Answer:
B. 2
Explanation:
The reaction expression is given as:
_S + 3O₂ → 2SO₃
Now let us balanced the expression;
On the product side we have 2 moles of S
On the reactant side we should have 2moles of S
So, we put the coefficient 2 to balance the expression;
We have 6 moles oxygen on both sides
Answer:
A) 3.13 m/s
B) 5.34 N
C) W = 26.9 J
Explanation:
We are told that the position as a function of time is given by;
x(t) = αt² + βt³
Where;
α = 0.210 m/s² and β = 2.04×10^(−2) m/s³ = 0.0204 m/s³
Thus;
x(t) = 0.21t² + 0.0204t³
A) Velocity is gotten from the derivative of the displacement.
Thus;
v(t) = x'(t) = 2(0.21t) + 3(0.0204t²)
v(t) = 0.42t + 0.0612t²
v(4.5) = 0.42(4.5) + 0.0612(4.5)²
v(4.5) = 3.1293 m/s ≈ 3.13 m/s
B) acceleration is gotten from the derivative of the velocity
a(t) = v'(t) = 0.42 + 2(0.0612t)
a(4.5) = 0.42 + 2(0.0612 × 4.5)
a(4.5) = 0.9708 m/s²
Force = ma = 5.5 × 0.9708
F = 5.3394 N ≈ 5.34 N
C) Since no friction, work done is kinetic energy.
Thus;
W = ½mv²
W = ½ × 5.5 × 3.1293²
W = 26.9 J
<h2>
Answer: 1.252</h2>
Explanation:
We are given this equation and we need to find the value of 
:
(1)
Firstly, we have to clear :
(2)
Applying<u> Natural Logarithm</u> on both sides of the equation (2):
(3)
(4)
According to the Natural Logarithm rules 
, so (4) can be written as:
(5)
Finally:
Answer:
<h3>473.8 m/s; 473.8 m/s</h3>
Explanation:
Given the initial velocity U = 670m/s
Horizontal velocity Ux = Ucos theta
Vertical component of the cannon velocity Uy = Usin theta
Given
U = 670m/s
theta = 45°
horizontal component of the cannonball’s velocity = 670 cos 45
horizontal component of the cannonball’s velocity = 670(0.7071)
horizontal component of the cannonball’s velocity = 473.757m/s
Vertical component of the cannonball’s velocity = 670 sin 45
Vertical component of the cannonball’s velocity = 670 (0.7071)
Vertical component of the cannonball’s velocity = 473.757m/s
Hence pair of answer is 473.8 m/s; 473.8 m/s
