2. Being a teenager can be fun, but it can also be difficult as your body shape changes. These physical
changes can be hard to deal with if they aren't what you are expecting. There can be pressure from friends to be or look a certain way, and this might affect the foods you eat. It's not a good time to crash diet, as you won't get enough nutrients, and you may not reach your full potential. Following a sensible, ell-balanced diet is a much better option, both for and the long term. It is important to eat small but frequent meals for adolescents. Why?
a) Because metabolism rate is very high in adolescents. Yes/No.
b) Rate of catabolism is low. Yes/No . c) Rate of anabolism is high. Yes/No.
changes can be hard to deal with if they aren't what you are expecting. There can be pressure from friends to be or look a certain way, and this might affect the foods you eat. It's not a good time to crash diet, as you won't get enough nutrients, and you may not reach your full potential. Following a sensible, ell-balanced diet is a much better option, both for and the long term. It is important to eat small but frequent meals for adolescents. Why?
a) Because metabolism rate is very high in adolescents. Yes/No.
b) Rate of catabolism is low. Yes/No . c) Rate of anabolism is high. Yes/No.
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Explanation:
any energy expended in moving (raising) an ob
ject against the field is recovered exactly when
the object is returned (lowered) to its original
position. Anonconservative gravitational field
could solve our energy problems forever.
Given a nonconservative field, it is of
course possible that the line integral may be
zero for certain closed paths. For example, con
sider the force field, F=sinpas. Around a cir
Figure 4.4 A simple de circuit problem that must
cular path of radius p=e, we have dead be solved by applying Yeo-o in the form of
Kirchholl's voltage law
$7.01-sin doa, - a sin de
2 mp, sin
and
The integral is zero if p=1,2,3,..., etc., but it is not vero for other values of pior for mos
her closed paths, and the given field is not conservative. A conservative field must yield a zeu
lue for the line integral around every possible closed path
D4.6. If we take the zero reference for potential at infinity, find the potential at (0, 0, 2) cause
by this charge configuration in free space (a) 12 nC/m on the line p 2.5m, 20; (b) point char
of 18 nCat (1, 2, -1); (c) 12 nC/m on the line y-2.5.20. -1.0<x<1.0.
Ans.(a) 529 V: (b) 43.2 V: (c) 66.3 V
do this on the top of the image