Answer:
a. The horizontal component of acceleration a₁ = 0.68 m/s²
The vertical component of acceleration a₂ = -0.11 m/s²
b. -9.19° = 350.81° from the the positive x-axis
Explanation:
The initial velocity v₁ of the fish is v₁ = 4.00i + 1.00j m/s. Its final velocity after accelerating for t = 19.0 s is v₂ = 17.0i - 1.00j m/s
a. The acceleration a = (v₂ - v₁)/t = [17.0i - 1.00j - (4.00i + 1.00j)]/19 = [(17.0 -4.0)i - (-1.0 -1.0)j]/19 = (13.0i - 2.0j)/19 = 0.68i - 0.11j m/s²
The horizontal component of acceleration a₁ = 0.68 m/s²
The vertical component of acceleration a₂ = -0.11 m/s²
b. The direction of the acceleration relative to the unit vector i,
tanθ = a₂/a₁ = -0.11/0.68 = -0.1618
θ = tan⁻¹(-0.1618) = -9.19° ⇒ 360 + (-9.19) = 350.81° from the the positive x-axis
Answer:
a = -1,33 m / s²
Explanation:
This problem is kinematic in one dimension, let's use the equations
v = v₀ + a t
v₀ = v - at
x = v₀ t + ½ a t²
we substitute the first in the second equation
x = (v-at) t + ½ a t²
x -v t = a t² (1 + ½)
a = (x -vt) / t² (1.5)
we calculate
a = (100 - 30 10) / 10² (1.5) =-200 / 150
a = -1,33 m / s²
The final answer is 51.6 microC which is answer choice b:
Answer:
The bond energy of F–F = 429 kJ/mol
Explanation:
Given:
The bond energy of H–H = 432 kJ/mol
The bond energy of H–F = 565 kJ/mol
The bond energy of F–F = ?
Given that the standard enthalpy of the reaction:
<u>H₂ (g) + F₂ (g) ⇒ 2HF (g)</u>
ΔH = –269 kJ/mol
So,
<u>ΔH = Bond energy of reactants - Bond energy of products.</u>
<u>–269 kJ/mol = [1. (H–H) + 1. (F–F)] - [2. (H–F)]</u>
Applying the values as:
–269 kJ/mol = [1. (432 kJ/mol) + 1. (F–F)] - [2. (565 kJ/mol)]
Solving for , The bond energy of F–F , we get:
<u>The bond energy of F–F = 429 kJ/mol</u>