Answer:
Answer:
Step-by-step explanation:
Qp = charge on proton = 1.6 x 10-19 C
Qs = charge on silicon = 14 x 1.6 x 10-19 C
rf = final distance from nucleus = ∞
ri = initial distance from nucleus = (3.6 x 10-15 + 1.2 x 10-15 ) = 4.8 x 10-15 m
initial Potential energy is given as
Ui = k Qp Qs / ri = (9 x 109) (1.6 x 10-19 ) (14 x 1.6 x 10-19 ) / (4.8 x 10-15 ) = 6.72 x 10-13 J
final Potential energy is given as
Uf = k Qp Qs / rf = (9 x 109) (1.6 x 10-19 ) (14 x 1.6 x 10-19 ) / (∞) = 0 J
Change in Potential energy = ΔU = Ui - Uf = 6.72 x 10-13 - 0 = 6.72 x 10-13 J
Let the Voltage through which proton is accelerated = V
Energy gained due to potential difference = Qp V
Using conservation of energy
Qp V = 6.72 x 10-13
(1.6 x 10-19 ) V = 6.72 x 10-13
V = 4.2 x 106 volts
Use rules of inference to prove that the following conclusion follows from these hypotheses:
C : ∃x (p(x) ∧ r(x))
Clearly label the inference rules used at every step of your proof.
2. Consider the following hypotheses:
H1 : ∀x (¬C(x) → ¬A(x)) H2 : ∀x (A(x) → ∀y B(y)) H3 : ∃x A(x)
Use rules of inference to prove that the following conclusion follows from these hypotheses:
C : ∃x (B(x) ∧ C(x))
Clearly label the inference rules used at every step of your proof.
3. Consider the following predicate quantified formula:
∃x ∀y (P (x, y) ↔ ¬P (y, y))
Prove the unsatisfiability of this formula using rules of inference.
Answer:
See deductions below
Step-by-step explanation:
1)
a) p(y)∧q(y) for some y (Existencial instantiation to H1)
b) q(y) for some y (Simplification of a))
c) q(y) → r(y) for all y (Universal instatiation to H2)
d) r(y) for some y (Modus Ponens using b and c)
e) p(y) for some y (Simplification of a)
f) p(y)∧r(y) for some y (Conjunction of d) and e))
g) ∃x (p(x) ∧ r(x)) (Existencial generalization of f)
2)
a) ¬C(x) → ¬A(x) for all x (Universal instatiation of H1)
b) A(x) for some x (Existencial instatiation of H3)
c) ¬(¬C(x)) for some x (Modus Tollens using a and b)
d) C(x) for some x (Double negation of c)
e) A(x) → ∀y B(y) for all x (Universal instantiation of H2)
f) ∀y B(y) (Modus ponens using b and e)
g) B(y) for all y (Universal instantiation of f)
h) B(x)∧C(x) for some x (Conjunction of g and d, selecting y=x on g)
i) ∃x (B(x) ∧ C(x)) (Existencial generalization of h)
3) We will prove that this formula leads to a contradiction.
a) ∀y (P (x, y) ↔ ¬P (y, y)) for some x (Existencial instatiation of hypothesis)
b) P (x, y) ↔ ¬P (y, y) for some x, and for all y (Universal instantiation of a)
c) P (x, x) ↔ ¬P (x, x) (Take y=x in b)
But c) is a contradiction (for example, using truth tables). Hence the formula is not satisfiable.
Answer:
No I do not agree. -2 is closer to 0 than -5
Step-by-step explanation:
-5 -----------> -2---------> 0
Answer: .833333
Step-by-step explanation:
5/6
You could get a 1, 2, 3, 4, 6
So 5 in total over 6 options
-12, 60, -300, 1500, ____, ____.
You are multiplying by -5 each time.
Multiply
1500 x -5 = -7500
-7500 x -5 = 37500
-12, 60, -300, 1500, -7500,37500
hope this helps