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Akimi4 [234]
3 years ago
15

A sailboat is traveling across the ocean with a speed of 4 m/s when large gusts of wind pick up that starts to accelerate the sa

ilboat at 3 m/s^2. how far will the boat have traveled one full minute after the gusts pick up.
Physics
2 answers:
Softa [21]3 years ago
7 0

With acceleration of 3 m/s^2, the boat gains 180 m/s of speed in 60 seconds.

Do it starts the minute at 4 m/s and ends it at 184 m/s. Its average speed for the whole minute is 92 m/s.

Averaging 92 m/s for 60 s, it covers (92•60)= 5.52 kilometers.

Notice that at the end of the minute, the speed is like 412 miles per hour ! The boat has been overtaken and carried along by the most monstrous tornado that ever existed on Earth, and it's definitely NOT the boat that covers 5.52 km in the next minute. It's a few shards and splinters of what WAS the boat a short time earlier, and we can totally fugedabout the poor guy who was sailing it and couldn't get out of the way fast enough. Poor guy. I only hope Royaltee dropped a decimal point out of that "3m/s^2" .

MatroZZZ [7]3 years ago
4 0
Initial velocity, u = 4 m/s

acceleration due to gusts of wind = 3 m/s^2

time, t = 1 min = 60 s

Let distance travelled = S

From equation of motion,

S = ut +  \frac{1}{2} a {t}^{2}  \\  \\ S = 4 \times 60 +   \frac{1}{2}  \times 3 \times  {60}^{2}  \\  \\ S = 240 + 5400 \\  \\ S = 5640 \: m

Thus, the boat would have traveled 5640m after gusts picked up.
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Answer:

A) since the density of the buoy is half the density of sea water when the buoy is at rest on an ocean half of the buoy will be submerged in water

B )angular frequency ( w ) = \sqrt{\frac{3g}{h} }

c )  h = 1.34 m ( 4\pi ^{2} / 3g )

Explanation:

A) since the density of the buoy is half the density of sea water when the buoy is at rest on an ocean half of the buoy will be submerged in water this is because the substances with lesser density floats when placed in a substance with a higher density

B ) when the Buoy is at rest ( t = 0 ) and is then pushed down calculate angular frequency ( w ) of small oscillations in terms of the given variables

volume of cone = hA /3.

h = height of cone, A = Base Area.

therefore the total volume of the Buoy above water level ( at rest )

= (\pi r^{2} * \frac{h}{3} ) * 2 = \frac{\pi r^{2} h  }{6}

The  part of the buoy immersed in water = x

then the net upward force will be

fb ( force of Buoy ) =  fg ( force of gravity )

note force = Mass * Acceleration

force of buoy = \frac{\alpha }{2} *\frac{\pi r^{2}  }{h}* a = ( force of gravity ) \alpha * T\frac{r^{2} }{4} * x * g

therefore a = \frac{3g}{h}  * x

angular frequency ( w ) = \sqrt{\frac{3g}{h} }

C ) height of the each cone

\frac{2\pi }{w}  = 1 therefore w = 2\pi

back to the angular frequency : \frac{3g}{h} = 4\pi ^{2}

therefore h = 1.34 m ( 4\pi ^{2} / 3g )

4 0
3 years ago
Explain whether a particle moving in a straight line with constant speed does or does not have an acceleration. b) Explain wheth
Lelechka [254]

Answer:

A: No because it is nor changing speed or direction

B: Yes because it changes direction even though the speed is constant

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2 years ago
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Firdavs [7]

Answer:

the heat of the light.

Explanation:

no matter the light, there's always heat being produced from it. and heat makes liuqid rise

4 0
3 years ago
True / False
devlian [24]
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8 0
2 years ago
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3. Una cuerda de guitarra tiene 60 cm de longitud y una masa de 0.05 kg de masa. Si se tensiona mediante una fuerza de 20 N. La
jok3333 [9.3K]

Answer:

f1 = 12.90 Hz

Explanation:

To calculate the first harmonic frequency you use the following formula for n = 1:

f_n=\frac{n}{2L}\sqrt{\frac{T}{M/L}}

f_1=\frac{1}{2L}\sqrt{\frac{T}{M/L}}    ( 1 )

It is necessary that the unist are in meters, then you have:

L: length of the string = 60cm = 0.6m

M: mass of the string = 0.05kg

T: tension on the string = 20 N

you replace the values of L, M and T in the expression (1) for getting f1:

f_1=\frac{1}{2(0.6m)}\sqrt{\frac{20N}{0.05kg/0.6m}}=12.90\ Hz

Hence, the first harmonic has a frequency of 12.90 Hz

4 0
3 years ago
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