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4 years ago

5

Look at the picture i even had a hard time getting the pictures on here please help

1 answer:

Luden [163]4 years ago

6 0

Here's my guesses.
1 sea stack ... A2 inlet ... E3 sand spit ... F4 sea arch ... B (looks like a land arch made by the sea ... ?)5 lagoon ... C6 barrier island ...A7 bay ... D8 sea cliff .... J9 sea cave .... G10 tombolo .... no idea
That's a lot of work for 9 points !!!!

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A dart leaves a toy dart gun with initial velocity of 7.76 m/s, regardless of the angle it is fired. What is the maximum horizon

In-s [12.5K]

Vf=vi plus 2 ad
0=7.76 + 2(9.8)d
d=0.395m

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Answer:

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3 years ago

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Explanation:

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3 years ago

You stand on a bridge above a river and drop a rock into the water below from a height of 25 m. (Assume no air resistance)

Ilia_Sergeevich [38]

PART a)

here when stone is dropped there is only gravitational force on it

so its acceleration is only due to gravity

so we will have

$a = g = 9.8 m/s^2$

Part b)

Now from kinematics equation we will have

$y = v_i t + \frac{1}{2} at^2$

now we have

y = 25 m

so from above equation

$25 = 0 + \frac{1}{2}(9.8 )t^2$

$t = 2.26 s$

Part c)

If we throw the rock horizontally by speed 20 m/s

then in this case there is no change in the vertical velocity

so it will take same time to reach the water surface as it took initially

So t = 2.26 s

Part D)

Initial speed = 20 m/s

angle of projection = 65 degree

now we have

$v_x = vcos\theta$

$v_x = 20 cos65 = 8.45 m/s$

$v_y = vsin\theta$

$v_y = 20 sin65 = 18.13 m/s$

PART E)

when stone will reach to maximum height then we know that its final speed in y direction becomes zero

so here we can use kinematics in Y direction

$v_f - v_y = at$

$0 - 18.13 = (-9.8) t$

$t = 1.85 s$

so it will take 1.85 s to reach the top

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3 years ago