Answer:
Explanation:
Given that,
Initially, the spaceship was at rest, u = 0
Final velocity of the spaceship, v = 11 m/s
Distance accelerated by the spaceship, d = 213 m
We need to find the acceleration experienced by the occupants of the spaceship during the launch. It is a concept based on the equation of kinematics. Using the third equation of motion to find acceleration.
So, the acceleration experienced by the occupants of the spaceship is 
.
Answer:
stove
Explanation:
The stove is the independent variable in this question as it does not depend on anything like other variables such as pot, water, temperature and of cause pasta.
The pot depends on the stove, the water depends on the pot and stove, the pasta depends on the water and its temperature.
Cheers.
Answer:
Explanation:
The maximum speed of the block occurs when spring has no deformation, that is, there is no elastic potential energy, which can be remarked from appropriate application of the Principle of Energy Conservation:
Answer:
a)11.6m
b)45.55s
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)\\
X=(Vf+Vo)T/2 (4)
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
a)
for this problem
Vo=0
Vf=319m/min=5.3m/s
a=1.2m/s^2
we can use the ecuation number 1 to calculate the time
t=(Vf-Vo)/a
t=(5.3-0)/1.2=4.4s
then we use the ecuation number 3 to calculate the distance
X=0.5at^2
X=0.5x1.2x4.4^2=11.6m
b)second part
We know that when the elevator starts to accelerate and decelerate, it takes a distance of 11.6m and a time of 4.4s, which means that if the distance is subtracted 2 times this distance (once for acceleration and once for deceleration)
we will have the distance traveled in with constant speed.
With this information we will find the time, and then we will add it with the time it takes for the elevator to accelerate and decelerate
X=218-11.6x2=194.8m
X=VT
T=X/v
t=194.8/5.3=36.75s
Total time=36.75+2x4.4=45.55s
