, K, m, M, and e for the charge of the electron.)

Answers

Answer:

E₁ / E₂ = M / m

Explanation:

Let the electric field be E₁ and E₂ for ions and electrons respectively .

Force on ions = E₁ e where e is charge on ions .

Acceleration on ions a = E₁ e / M . Let initial velocity of both be u . Final velocity v = 0

v² = u² - 2as

0 = u² - 2 x E₁ e d / M

u² = 2 x E₁ e d / M

Similarly for electrons

u² = 2 x E₂ e d / m

Hence

2 x E₁ e d / M = 2 x E₂ e d / m

E₁ / E₂ = M / m

Final answer:

The ratio of the magnitude of the electric field the ions travel through to the magnitude of the electric field found in part (a) is M/m.

Explanation:

The ratio of the magnitude of the electric field the ions travel through to the magnitude of the electric field found in part (a) can be determined using the concept of mechanical energy conservation. Since the ions come to a stop, their initial kinetic energy must be equal to the work done by the electric field on them. The work done is given by the equation:

Work = Change in kinetic energy

The change in kinetic energy can be calculated using the formula:

Change in kinetic energy = (1/2)Mv2 - (1/2)mv2

where M and m are the masses of the ions and electrons respectively, and v is their initial speed. Solving for the ratio, we get:

Ratio = (1/2)M/(1/2)m = M/m

So, the ratio of the magnitude of electric field the ions travel through to the magnitude of the electric field found in part (a) is M/m.

Learn more about Electric field here:

brainly.com/question/33547143

#SPJ3

5. (Serway 9th ed., 7-3) In 1990, Walter Arfeuille of Belgium lifted a 281.5-kg object through a distance of 17.1 cm using only his teeth. (a) How much work was done on the object by Arfeuille in this lift, assuming the object was lifted at constant speed? (b) What total force was exerted on Arfeuille’s teeth during the lift? (Ans. (a) 472 J; (b) 2.76 kN)

Answers

Para resolver este problema es necesario aplicar los conceptos de Fuerza, dados en la segunda Ley de Newton y el concepto de Trabajo, como expresión de la fuerza necesaria para realizar una actividad en una distancia determinada.

El trabajo se define como

W = F*d

Where,

F = Force

d = Distance

At the same time we have that the Force by second's Newton law is equal to

F = mg

Where,

m = mass

g = Gravitational acceleration

PART A) Using our values and replacing we have that

$W = F*d\nW = mg*d\nW=281.5*9.8(17.1*10^(-2)\nW = 471.738 J\approx 472J$

PART B) Using Newton's Second law we have that,

$F = mg \nF= 281.5*9.8\nF= 2758.7 N \approx 2.76kN$

"At time t = 0 a 2330-kg rocket in outer space fires an engine that exerts" an increasing force on it in the +x-direction. This force obeys the equation Fx=At2, where t is time, and has a magnitude of 781.25 N when t = 1.27 s .What impulse does the engine exert on the rocket during the 1.50- s interval starting 2.00 s after the engine is fired?

Answers

Answer:

$Imp = 5626.488\,(kg\cdot m)/(s)$

Explanation:

First, it is required to model the function that models the increasing force in the +x direction:

$a =(781..25\,N)/((1.27\,s)^(2))$

$a = 484 (N)/(s^(2))$

The equation is:

$F_(x) = 484\,(N)/(s^(2))\cdot t^(2)$

The impulse done by the engine is given by the following integral:

$Imp=484\,(N)/(s^(2)) \int\limits^(3.50\,s)_(2\,s) {t^(2)} \, dt$

$Imp = 161.333\,(N)/(s^(2))\cdot [(3.50\,s)^(3)-(2\,s)^(3)]$

$Imp = 5626.488\,(kg\cdot m)/(s)$

Ted Clubber Lang. A hook in boxing primarily involves horizontal flexion of the shoulder while maintaining a constant angle at the elbow. During this punch, the horizontal flexor muscles of the shoulder contract and shorten at an average speed of 75 cm/s. They move through an arc length of 5 cm during the hook, while the first moves through an arc length of 100 cm. What is the average speed of the first during the hook?

