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>
Aqueous humor and vitreous humor are the liquids present in the human eye.
<em>Hope </em><em>it</em><em> helped</em><em> you</em><em>.</em><em>.</em><em>.</em><em> </em><em>pls </em><em>mark</em><em> brainliest</em>
Answer:
A mass of 4 Kg rest on the horizontal plane. The plane is gradually inclined until at an angle of θ=15
∘
with the horizontal,the mass just being to slide what is the coffeficient of static friction between the block & the surface.
Explanation:
Explanation:
Given that,
Length of gold wire, l = 4 m
Voltage of battery, V = 1.5 V
Current, I = 4 mA
The resistivity of gold, 
Resistance in terms of resistivity is given by :

Also, V = IR
So,

A is area of wire,
, r is radius, r = d/2 (diameter=d)

Out of four option, near option is (C) 17 μm.