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ryzh [129]
3 years ago
6

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?
Physics
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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A string with a mass density of 3 * 10^-3 kg/m is under a tension of 380 N and is fixed at both ends. One of its resonance frequ
Delvig [45]

Answer:

(a) the fundamental frequency of this string is 65 Hz

(b) the harmonics of the given frequencies are third and fourth respectively.

(c) the length of the string is 2.74 m

Explanation:

Given;

mass density of the string, μ = 3 x 10⁻³ kg/m

tension of the string, T = 380 N

resonating frequencies, 195 Hz and 260 N

For the given resonant frequencies;

195 = \frac{n}{2l} \sqrt{\frac{T}{\mu} } ---(1)\\\\260 = \frac{n+1}{2l} \sqrt{\frac{T}{\mu} } ---(2)\\\\divide \ (2) \ by (1)\\\\\frac{260}{195} = \frac{n+1 }{n} \\\\260n = 195(n+1)\\\\260 n = 195 n + 195\\\\260n - 195n = 195\\\\65n = 195\\\\n = \frac{195}{65} \\\\n = 3

(c) From any of the equations, solve for Length of the string (L);

195 = \frac{n}{2l} \sqrt{\frac{T}{\mu} } \\\\195 = \frac{3}{2l}\sqrt{\frac{380}{3\times 10^{-3}} } \\\\l = \frac{3}{2\times 195}\sqrt{\frac{380}{3\times 10^{-3}} }\\\\l = 2.74 \ m

(a) the fundamental frequency is calculated as;

f_o = \frac{1}{2l} \sqrt{\frac{T}{\mu} } \\\\f_o = \frac{1}{2\times 2.74} \sqrt{\frac{380}{3\times 10^{-3} } }\\\\f_o =  65 \ Hz

(b) harmonics of the given frequencies;

the first harmonic (n = 1) = f₀ = 65 Hz

the second harmonic (n = 2) = 2f₀ = 130 Hz

the third harmonic (n = 3) = 3f₀ = 195 Hz

the fourth harmonic (n = 4) = 4f₀ = 260 Hz

Thus, the harmonics of the given frequencies are third and fourth respectively.

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Answer:

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