Explanation:
Velocity = displacement / time
v = √((58 m)² + (135 m)²) / (12 min × 60 s/min)
v = 0.20 m/s
To solve the exercise it is necessary to keep in mind the concepts about the ideal gas equation and the volume in the cube.
However, for this case the Boyle equation will not be used, but the one that corresponds to the Boltzmann equation for ideal gas, in this way it is understood that
Where,
N = Number of molecules
k = Boltzmann constant
V = Volume
T = Temperature
P = Pressure
Our values are given as,
Rearrange the equation to find V we have,
We know that length of a cube is given by
Therefore the Length would be given as,
Therefore each length of the cube is 3.44nm
Answer:
<h2>The answer is planetary motion</h2>
Explanation:
According to Johannes Kepler, the laws governing planetary motion
states that:
1. The orbit of a planet is an ellipse with the Sun at one of the two foci.
2. A line segment joining a planet and the Sun sweeps out equal areas
during equal intervals of time.
3. The square of a planet's orbital period is proportional to the cube of the semi-major of its orbit.
Johannes Kepler was a German astronomer, mathematician, and astrologer
Born: 27 December 1571, Weil der Stadt, Germany
Died: 15 November 1630
We can use the equation vf (the final velocity) =vi (the initial velocity) +at (aceleration times time)
We know the final velocity 100m/s, the initial velocity 0, and the acceleration (gravity) 9.8m/s^2. So, 100=0+9.8t. t=100/9.8
Answer:
solution:
to find the speed of a jogger use the following relation:
V
=
d
x
/d
t
=
7.5
×m
i
/
h
r
...........................(
1
)
in Above equation in x and t. Separating the variables and integrating,
∫
d
x
/7.5
×=
∫
d
t
+
C
or
−
4.7619
=
t
+
C
Here C =constant of integration.
x
=
0 at t
=
0
, we get: C
=
−
4.7619
now we have the relation to find the position and time for the jogger as:
−
4.7619 =
t
−
4.7619
.
.
.
.
.
.
.
.
.
(
2
)
Here
x is measured in miles and t in hours.
(a) To find the distance the jogger has run in 1 hr, we set t=1 in equation (2),
to get:
= −
4.7619
=
1
−
4.7619
= −
3.7619
or x
=
7.15
m
i
l
e
s
(b) To find the jogger's acceleration in m
i
l
/
differentiate
equation (1) with respect to time.
we have to eliminate x from the equation (1) using equation (2).
Eliminating x we get:
v
=
7.5×
Now differentiating above equation w.r.t time we get:
a
=
d
v/
d
t
=
−
0.675
/
At
t
=
0
the joggers acceleration is :
a
=
−
0.675
m
i
l
/
=
−
4.34
×
f
t
/
(c) required time for the jogger to run 6 miles is obtained by setting
x
=
6 in equation (2). We get:
−
4.7619
(
1
−
(
0.04
×
6 )
)^
7
/
10=
t
−
4.7619
or
t
=
0.832
h
r
s
