Explanation:
The table is level and there are no other forces on the book, so the normal force is equal to the weight.
N = mg
N = (2.3 kg) (9.8 m/s²)
N = 22.5 N
Solution:
According to the equations for 1-D kinematics. The only change to them is that instead one equation that describes general motion.
So we will have to use the equations twice: once for motion in the x direction and another time for the y direction.
v_f=v_o + at ……..(a)
[where v_f and v_o are final velocity and initial velocity, respectively]
Now ,
Initially, there was y velocity, however gravity began to act on the football, causing it to accelerate.
Applying this value in equation (a)
v_yf = at = -9.81 m/s^s * 1.75 = -17.165 m/s in the y direction
For calculating the magnitude of the equation we have to square root the given value
(16.6i - 17.165y)
\\
\left | V \right |=sqrt{16.6^{2}+17.165^{2}}\\ =
\sqrt{275.56+294.637225}\\=
\sqrt{570.197225}\\=
23.87[/tex]
<span>The distance between two objects is increased by three times the oringinal distance. Since they were already separated by one time the original distance,
the additional three times the oringinal distance now puts them four times the original distance apart.
Whether we're talking about the gravitational forces of attraction or
the electrical forces of attraction, either one is inversely proportional
to the square of the distance between the objects.
So changing the distance to four times the original distance causes
the forces to become 1/4</span>² as strong as they were originally.
The forces become 1/16 of their original magnitude.<span>
</span>
Answer:
Atom - the basic particle of matter
Density - calculated from measurements of mass and volume
Motion - calculated from measurements of distance and time
Energy - can change form and move matter
Matter - the scientific word for <em>stuff</em>
<em />
Hope this helps! Please mark brainliest if correct :D
Objects dropped straight or thrown horizontally from the same height
change their vertical velocity at the same rate, and fall through equal
vertical distances in equal time intervals.
The statement is false.
