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shepuryov [24]
2 years ago
9

A small boat sailed straight north out of a harbor in strong east wind (blowing from west to east). After sailing for 120 minute

s, it ended up hitting a buoy 60^\circ60 ∘ to the north-east of the harbor. If the straight-line distance between the buoy and the harbor is 3 km, what is the speed of the east wind?.
Physics
1 answer:
Amiraneli [1.4K]2 years ago
7 0

it can be said that  the speed of the east wind is

v=0.3608m/s

From the question we are told

A small boat sailed <u>straight </u>north out of a harbor in <em>strong </em>east wind (blowing from west to east).

After sailing for 120 minutes, it ended up hitting a buoy 60^\circ60 ∘ to the north-east of the harbor. If the straight-line distance between the buoy and the harbor is 3 km,

  • what is the speed of the east wind?.

<h3> the speed of the east wind</h3>

Generally the equation for the distance  is mathematically given as

BA=3000sin60

BA=2598.07m

Therefore

the speed of the east wind

V_w=\frac{BA}{120*60}\\\\V_w=\frac{2598.07}{120*60}

v=0.3608

For more information on this visit

brainly.com/question/22568180

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Alekssandra [29.7K]

Answer:

0.53m/s^2

Explanation:

We are given that

Mass of wire=m=3.6 g=3.6\times 10^{-3} kg

1 kg=1000 g

Length of wire=l=1.6 m

Mass of object=m'=3 kg

Time,t=60.1 ms=60.1\times 10^{-3} s

1 ms=10^{-3} s

Speed,v=\frac{distance}{time}=\frac{1.6}{60.1\times 10^{-3}}=26.62 m/s

g=\frac{v^2m}{m'l}

Using the formula

g=\frac{(26.62)^2\times 3.6\times 10^{-3}}{3\times 1.6}=0.53m/s^2

8 0
4 years ago
What can always be said about a negatively-charged ion?
KiRa [710]

Answer: A negatively-charged ion always has more electrons than protons

Explanation:

First, we know that the elementary negative charge is the electron, while the positive one is the proton. Such that both have the same charge in magnitude, but a different sign. Such that if we have the same number of electrons and protons in an atom, the charge of this atom will be neutral.

And an ion is an atom with a different number of electrons and protons, so the charge of the atom is not neutral.

Then if we have a negatively-charged ion, the charge of this atom is negative. Then we must have a larger number of electrons (the negative ones) than protons (the positive ones)

Then the correct option is:

A negatively-charged ion always has more electrons than protons

5 0
3 years ago
A moving van travels 10km North, then 4 km east, drops off some furniture and then drives 8 km south. (a) Sketch the path of the
Juli2301 [7.4K]

Answer:

4.47 km

Explanation:

If we draw the path of the van then we get a shape with two exposed points A and D. If we draw a line from point D perpendicular to BA we get point E. This gives us a right angled triangle ADE.

From Pythagoras theorem

AD² = AE² + ED²

AD=\sqrt{AE^2+ED^2}\\\Rightarrow AD=\sqrt{2^2+4^2}\\\Rightarrow AD=\sqrt{20}\\\Rightarrow AD=4.47\ km

Hence, the van is 4.47 km from its initial point

3 0
3 years ago
Water is pumped steadily out of a flooded basement at a speed of 5.4 m/s through a uniform hose of radius 0.83 cm. The hose pass
Gala2k [10]

To solve this problem it is necessary to apply the concepts related to the flow as a function of the volume in a certain time, as well as the potential and kinetic energy that act on the pump and the fluid.

The work done would be defined as

\Delta W = \Delta PE + \Delta KE

Where,

PE = Potential Energy

KE = Kinetic Energy

\Delta W = (\Delta m)gh+\frac{1}{2}(\Delta m)v^2

Where,

m = Mass

g = Gravitational energy

h = Height

v = Velocity

Considering power as the change of energy as a function of time we will then have to

P = \frac{\Delta W}{\Delta t}

P = \frac{\Delta m}{\Delta t}(gh+\frac{1}{2}v^2)

The rate of mass flow is,

\frac{\Delta m}{\Delta t} = \rho_w Av

Where,

\rho_w = Density of water

A = Area of the hose \rightarrow A=\pi r^2

The given radius is 0.83cm or 0.83 * 10^{-2}m, so the Area would be

A = \pi (0.83*10^{-2})^2

A = 0.0002164m^2

We have then that,

\frac{\Delta m}{\Delta t} = \rho_w Av

\frac{\Delta m}{\Delta t} = (1000)(0.0002164)(5.4)

\frac{\Delta m}{\Delta t} = 1.16856kg/s

Final the power of the pump would be,

P = \frac{\Delta m}{\Delta t}(gh+\frac{1}{2}v^2)

P = (1.16856)((9.8)(3.5)+\frac{1}{2}5.4^2)

P = 57.1192W

Therefore the power of the pump is 57.11W

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Inessa05 [86]
90+66=156
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Reply:78kilometers in 2 hours.
3 0
4 years ago
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