A is pulling the block straight down toward the center of the Earth, no matter what the slope of the plane may be. A is the force of gravity.
The directions of B and C both depend on the slope of the plane.
B is a force that's parallel to the plane, pulling the block UP the plane. B is the force of friction.
C is a force perpendicular to the plane, preventing the block from falling down through the plane. C is the normal force.
A decrease in it's operating temperature would make a heat engine less efficient. This is because in order to operate, a heat engine needs to be hot and maintain that temperature.
Answer:
The results have not been through the rigorous process of peer review
Explanation:
When a scientist conducts a study and obtains results, those results ought to be submitted to a reputable journal where the results would go through the rigorous protocol of peer review.
During this process, the reliability of the data presented is ascertained before the results are published for other scientists to see.
If the results are hurriedly published on the internet, many researchers who come in contact with the work may be fed with inaccurate information.
1) The distance travelled by the rocket can be found by using the basic relationship between speed (v), time (t) and distance (S):
Rearranging the equation, we can write
In this problem, v=14000 m/s and t=150 s, so the distance travelled by the rocket is
2) We can solve the second part of the problem by using the same formula we used previously. This time, t=300 s, so we have:
Answer:
x=4.06m
Explanation:
A body that moves with constant acceleration means that it moves in "a uniformly accelerated movement", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.
When performing a mathematical demonstration, it is found that the equations that define this movement are as follows.
Vf=Vo+a.t (1)\\\\
{Vf^{2}-Vo^2}/{2.a} =X(2)\\\\
X=Xo+ VoT+0.5at^{2} (3)\\
Where
Vf = final speed
Vo = Initial speed
T = time
A = acceleration
X = displacement
In conclusion to solve any problem related to a body that moves with constant acceleration we use the 3 above equations and use algebra to solve
for this problem
Vf=7.6m/s
t=1.07
Vo=0
we can use the ecuation number one to find the acceleration
a=(Vf-Vo)/t
a=(7.6-0)/1.07=7.1m/s^2
then we can use the ecuation number 2 to find the distance
{Vf^{2}-Vo^2}/{2.a} =X
(7.6^2-0^2)/(2x7.1)=4.06m
