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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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S GP A projectile of mass m moves to the right with a speed vi (Fig. P11.51a). The projectile strikes and sticks to the end of a
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K_f=\frac{3}{2} \frac{m^{2}v_i^{2}  }{(M+3m)}

How this is calculated?

Given:

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mass of rod=M

Let, Initial kinetic energy =K_i

Final kinetic energy=K_f

Moment of inertia =I

What is the moment of inertia?

I=(I_p)_0+(I_{rod})_0\\I=m(\frac{d}{2})^{2}  +\frac{Md^{2} }{12} \\I=\frac{(M+3m)d^{2} }{12}

What is the angular momentum?

By conservation of angular momentum,

L_i=L_f

mv_i\frac{d}{2}=\frac{(M+3m)d^{2}\omega }{12}  \\\omega=\frac{6mv_i^{2} }{d(M+3m)}

We know that, the final kinetic energy is given by,

K_f=I\omega^{2}\\K_f=\frac{1}{2} *\frac{(M+3m)d^{2} }{12} *\frac{36m^{2}v_i^{2}}{d^{2}(M+3m)^{2}}\\ K_f=\frac{3}{2} \frac{m^{2}v_i^{2}  }{(M+3m)}

What is the kinetic energy?

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To know more about kinetic energy, refer:

brainly.com/question/114210

#SPJ4

8 0
1 year ago
Coherent monochromatic light of wavelength l passes through a narrow slit of width a, and a diffraction pattern is observed on a
geniusboy [140]

Answer:

λ = a

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This is a diffraction exercise that is described by the expression

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         sin θ  = m λ/ a

the first zero of the diffraction occurs for m = 1

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angles are generally very small and are measured in radians

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we substitute

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the width of the central maximum is twice the distance to zero

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in the exercise indicate that this width is equal to twice the distance to the screen (2x)

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we substitute

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we see that the width of the slit is equal to the wavelength used.

3 0
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