Answer:
The maximum height the pebble reaches is approximately;
A. 6.4 m
Explanation:
The question is with regards to projectile motion of an object
The given parameters are;
The initial velocity of the pebble, u = 19 m/s
The angle the projectile path of the pebble makes with the horizontal, θ = 36°
The maximum height of a projectile,
, is given by the following equation;

Therefore, substituting the known values for the pebble, we have;

Therefore, the maximum height of the pebble projectile,
≈ 6.4 m.
Answer:
The universal sign for choking is __________.
A.
two balled fists pressing the abdomen
B.
pointing at an open mouth
C.
two hands grasping the neck
D.
pretending to cough
Explanation:
Answer: Frequency factor A = 8 × 10⁹
activation energy Ea = 15.5 KJ/Mol
Explanation: to begin, let us first define the parameters given;
K₁ = 1.44 × 10⁷dm³mol⁻¹s⁻¹
K₂ = 3.03 × 10⁷ dm³mol⁻¹s⁻¹
K₃ = 6.9 × 10 dm³mol⁻¹s⁻¹
also T₁ = 300.3 K
T₂ = 341.2 K
T₃ = 392.2 K
we know that;
㏑ K₂ / K₁ = Ea/R [1/T₁ -1/T₂]
where R is given as 8.314 J/mol-k
Ea = activation energy
K₁, K₂ = rate constant
T₁, T₂ = Temperature
therefore, ㏑ (3.03 × 10⁷/ 1.44 × 10⁷) = Ea / 8.314 [1/300.3 - 1/341.2]
this gives Ea = 15496.16 J/Mol ≈ 15.5 KJ /Mol
∴ Ea = 15.5 KJ/ Mol
also given that K = A e⁻∧Ea/RT
here A = frequency factor
∴ 6.9 × 10⁷ = A e⁻ ∧(15496.16/8.314 × 392.2)
A = 7.99 × 10⁹ = 8 × 10⁹