Given acceleration a = 5-3t, and its velocity is 7 at time t = 2, the value of s2 - s1 = 7
<h3>How to solve for the value of s2 - s1</h3>
We have
=
v2 = 5x2 - 3x2 + c
= 10-6+c
= 4+c
S2 - S1
= 6 + 6+c - 2+3+c
12+c-5+c = 0
7 = c
Read more on acceleration here: brainly.com/question/605631
Answer:
The given grammar is :
S = T V ;
V = C X
X = , V | ε
T = float | double
C = z | w
1.
Nullable variables are the variables which generate ε ( epsilon ) after one or more steps.
From the given grammar,
Nullable variable is X as it generates ε ( epsilon ) in the production rule : X -> ε.
No other variables generate variable X or ε.
So, only variable X is nullable.
2.
First of nullable variable X is First (X ) = , and ε (epsilon).
L.H.S.
The first of other varibles are :
First (S) = {float, double }
First (T) = {float, double }
First (V) = {z, w}
First (C) = {z, w}
R.H.S.
First (T V ; ) = {float, double }
First ( C X ) = {z, w}
First (, V) = ,
First ( ε ) = ε
First (float) = float
First (double) = double
First (z) = z
First (w) = w
3.
Follow of nullable variable X is Follow (V).
Follow (S) = $
Follow (T) = {z, w}
Follow (V) = ;
Follow (X) = Follow (V) = ;
Follow (C) = , and ;
Explanation:
Answer:
a)Are generally associated with factor.
Explanation:
We know that losses are two types
1.Major loss :Due to friction of pipe surface
2.Minor loss :Due to change in the direction of flow
As we know that when any hindrance is produced during the flow of fluid then it leads to generate the energy losses.If flow is along uniform diameter pipe then there will not be any loss but if any valve and fitting placed is the path of fluid flow due to this direction of fluid flow changes and it produce losses in the energy.
Lot' of experimental data tell us that loss in the energy due to valve and fitting are generally associated with K factor.These losses are given as