Answers

Answer:

15 m/s or 1500 cm/s

Explanation:

Given that

Speed of the shoulder, v(h) = 75 cm/s = 0.75 m/s

Distance moved during the hook, d(h) = 5 cm = 0.05 m

Distance moved by the fist, d(f) = 100 cm = 1 m

Average speed of the fist during the hook, v(f) = ? cm/s = m/s

This can be solved by a very simple relation.

d(f) / d(h) = v(f) / v(h)

v(f) = [d(f) * v(h)] / d(h)

v(f) = (1 * 0.75) / 0.05

v(f) = 0.75 / 0.05

v(f) = 15 m/s

Therefore, the average speed of the fist during the hook is 15 m/s or 1500 cm/s

As the Moon revolves around the Earth, it also rotates on its axis. Why is it that the same side of the Moon is always visible from Earth?

Answers

Answer: The speed of the moon's rotation keeps the same side always facing Earth.

Explanation: Please mark me brainiest

Answer:

The speed of the Moon's rotation keeps the same side always facing Earth.

Explanation:

got it right on study island :)

Determine the tension in the string that connects M2 and M3.

Answers

thereforemass m1=4.8kg and the tension

in the horizontalspring T2=10N.

HOPEIT HELPS YOU

PLEASEMarkme as brainliest.

☺️☺️☺️☺️☺️☺️☺️☺️☺️☺️✌️✌️✌️✌️✌️✌️✌️✌️❤️❤️❤️

To determine the tension in the string that connects M2 and M3, we can follow these steps:

Step 1: Identify the necessary variables. Given data (for example) could be:
- Mass of M2, which is 5 kg
- Mass of M3, which is 10 kg
- The acceleration due to gravity, which is approximately 9.8 m/s²
- The angle at which the string pulls on M2, which is 30 degrees
- Assume the system is in equilibrium, meaning there is no net acceleration, so the acceleration is 0 m/s²

Step 2: Calculate the weight of M3, which is its mass times the acceleration due to gravity. This is because weight is the force exerted by gravity on an object, which equals the object's mass times the acceleration due to gravity.

For M3, this calculation would be M3 * g = 10 kg * 9.8 m/s² = 98 N (Newtons).

Step 3: Determine the force exerted by M2 that acts along the line of the string. This won't be the full weight of M2, because the string pulls at an angle. This component of the force can be calculated using the sine of the angle, because sine gives us the ratio of the side opposite the angle (here, the force along the string) to the hypotenuse (here, the full weight of M2) in a right triangle.

The horizontal component of the force of M2 is then M2 * g * sin(30deg) = 5 kg * 9.8 m/s² * sin(30deg) = 24.5 N.

Step 4: The tension in the string is the force M3 exerts on it, which is its weight, minus the component of M2's weight that acts along the string. This is because M2 and M3 are pulling in opposite directions, so they subtract from each other.

The tension in the string is then the weight of M3, 98 N, minus the horizontal (along the string) component of M2's weight, 24.5 N.

So, the tension in the string is 98 N - 24.5 N = 73.5 N.

This is the force that the string needs to exert in order to keep M2 and M3 connected and in equilibrium.

Learn more about Tension in a string here:

brainly.com/question/34111688

#SPJ12

Other Questions

Some amount of ideal gas with internal energy U and initial temperature 1000C was compressed to half of volume meanwhile absolute pressure inside of a container increased twice. We can say that internal energy of this gas after compression in terms of U is (20.2, 20.1, 19.4, 19.5) Group of answer choices

What If? What would be the new angular momentum of the system (in kg · m2/s) if each of the masses were instead a solid sphere 15.0 cm in diameter? (Round your answer to at least two decimal places.)

John and Linda are arguing about the definition of density. John says the density of an object is proportional to itsmass. Linda says the object's mass is proportional to its density and to its volume. Which one, if either, is correct?A. They are both wrong B. John is correct, but Linda is wrong C. John is wrong, but Linda is correct D. They are both correct. E. John must be wrong, because Linda always wins these arguments.

A ball with a mass of 170 g which contains 3.80×108 excess electrons is dropped into a vertical shaft with a height of 145 m . At the bottom of the shaft, the ball suddenly enters a uniform horizontal magnetic field that has a magnitude of 0.250 T and direction from east to west.A)If air resistance is negligibly small, find the magnitude of the force that this magnetic field exerts on the ball just as it enters the field.Use 1.602×10−19 C for the magnitude of the charge on an electron.B)Find the direction of the force that this magnetic field exerts on the ball just as it enters the field.a-from north to southb-from south to north