do not obey ohm's law so it's a I believe
Answer:
True
Explanation
When an object slows down the Acceleration is in the other direction which “ slows it down
Answer:
5.56 A
Explanation:
From the question,
Q = it.............. Equation 1
Where Q = charges, i = current, t = time.
Make i the subject of the equation
i = Q/t.............. Equation 2
Given: Q = 200 coulombs, t = 0.6 minutes = (0.6×60) seconds
Substitite these values into equation 2
i = 200/(0.6×60)
i = 5.56 A
Hence the magnitude of the current flowing through the circuit is 5.56 A
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
Answer:
5.95 m
Explanation:
Given that the biggest loop is 40.0 m high. Suppose the speed at the top is 10.8 m/s and the corresponding centripetal acceleration is 2g
For the car to stick to the loop without falling down, at the top of the ride, the centripetal force must be equal to the weight of the car. That is,
(MV^2) / r = mg
V^2/ r = centripetal acceleration which is equal to 2g
2 × 9.8 = 10.8^2 / r
r = 116.64 /19.6
r = 5.95 m
