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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:

m = 8

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            m = - f₀ / f_{e}

Where f₀ is the focal length of the lens and f_{e} is the false distance of the eyepiece.

It is this problem that gives us the diopter of each lens, these are related to the focal length in meters

           D = 1 / f

Let's find the focal length

       f₁ = 1 / D₁

       f₁ = 1 / 1.16

       f₁ = 0.862 m

     

        f₂ = 1 / 9.37

        f₂ = 0.1067 m

Therefore, the lens with f₂ is the eyepiece and the slow one with the  

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8 0
3 years ago
A cube is 4.4 cm on a side, with one corner at the origin. Part 1 (a) What is the unit vector pointing from the origin to the di
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Answer:

(a) \hat{A} = \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}}

(b) \theta = 85.44^{\circ}

Solution:

As per the question:

Side of the cube, a = 4.4 cm

Coordinates of the diagonally opposite corner, A = <4.4, 4.4, 4.4> cm

Now,

(a) To calculate the unit vector:

\hat{A} = \frac{\vec{A}}{|A|}

\hat{A} = \frac{4.4\hat{i} + 4.4\hat{j} + 4.4\hat{k}}{\sqrt{()4.4}^{2} + (4.4)^{2} + (4.4)^{2}}

\hat{A} = \frac{4.4(\hat{i} + \hat{j} + \hat{k})}{4.4\sqrt{3}}

\hat{A} = \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}}

(b) To calculate the angle between the two vectors say A and A' is given by:

\vec{A}\cdot \vec{A'} = \vec{A}\vec{A'}cos\theta                      

\theta = cos^{- 1}(\frac{\vec{A}\cdot \vec{A'}}{\vec{A}\vec{A'}})        (1)

Now,

The coordinates of the diagonally opposite corner, A' is <0, 0, 1> cm

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\vec{A'} = 0\hat{i} + 0\hat{j} + 1\hat{k} = \hat{k}

Now,

Using equation (1) :

\theta = cos^{- 1}(\frac{(\frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}})\cdot \hat{k}}{|A||A'|})

|A||A'| = (\sqrt{4.4^{2} +4.4^{2} + 4.4^{2}})(\sqrt{0^{2} + 0^{2} + 0^{2}}) = 7.261

Thus

\theta = cos^{- 1}(\frac{\frac{1}{\sqrt{3}}}{7.261})

\theta = cos^{- 1}(0.07946) = 85.44^{\circ}

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