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3 years ago

A Lemon falls off a blimp and hits the ground 14 seconds later. How fast was it moving before it hits the ground?

1 answer:

5 0

Answer: Lemon blossoms falling off a potted lemon tree may also be caused by cool drafts, as well as under or over watering. A lemon tree dropping flowers may be a sign of drought or other changes in watering. When water is scarce, a lemon tree will drop flowers or fruit to conserve energy. Flooding, waterlogged soil or over-watering can also cause ...

Explanation:

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Which statements describe properties of stars?

Ede4ka [16]

Statements 1, 3, and 5 are true.

(A, C, and E)

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3 years ago

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A diver springs upward from a board that is 2.70 m above the water. At the instant she contacts the water her speed is 10.9 m/s

TiliK225 [7]

Answer:

vo=5.87m/s

Explanation:

Hello! In this problem we have a uniformly varied rectilinear movement.

Taking into account the data:

α =69.2

vf = 10m / s

h=2.7m

g=9.8m/s2

We know we want to know the speed on the y axis.

We calculate vfy

vfy = 10m / s * (sen69.2) = 9.35m / s

We can use the following equation.

$vf^{2} =vo^{2}+2*g*h\\$

We clear the vo (initial speed)

$vo=\sqrt{vf^{2}-2*g*h }$

$v0=\sqrt{(9.35m/s)^{2}-2*9.8m/s^{2} *2.7m}$

vo=5.87m/s

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3 years ago

ShEs jUsT a FrIenD (Answer if you understand)

GarryVolchara [31]

I hate it so muchhhhhhhh

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3 years ago

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An object islaunched at velocity of 20 m/s in a direction making an angle of 25 degree upward with the horizontal. what is the m

Aneli [31]

Answer:

$\displaystyle y_m=3.65m$

Explanation:

<u>Motion in The Plane</u>

When an object is launched in free air with some angle respect to the horizontal, it describes a known parabolic path, comes to a maximum height and finally drops back to the ground level at a certain distance from the launching place.

The movement is split into two components: the horizontal component with constant speed and the vertical component with variable speed, modified by the acceleration of gravity. If we are given the values of $v_o$ and $\theta\\$ as the initial speed and angle, then we have

$\displaystyle v_x=v_o\ cos\theta$

$\displaystyle v_y=v_o\ sin\theta-gt$

$\displaystyle x=v_o\ cos\theta\ t$

$\displaystyle y=v_o\ sin\theta\ t -\frac{gt^2}{2}$

If we want to know the maximum height reached by the object, we find the value of t when $v_y$ becomes zero, because the object stops going up and starts going down