According to Newton's second law, the force applied to an object is equal to the product between the mass of the object and its acceleration:

where F is the magnitude of the force, m is the mass of the object and a its acceleration.
In this problem, the object is the insect, with mass

. The acceleration of the insect is

, therefore we can calculate the force exerted by the car on the insect:

How do we find the force exerted by the insect on the car?
According to Newton's third law (known as action-reaction law), when an object A exerts a force on an object B, object B also exerts a force equal and opposite on object A. Therefore, the force exerted by the insect on the car is equal to the force exerted by the car on the object, so it is 0.01 N.
Answer:
Read below!
Explanation:
You can watch the sun wheel across the sky during the day, and the stars at night. Focus a telescope on any star besides the north star--especially southern stars--and you can watch them drift across your field of view.
An alternative explanation is that all the stars are painted on (or holes in) some canopy that rotates around the earth. This explanation does not account for the motion of the "wanderers," or planets, as the Greeks called them, or for the path of the moon among the stars.
As we know the stars are massive bodies of significant and varying distance to the earth, the notion they all swing around us in unison seems highly implausible
The force needed to overcome sliding friction is more than the force needed to overcome rolling friction or static or even fluid
Answer:
x ’= 1,735 m, measured from the far left
Explanation:
For the system to be in equilibrium, the law of rotational equilibrium must be fulfilled.
Let's fix a reference system located at the point of rotation and that the anticlockwise rotations have been positive
They tell us that we have a mass (m1) on the left side and another mass (M2) on the right side,
the mass that is at the left end x = 1.2 m measured from the pivot point, the mass of the right side is at a distance x and the weight of the body that is located at the geometric center of the bar
x_{cm} = 1.2 -1
x_ {cm} = 0.2 m
Σ τ = 0
w₁ 1.2 + mg 0.2 - W₂ x = 0
x =
x = 
let's calculate
x =
2.9 1.2 + 4 0.2 / 8
x = 0.535 m
measured from the pivot point
measured from the far left is
x’= 1,2 + x
x'= 1.2 + 0.535
x ’= 1,735 m
Height (y) = 36t - 16t^2, where t = time in seconds (s).
Our height (y) after 1s = 36(1) - 16(1)^2
y = 36 - 16 = 20 ft
So it reached a height of 20 ft during that 1 second, which means that at that 1 second it had a velocity of 20ft/s, since v = d(distance)/t = 20ft/1s