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3 years ago

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Two froghoppers sitting on the ground aim at the same leaf, located 35 cm above the ground. Froghopper A jumps straight up while

froghopper B jumps at a takeoff angle of 58o above the horizontal. Which froghopper experiences the greatest change in kinetic energy from the start of the jump to when it reaches the leaf?

2 answers:

IRINA_888 [86]3 years ago

6 0

Answer:

change in kinetic energy of both froghopper will be same

Explanation:

By work energy theorem we know that

Work done by all forces = change in kinetic energy

Here we know that two froghopper are sitting at same height and both jumps with same kinetic energy at different angles

1) First jump vertically upwards

2) Second jumps at an angle of 58 degree

so here while they are in air only gravitational force will work on it

so work done by gravity = change in kinetic energy

since both are initially at same height so work done by gravity on both of them will be same

So change in kinetic energy of both froghopper will be same

evablogger [386]3 years ago

3 0

Frog hopper B❤️recent emojis

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Answer:

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$4.85x10^{-6}pc$

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$9.46x10^{15}m$

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The parallax angle can be used to find out the distance using triangulation. Making a triangle between the nearby star, the Sun and the Earth, knowing that the distance between the Earth and the Sun ( $1.496x10^{11} m$ ) is defined as 1 astronomical unit:

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$p('') = p(rad)\frac{180^\circ}{\pi rad}.\frac{60'}{1^\circ}.\frac{60''}{1'}$

$p('') = 2.06x10^5 p(rad)$

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<em>a) How many parsecs are there in one astronomical unit? </em>

$1AU . \frac{1pc}{2.06x10^5AU}$ ⇒ $4.85x10^{-6}pc$

<em>(b) How many meters are in a parsec? </em>

$2.06x10^{5}AU . \frac{1.496x10^{11}m}{1AU}$ ⇒ $3.081x10^{16}m$

<em>(c) How many meters in a light-year? </em>

To determine the number of meters in a light-year it is necessary to use the next equation:

$x = c.t$

Where c is the speed of light ( $c = 3x10^{8}m/s$ ) and x is the distance that light travels in 1 year.

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$x = 9.46x10^{15}m$

<em>(d) How many astronomical units in a light-year?</em>

$9.46x10^{15}m . \frac{1AU}{1.496x10^{11}m}$ ⇒ $63325AU$

<em>(e) How many light-years in a parsec?</em>

$2.06x10^{5}AU . \frac{1ly}{63235AU}$ ⇒ $3.26ly$

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Answer:

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