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3 years ago

11

A rock is thrown downward from the top of a cliff with an initial speed of 20 m/s. If the rock hits the ground after 5.0 s, what

is the height of the cliff? (Disregard air resistance. a =-g=-9.81 m/s2.)
（i want step by step)

1 answer:

Vinil7 [7]3 years ago

7 0

Answer

I hope it's helps you

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Which of the following is the same in all frames of reference?

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The correct option is B.
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4 years ago

Hooke's Law Practice

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Answer:sheeExplanation:

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3 years ago

A traffic light is weighing 200N hangs from a vertical cable tied to two other cables that are fastened to a support. The upper

Levart [38]

Answer:

T₁ = 93.6 N , T₂ = 155.6 N , T₃ = 200 N

Explanation:

This is a balance exercise where we must apply the expressions for translational balance in the two axes

∑ F = 0

Suppose that cable t1 goes to the left and the angles are 41º with respect to the horizontal and cable t2 goes to the right with angles of 63º

decompose the tension of the two upper cables

cos 41 = T₁ₓ / T1

sin 41 = T₁y / T1

T₁ₓ = T₁ cos 41

T₁y= T₁ sin 41

for cable gold

cos 63 = T₂ / T₂

sin 63 = $T_{2y}$ / T₂

We apply the two-point equilibrium equation: The junction point of the three cables and the point where the traffic light joins the vertical cable.

Let's start by analyzing the point where the traffic light meets the vertical cable

T₃ - W = 0

T₃ = W

T₃ = 200 N

now let's write the equations for the single point of the three wires

X axis

- T₁ₓ + T₂ₓ = 0

T₁ₓ = T₂ₓ

T1 cos 41 = T2 cos 63

T1 = T2 cos 63 / cos 41 (1)

y Axis

$T_{1y}$ + T_{2y} - T3 = 0

T₁ sin 41 + T₂ sin 63 = T₃ (2)

to solve the system we substitute equation 1 in 2

T₂ cos 63 / cos 41 sin 41 + T₂ sin 63 = W

T₂ (cos 63 tan 41 + sin 63) = W

T₂ = W / (cos 63 tan 41 + sin 63)

We calculate

T₂ = 200 / (cos 63 tan 41 + sin 63)

T₂ = 200 / 1,2856

T₂ = 155.6 N

we substitute in 1

T₁ = T₂ cos 63 / cos 41

T₁ = 155.6 cos63 / cos 41

T₁ = 93.6 N

therefore the tension in each cable is

T₁ = 93.6 N

T₂ = 155.6 N

T₃ = 200 N

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3 years ago

A tool for studying motion is called?

kakasveta [241]

Answer:

phytochemical

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3 years ago

Two circus performers rehearse a trick in which a ball and a dart collide. Horatio stands on a platform 9.8 m above the ground a

xenn [34]

Answer:

Time = 0.55 s

Height = 8.3 m

Explanation:

The ball is dropped and therefore has an initial velocity of 0. Its acceleration, g, is directed downward in the same direction as its displacement, $h_b$ .

The dart is thrown up in which case acceleration, g, acts downward in an opposite direction to its displacement, $h_d$ . Both collide after travelling for a time period, t. Let the height of the dart from the ground at collision be $h_d$ and the distance travelled by the ball measured from the top be $h_b$ .

It follows that $h_d+h_b=9.8$ .

Applying the equation of motion to each body (h = v_0t + 0.5at^2),

Ball:

$h_b=0\times t + 0.5\times 9.8t^2$ (since $v_{b0} =0$ .)

$h_b=4.9t^2$

Dart:

$h_d=17.8\times t - 0.5\times9.8t^2$ (the acceleration is opposite to the displacement, hence the negative sign)

$h_d=17.8\times t - 4.9t^2$

But

$h_b+h_d =9.8$

$17.8\times t - 4.9t^2+4.9t^2 =9.8$

$17.8\times t = 9.8$

$t = 0.55$

The height of the collision is the height of the dart above the ground, $h_d$ .

$h_d=17.8\times t - 4.9t^2$

$h_d=17.8\times 0.55 - 4.9\times(0.55)^2$

$h_d=9.79 - 1.48225$

$h_d=8.3$

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4 years ago