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Mila [183]
3 years ago
11

A baseline for an experimental investigation is provided by the ??

Physics
1 answer:
ki77a [65]3 years ago
4 0

Hypothesis

I think

I hope this helps tho


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A bicycle takes 8.0 seconds to accelerate at a constant rate from rest to a speed of 4.0 m/s. If the mass of the bicycle and rid
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6 0
3 years ago
A diver jumps off a cliff 50m high and needs to clear the rock that extend outward 5.0m from the base of the cliff. The diver ju
igor_vitrenko [27]

Answer:

He should run at least at 1.5 m/s

The diver will enter the water at an angle of 87° below the horizontal.

Explanation:

Hi there!

The position and velocity of the diver are given by the following vectors:

r = (x0 + v0x · t, y0 + v0y · t + 1/2 · g · t²)

v = (v0x, v0y + g · t)

Where:

r = position vector at time t

x0 = initial horizontal position

v0x = initial horizontal velocity

t = time

y0 = initial vertical position

v0y = initial vertical velocity

g = acceleration due to gravity (-9.8 m/s² considering the  upward direction as positive)

v = velocity vector at time t

Please, see the attached figure for a description of the problem. Notice that the origin of the frame of reference is located at the jumping point so that x0 and y0 = 0.

We know that, to clear the rocks, the position vector r final (see figure) should be:

r final = ( > 5.0 m, -50 m)

So let´s find first at which time the y-component of the vector r final is - 50 m:

y = y0 + v0y · t + 1/2 · g · t²

-50 m = 2.1 m/s · t - 1/2 · 9.8 m/s² · t²

0 = -4.9 m/s² · t² + 2.1 m/s · t + 50 m

Solving the quadratic equation

t = 3.4 s

Now, we can calculate the initial horizontal velocity using the equation of the x-component of the position vector knowing that at t =3.4 the horizontal component should be greater than 5.0 m:

x = x0 + v0x · t      (x0 = 0)

5.0 m < v0x · 3.4 s

v0x > 5.0 m / 3.4 s

v0x > 1.5 m/s

The initial horizontal velocity should be greater than 1.5 m/s

To find the angle at which the diver enters the water, we have to find the magnitude of the final velocity (vector vf in the figure). We already know the magnitude of the x-component of the vector vf, since the horizontal velocity is constant. So:

vfx > 1.5 m/s

Now, let´s calculate vfy:

vfy = v0y + g · t

vfy = 2.1 m/s - 9.8 m/s² · 3.4 s

vfy = -31 m/s

Let´s calculate the minimum magnitude that the final velocity will have if the diver safely clears the rocks. Let´s consider the smallest value allowed for vfx: 1.5 m/s. Then:

|v| = \sqrt{(1.5 m/s)^{2} + (31m/s)^{2}} = 31 m/s

Then the final velocity of the diver will be greater or equal than 31 m/s.

To find the angle, we have to use trigonometry. Notice in the figure that the vectors vf, vfx and vy form a right triangle in which vf is the hypotenuse, vfx is the adjacent side and vfy is the opposite side to the angle. Then:

cos θ = adjacent / hypotenuse = vfx / vf = 1.5 m/s / 31 m/s

θ = 87°

The diver will enter the water at an angle of 87° below the horizontal.

8 0
3 years ago
How does a metamorphic rock become an igneous rock?
Oksana_A [137]

Answer:

If the newly formed metamorphic rock continues to heat, it can eventually melt and become molten (magma). When the molten rock cools it forms an igneous rock. Metamorphic rocks can form from either sedimentary or igneous rocks.

7 0
2 years ago
Determine the critical crack length for a through crack contained within a thick plate of 7150-T651 aluminum alloy that is in un
Mila [183]

Explanation:

Formula to determine the critical crack is as follows.

          K_{IC} = \gamma \sigma_{f} \sqrt{\pi \times a}

  \gamma = 1,     K_{IC} = 24.1

  [/tex]\sigma_{y}[/tex] = 570

and,   \sigma_{f} = 570 \times \frac{3}{4}

                       = 427.5

Hence, we will calculate the critical crack length as follows.

      a = \frac{1}{\pi} \times (\frac{K_{IC}}{\sigma_{f}})^{2}

        = \frac{1}{3.14} \times (\frac{24.1}{427.5})^{2}

       = 10.13 \times 10^{-4}

Therefore, largest size is as follows.

            Largest size = 2a

                                 = 2 \times 10.13 \times 10^{-4}

                                 = 20.26 \times 10^{-4}

Thus, we can conclude that the critical crack length for a through crack contained within the given plate is 20.26 \times 10^{-4}.

4 0
3 years ago
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