If the car travels
300 km.........................5 hrs
?km ..............................2h
to find out we do (300*2)/5=600/5=120km
also
300km/5h=60km/h
in 2 hours
120km
Answer:
hello your question lacks some data and required diagram
G = 77 GPa, т all = 80 MPa
answer : required diameter = 252.65 * 10-^3 m
Explanation:
Given data :
force ( P ) = 660 -N force
displacement = 15 mm
G = 77 GPa
т all = 80 MPa
i) Determine the required diameter of shaft BC
considering the vertical displacement ( looking at handle DC from free body diagram )
D' = 0.3 sin∅ , where D = 0.015
hence ∅ = 2.8659°
calculate the torque acting at angle ∅ of CD on the shaft BC
Torque = 660 * 0.3 cos∅
= 660 * 0.3 * cos 2.8659 = 198 * -0.9622 = 190.5156 N
hello attached is the remaining part of the solution
Answer:
brand name, or company
Explanation:
a brand name is the kind of product they are selling. a company is the group of peaple that wok together.
Use Newton's second law F = mass * acceleration
In your problem F = 500, and we know gravity is working on it so use a = 9.81
Substitute into the equation
500 = m * 9.81
m = 50.97 kg
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
