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GalinKa [24]
3 years ago
6

A small bug is lying resting when nine ants simultaneously ambush it and begin pulling it in different directions. They are each

applying force to the meatloaf of a different magnitude and direction. The resultant of all these forces is zero, and the bug amazingly does not rip apart under the strain. (a) Will the bug move or remain motionless? Why?
Physics
1 answer:
xxMikexx [17]3 years ago
4 0

Answer: The bug will remain motionless

Explanation:

According to Newton's first Law of Motion (sometimes called Law of Inertia):

<em>An object at rest or describing a uniform straight line motion (moving at constant velocity), will remain at rest or moving unless an external force is applied to it and changes its state of rest or motion. </em>

In other words:

An object or body will keep its state of motion until an external force changes its state

This means that objects tend to remain in its state of motion, and is the definition of the inertia, as well.

In addition, according to his law, an object in rest can be in equilibrium (net force equals to zero), and a moving object can also be in equilibrium, as long as it keeps a constant velocity.

<h2>This is why the bug, which is at rest will remain at rest, although the ants are simultaneously pulling it in different directions, since the resultant of all these forces is zero.</h2>
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A 1-kilogram mass is attached to a spring whose constant is 21 N/m, and the entire system is then submerged in a liquid that imp
amm1812

Answer:

the required value is x(t) = \frac{7}{4} e^{-3t}-\frac{3}{4} e^{-7t}

Explanation:

Given that,

mass, m = 1kg

spring constant k = 21N/M

damping force = -\beta\frac{dx}{dt} = \frac{-10dx}{dt}

\beta = 10

By Newtons second law ,

The diffrential equation of motion with damping is given by

m\frac{d^2x}{dt^2} = -kx-\beta\frac{dx}{dt}

substitute the value of m =1kg, k = 21N/M, and \beta = 10

1\frac{d^2x}{dt^2} = -21x=10\frac{dx}{dt}

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suppose the equation of the form x =e^m^t,

and the auxilliary equation is given by

m^2 + 10m + 21 = 0\\\\m^2 + 7m+3m+21=0\\\\m(m+7)+3(m+7)=0\\\\(m+7)(m+3)=0\\\\m=-7\\m=-3

The general solution for the above differential equation is

x(t) =C_1e^{-3t}+C_2e^{-7t}

Derivate with respect to t

x'(t)=-3C_1e^{-3t}-7C_2e^{-7t}

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since time is 0 then mass is one meter below

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substitute the initial condition

C_1 +C_2 = 1

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Solve the above equation to get C₁ and C₂

C_1 =\frac{7}{4} and C_2 = -\frac{3}{4}

substitute for C₁ and C₂ in general solution

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Thus the required value is x(t) = \frac{7}{4} e^{-3t}-\frac{3}{4} e^{-7t}

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