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77julia77 [94]
3 years ago
7

The trap-jaw ant can snap its mandibles shut in as little as 1.3 10-4 s. In order to shut, both mandibles rotate through a 90° a

ngle. What is the average angular velocity of one of the mandibles of the trap-jaw ant when the mandibles snap shut in 2.20 10-4 s?
Physics
1 answer:
arsen [322]3 years ago
5 0

Answer

given,

Time,t = 1.3 x 10⁻⁴ s

angle of rotation = 90° = \dfrac{\pi}{2}\ rad

Average angular velocity of the mandible

\omega =\dfrac{\Delta \theta}{\Delta t}

\omega =\dfrac{\dfrac{\pi}{2}}{1.3\times 10^{-4}}

        ω = 1.21 x 10⁴ rad/s

average angular velocity is equal to 1.21 x 10⁴ rad/s

if the time given is = 2.20 x 10⁻⁴ s

\omega =\dfrac{\Delta \theta}{\Delta t}

\omega =\dfrac{\dfrac{\pi}{2}}{2.20\times 10^{-4}}

        ω = 7.14 x 10³ rad/s

average angular velocity is equal to 7.14 x 10³ rad/s

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1.60 kg frictionless block is attached to an ideal spring with force constant 315 N/m . Initially the spring is neither stretche
Tatiana [17]

Answer:

(a). The amplitude of the motion is 0.926 m.

(b). The block’s maximum acceleration is 182.31 m/s².

(c). The maximum force the spring exerts on the block is 291.69 N.

Explanation:

Given that,

Mass of block =1.60 kg

Force constant = 315 N/m

Speed = 13.0 m/s

(a). We need to calculate the amplitude of the motion

Using conservation of energy

\dfrac{1}{2}mv^2=\dfrac{1}{2}kA^2

A^2=\dfrac{mv^2}{k}

Put the value into the relation

A^2=\dfrac{1.60\times13.0^2}{315}

A=\sqrt{0.858}

A=0.926\ m

(b). We need to calculate the block’s maximum acceleration

Using formula of acceleration

a=A\omega^2

a=A\times\dfrac{k}{m}

Put the value into the formula

a=0.926\times\dfrac{315}{1.60}

a=182.31\ m/s^2

(c). We need to calculate the maximum force the spring exerts on the block

Using formula of force

F=ma

Put the value into the formula

F= 1.60\times182.31

F=291.69\ N

Hence, (a). The amplitude of the motion is 0.926 m.

(b). The block’s maximum acceleration is 182.31 m/s².

(c). The maximum force the spring exerts on the block is 291.69 N.

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