Answer:
vB = - 0.176 m/s (↓-)
Explanation:
Given
(AB) = 0.75 m
(AB)' = 0.2 m/s
vA = 0.6 m/s
θ = 35°
vB = ?
We use the formulas
Sin θ = Sin 35° = (OA)/(AB) ⇒ (OA) = Sin 35°*(AB)
⇒ (OA) = Sin 35°*(0.75 m) = 0.43 m
Cos θ = Cos 35° = (OB)/(AB) ⇒ (OB) = Cos 35°*(AB)
⇒ (OB) = Cos 35°*(0.75 m) = 0.614 m
We apply Pythagoras' theorem as follows
(AB)² = (OA)² + (OB)²
We derive the equation
2*(AB)*(AB)' = 2*(OA)*vA + 2*(OB)*vB
⇒ (AB)*(AB)' = (OA)*vA + (OB)*vB
⇒ vB = ((AB)*(AB)' - (OA)*vA) / (OB)
then we have
⇒ vB = ((0.75 m)*(0.2 m/s) - (0.43 m)*(0.6 m/s) / (0.614 m)
⇒ vB = - 0.176 m/s (↓-)
The pic can show the question.
Straight lines show the conversion between a photon of light and the energy of an electron. Curved lines show transitions of electrons without any interaction with light.
Answer:
first step here is to substitute the 3 of your two equations into the second;
3 Ne^(-Q_v/k(1293)) = Ne^(-Q_v/k(1566))
Since 'N' is a constant, we can remove it from both sides.
We also want to combine our two Q_v values, so we can solve for Q_v, so we should put them both on the same side:
3 = e^(-Q_v/k(1293)) / e^(-Q_v/k(1566))
3 = e^(-Q_v/k(1293) + Q_v/k(1566) ) (index laws)
ln (3) = -Q_v/k(1293) + Q_v/k(1566) (log laws)
ln (3) = -0.13Q_v / k(1566) (addition of fractions)
Q_v = ln (3)* k * 1566 / -0.13 (rearranging the equation)
Now, as long as you know Boltzmann's constant it's just a matter of substituting it for k and plugging everything into a calculator.
Answer:
The method will be explained below
Explanation:
The APWA methodology will be used to compute the calculation.
Answer:
the engine is burning excessive oil
Explanation: