Answer:
The net force acting on this object is 180.89 N.
Explanation:
Given that,
Mass = 3.00 kg
Coordinate of position of 
Coordinate of position of 
Time = 2.00 s
We need to calculate the acceleration

For x coordinates

On differentiate w.r.to t

On differentiate again w.r.to t

The acceleration in x axis at 2 sec

For y coordinates

On differentiate w.r.to t

On differentiate again w.r.to t

The acceleration in y axis at 2 sec

The acceleration is

We need to calculate the net force



The magnitude of the force


Hence, The net force acting on this object is 180.89 N.
Answer:
The magnitude of the force of friction equals the magnitude of my push
Explanation:
Since the crate moves at a constant speed, there is no net acceleration and thus, my push is balanced by the frictional force on the crate. So, the magnitude of the force of friction equals the magnitude of my push.
Let F = push and f = frictional force and f' = net force
F - f = f' since the crate moves at constant speed, acceleration is zero and thus f' = ma = m (0) = 0
So, F - f = 0
Thus, F = f
So, the magnitude of the force of friction equals the magnitude of my push.
Answer:
C. Horizontal component of velocity
Explanation:
Object in motion stays in motion,
nothing works against its motion in the horizontal direction, unlike in the vertical direction, gravity pulls object down.
The equation of motion of a pendulum is:

where
it its length and
is the gravitational acceleration. Notice that the mass is absent from the equation! This is quite hard to solve, but for <em>small</em> angles (
), we can use:

Additionally, let us define:

We can now write:

The solution to this differential equation is:

where
and
are constants to be determined using the initial conditions. Notice that they will not have any influence on the period, since it is given simply by:

This justifies that the period depends only on the pendulum's length.
<span>Visible satellite images are like photos which are dependent on visible
light from the sun so they work best during the day. The sensor works by
detecting radiation within the range that wavelength is visible. Because of
this, the rays is usually seen as reaching earth from the East. </span>