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3 years ago

A stone is dropped from a cliff and falls 9.44 meters. What is the speed of the stone when it reaches the ground?

Physics

1 answer:

Lunna [17]3 years ago

3 0

Answer:

option A

Explanation:

given,

height of the drop of stone = 9.44 m

speed of the stone = ?

As the stone is dropped the energy of the stone will be conserved.

using conservation of energy.

Potential energy = Kinetic energy

$m g h = \dfrac{1}{2} m v^2$

$v = \sqrt{2gh}$

$v = \sqrt{2\times 9.8 \times 9.44}$

$v = \sqrt{185.024}$

v = 13.60 m/s

Hence, the correct answer is option A

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3 years ago

The froghopper, Philaenus spumarius, holds the world record for insect jumps. When leaping at an angle of 58.0° above the horizo

Jobisdone [24]

(a) 4.0 m/s

We can solve this part just by analyzing the vertical motion of the froghopper.

The initial vertical velocity of the froghopper as it jumps from the ground is given by

$u_y = u_0 sin \theta$ (1)

where

$u_0$ is the takeoff speed

$\theta=58.0^{\circ}$ is the angle of takeoff

The maximum height reached by the froghopper is

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$v_y = 0$

Since the vertical motion is an accelerated motion with constant (de)celeration $g=-9.8 m/s^2$ , we can use the following SUVAT equation:

$v_y^2 - u_y^2 = 2gh$

Solving for $u_y$ ,

$u_y = \sqrt{v_y^2-2gh}=\sqrt{-2(-9.8)(0.587)}=3.4 m/s$

And using eq.(1), we can now find the initial takeoff speed:

$u_0 = \frac{u_y}{sin \theta}=\frac{3.4}{sin 58.0^{\circ}}=4.0 m/s$

(b) 1.47 m

For this part, we have to analyze the horizontal motion of the froghopper.

The horizontal velocity of the froghopper is

$u_x = u_0 cos \theta = (4.0) cos 58.0^{\circ} =2.1 m/s$

And this horizontal velocity is constant during the entire motion.

We now have to calculate the time the froghopper takes to reach the ground: this is equal to twice the time it takes to reach the maximum height.

The time needed to reach the maximum height can be found through the equation

$v_y = u_y + gt$

Solving for t,

$t=-\frac{u_y}{g}=-\frac{3.4}{9.8}=0.35 s$

So the time the froghopper takes to reach the ground is

$T=2t=2(0.35)=0.70 s$

And since the horizontal motion is a uniform motion, we can now find the horizontal distance covered:

$d=u_x T = (2.1)(0.70)=1.47 m$

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Answer:

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Similarity between mass and charge

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Answer:

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