Looks like you need to review through the lesson and take notes as it tells you in the lesson what each of these are.
I think the answer should be: “100.4957 N”
What is the longest the bolt can be and still be acceptable
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
Under the assumption that the tires do not change in volume, apply Gay-Lussac's law:
P/T = const.
P = pressure, T = temperature, the quotient of P/T must stay constant.
Initial P and T values:
P = 210kPa + 101.325kPa
P = 311.325kPa (add 101.325 to change gauge pressure to absolute pressure)
T = 25°C = 298.15K
Final P and T values:
P = ?, T = 0°C = 273.15K
Set the initial and final P/T values equal to each other and solve for the final P:
311.325/298.15 = P/273.15
P = 285.220kPa
Subtract 101.325kPa to find the final gauge pressure:
285.220kPa - 101.325kPa = 183.895271kPa
The final gauge pressure is 184kPa or 26.7psi.
