Answer:
Explanation:
We can see from the question that
and
% of
=>
from the question % of
Substituting this into the equation
=>
From the equation above
Its fake some of its real like the space part and the traveling through space part but not the multi dimensional stuff
Answer:
t = 5.55 mm
Explanation:
given,
Gauge pressure (P)=640 kpa
Allowable normal stress =150 Mpa.
Inner diameter of tank =2.6 m.
Minimum thickness of wall. = ?
In case of closed cylinders there are two cases.
one is Circumferential stress and another one is longitudinal stress.
According to circumferential stress
Where
p=pressure, d=diameter, t= thickness,
by substituting all values we get,
t = 5.55 mm
minimum thickness of the wall is equal to t = 5.55 mm
Answer:
Coccus is suitable for dry environment and a rod is better adapted for moist environment.
Explanation:
The probability of Coccus would be better suitable for a environment which is dry. A sphere posses less surface area for the volume, as such, moisture which is lost through osmosis is less in dry environment conditions.
During a moist environment, the ratio of the surface area to volume of a rod shaped bacterium is high, which allow higher efficiency to transfer water and solutes into and outside of the cell, and to make possible for a cell which is in rod shaped, have a metabolic rate higher.
(a) we can solve the problem by using the mirror equation:
where
f is the focal length of the mirror
is the distance of the object from the mirror
is the distance of the image from the mirror.
For the sign convention, f is taken as negative for a convex mirror and
is taken as negative if the image is located behind the mirror, as in this problem. So we have
and re-arranging the mirror equation we can find the distance of the object from the mirror:
from which we find
(b) The magnitude is defined as the ratio between the size of the image and the size of the object, which is also equal to the negative of the ratio between the distance of the image and the distance of the object from the mirror:
Using what we found at point (a), the magnification in this problem is
where the positive sign means the image is upright.