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3 years ago

Kara drops a rock from the edge of a cliff and it takes 7.00 s for the rock to reach the ground. What is the cliff’s height?

Physics

1 answer:

stepan [7]3 years ago

5 0

Answer:

h = 204.1 m

Explanation:

Given that,

Time taken to reach the ground, t = 7 s

We need to find the height of the cliff. We can find it using second equation of kinematics as follows :

$h=ut+\dfrac{1}{2}at^2$

Here,

u = 0 (at rest initially)

a = g

So,

$h=\dfrac{1}{2}gt^2\\\\h=\dfrac{1}{2}\times 9.8\times 7^2\\\\h=204.1\ m$

So, the cliff's height is 204.1 m.

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Sveta_85 [38]

The block's speed at the point where x=0.25A is v = 31.95 cm/s.

<h3>What is Spring constant?</h3>

The spring stiffness is quantified by the spring constant, or k. For various springs and materials, it varies. The stiffer the spring is and the harder it is to stretch, the bigger the spring constant.

question is incomplete, this is the remaining statement

What is the amplitude of the subsequent oscillations? And What is the block's speed at the point where x=0.25A?

x = Asin(wt)

v = Aw coswt

at t = 0

w = sqrt(k/m)

v = Aw

A = v/w

A = 7.17 cm

part b )

E = 1/2mv^2 + 1/2kx^2 = 1/2kA^2

mv^2 + k(1/4A)^2 = 1/2kA^2

mv^2 + kA^2/16 = kA^2

mv^2 = kA^2 - kA^2/16

mv^2 = 15kA^2/16

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v = sqrt(15/16) * wA

v = 31.95 cm/s

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2 years ago

Assume that the energy lost was entirely due to friction and that the total length of the PVC pipe is 1 meter. Use this length t

gladu [14]

The question is incomplete. The complete question is :

Assume that the energy lost was entirely due to friction and that the total length of the PVC pipe is 1 meter. Use this length to compute the average force of friction (for this calculation, you may neglect uncertainties).

Mass of the ball : 16.3 g

Predicted range : 0.3503 m

Actual range : 1.09 m

Solution :

Given that :

The predicted range is 0.3503 m

Time of the fall is :

$t=\sqrt{\frac{2H}{g}}$

$v_1t= 0.35$ ...........(i)

$v_0t= 1.09$ ...........(ii)

Dividing the equation (ii) by (i)

$\frac{v_0t}{v_1t}=\frac{1.09}{035} = 3.11$

∴ $v_0=3.11 \ v_1$

Now loss of energy = change in the kinetic energy

$W=\frac{1}{2} m [v_0^2-v_1^2]$

$W=\frac{1}{2} \times (16.3 \times 10^{-3}) \times [v_0^2-\left(\frac{v_0}{3.11}\right)^2]$

$W=7.307\times 10^{-3} \ v_0^2$

If f is average friction force, then

(f)(L) = W

(f) (1) = $7.307\times 10^{-3} \ v_0^2$

(f) = $7.307\times 10^{-3} \ v_0^2$

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Answer:

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