Answer:
<h3>0.69</h3>Explanation:
Using the Newtons law of motion;
Fm is the moving force = 400N
Ff is the frictional force = μR
μ is the coefficient of kinetic friction
R is the reaction = mg
m is the mass
a is the acceleration
The equation becomes;
Hence the coefficient of kinetic friction between the box and floor is 0.69
1) Physical principles:
a) Total mechanical energy = kinetic energy + potential energy.
b) Total mechanical energy is conserved (neglecting external forces, like drag and friction)
2) Notation:
a) Total mechanical energy: ME.
b) Kinetic energy: KE
c) Gravitational potential energy: PE
∴ ME = KE + PE = constant
3) Solution:
a) Since, ME is conserved, it is constant and would be represented in the graph by a horizontal line.
b) At start (t = 0), the ball has only KE, so KE =ME = E and PE = 0
c) As the time goes, the ball gains altitude (PE increases) and loses speed (KE decreases).
d) PE increases from 0 to a maximum value. In the graph that happens at t = 2s.
At that point, KE = 0, and PE = ME.
That is the point of highest altitude and where the speed is zero.
d) From t = 2 seg, the ball starts to lose altitude, then the ball loses PE, and gains KE.
Just before reaching the ground, at t = 4s, the ball has the same initial KE and PE as at t = 0: KE = ME and PE = 0.
The PE may be sketched on the same graph along with the KE and the ME.
The graph is attached. The red line is the ME and the blue line is the PE.
Note that at any point in the graph PE + KE = ME.
Answer:
<em>1.49 x </em><em></em>
<em></em>
Explanation:
Kepler's third law states that <em>The square of the orbital period of a planet is directly proportional to the cube of its orbit.</em>
Mathematically, this can be stated as
∝
<em>to remove the proportionality sign we introduce a constant</em>
= k
k =
Where T is the orbital period,
and R is the orbit around the sun.
For mars,
T = 687 days
R = 2.279 x
for mars, constant k will be
k = = 3.987 x
For Earth, orbital period T is 365 days, therefore
= 3.987 x
x
= 3.34 x
R =<em> 1.49 x </em><em></em>