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4 years ago

7

A skateboarder shoots off a ramp with a velocity of 6.6 m/s, directed at an angle of 58 above the horizontal. The end of the ram

p is 1.2 m above the ground. Let the x axis be parallel to the ground, the +y direction be vertically upward, and take as the origin the point on the ground directly below the top of the ramp. (a) How high above the ground is the highest point that the skateboarder reaches? (b) When the skateboarder reaches the highest point, how far is this point horizontally from the end of the ramp?

1 answer:

sweet-ann [11.9K]4 years ago

8 0

Answer:

a) Maximum height reached above ground = 2.8 m

b) When he reaches maximum height he is 2 m far from end of the ramp.

Explanation:

a) We have equation of motion v²=u²+2as

Considering vertical motion of skateboarder.

When he reaches maximum height,

u = 6.6sin58 = 5.6 m/s

a = -9.81 m/s²

v = 0 m/s

Substituting

0²=5.6² + 2 x -9.81 x s

s = 1.60 m

Height above ground = 1.2 + 1.6 = 2.8 m

b) We have equation of motion v= u+at

Considering vertical motion of skateboarder.

When he reaches maximum height,

u = 6.6sin58 = 5.6 m/s

a = -9.81 m/s²

v = 0 m/s

Substituting

0= 5.6 - 9.81 x t

t = 0.57s

Now considering horizontal motion of skateboarder.

We have equation of motion s =ut + 0.5 at²

u = 6.6cos58 = 3.50 m/s

a = 0 m/s²

t = 0.57

Substituting

s =3.5 x 0.57 + 0.5 x 0 x 0.57²

s = 2 m

When he reaches maximum height he is 2 m far from end of the ramp.

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Answer:

a) 404 m² b) apparent height = 7.5 m

Explanation:

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Here I will take refractive index of air and water

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Now let's look at the diagram I have attached here

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According to snell's law

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At critical angle B = 90°

$\frac{3)}{4}sin(B) = [tex]\frac{3}{4} sin(90^\circ ) = 0.75 = sin(A)$

Therefore

$A = 48.6^\circ$

With this, we can find the radius of the circle (refer to my diagram)

$h* tan (A) = R\\R =11.3 m$

And with that we can find the area

$A = \pi R^2=404\ m^2$

Additional Problem

For apparent depth from above, we can think that, since we are accustomed to seeing light at the speed of c in air, our brain interpret light from <em>any</em> source to be traveling at c. This causes light that originated under water, which has the speed of

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to appear as if it has traveled with the same duration as light with speed c

In order for this to happen our brain perceive shortened length which is the apparent depth.

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According to the directions on a can of frozen orange juice concentrate, 1 can of concentrate is to be mixed with 3 cans of wate

WARRIOR [948]

Answer:

25

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Given:

1 can of concentrate requires 3 cans of water

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for 1 can of concentrate requires 3 cans of water

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for 12 ounces can water can required = 3 × 12 ounces = 36 ounces of cans

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In this example, the original nucleus is Ac (Actinium), with

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After the decay, it must be

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B) Polonium-211

In a beta-minus decay, a neutron in the nucleus turns into a proton, emitting a fast-moving electron (the beta particle) and an anti-neutrino.

$n \rightarrow p + e^- + \bar{\nu}$

Therefore, in this process:

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