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Leona [35]
3 years ago
10

Gulls are often observed dropping clams and other shellfish from a height to the rocks below, as a means of opening the shells.

If a seagull drops a shell from rest at a height of 16 m, how fast is the shell moving when it hits the rocks
Physics
1 answer:
kumpel [21]3 years ago
8 0

Answer:

v = 17.71 m / s

Explanation:

We can work this exercise with the kinematics equations. In general the body is released so that its initial velocity is zero, the acceleration of the acceleration of gravity

                v² = v₀² - 2 g (y -y₀)

                v² = 0 - 2g (y -y₀)

when it hits the stone the height is zero and part of the height of the seagull I

              v² = 2g y₀

              v = Ra (2g i)

let's calculate

              v =√ (2 9.8 16)

              v = 17.71 m / s

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Karla Ayala pulls a sled on an icy road (dangerous!). Because of Karla's pull, the tension force is 151 N, and the rope makes a
skelet666 [1.2K]

Answer:

W = 1418.9 J = 1.418 KJ

Explanation:

In order to find the work done by the pull force applied by Karla, we need to can use the formula of work done. This formula tells us that work done on a body is the product of the distance covered by the object with the component of force applied in the direction of that displacement:

W = F.d

W = Fd Cosθ

where,

W = Work Done = ?

F = Force = 151 N

d = distance covered = 10 m

θ = Angle with horizontal = 20°

Therefore,

W = (151 N)(10 m) Cos 20°

<u>W = 1418.9 J = 1.418 KJ</u>

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pychu [463]

The global warming is the uniform heating of the globe due to the effect of greenhouse gases.

Answer: Option B

<u>Explanation:</u>

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6 0
3 years ago
0/2 File Limit
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Answer:

Speed at which it will reach the ground is given as

v_f = 46.8 m/s

Total time for which it will remain in air is given as

t = 6.3 s

Explanation:

As we know that the object is projected upwards with speed

v_i = 15 m/s

g = - 9.81 m/s^2

now when it will reach the ground then we have

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

so we have

-100 = 15 t - \frac{1}{2}(-9.81) t^2

4.905 t^2 - 15 t - 100 = 0

so we have

t = 6.3 s

Now speed of the object when it reaches the ground is given as

v_f = v_i + at

v_f = -15 + (9.81)(6.3)

v_f = 46.8 m/s

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wolverine [178]

Answer:

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Explanation:

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5 0
3 years ago
Read 2 more answers
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