A, Mechanical and A, Chemical.
Answer:
The car will travel 30 miles during the 30-minutes period of acceleration.
Explanation:
Given data :
Initial velocity = v₁ = 50 miles/hour
Final velocity = v₂ = 70 miles/hour
Time = t = 30 min = 0.5 hour
Using the definition of acceleration, we find the acceleration (a)
a = (v₂ - v₁) ÷ t
a = (70 - 50) ÷ 0.5
a = 20 ÷ 0.5
a = 40 miles/hour²
Using 3rd equation of motion, we find the distance travel (s)
2as = v₂² - v₁²
2(40)s = 70² - 50²
80 × s = 4900 - 2500
s = 2400 ÷ 80
s = 30 miles
Answer:
What Electricity questions?
Explanation:
Answer:
Time of flight A is greatest
Explanation:
Let u₁ , u₂, u₃ be their initial velocity and θ₁ , θ₂ and θ₃ be their angle of projection. They all achieve a common highest height of H.
So
H = u₁² sin²θ₁ /2g
H = u₂² sin²θ₂ /2g
H = u₃² sin²θ₃ /2g
On the basis of these equation we can write
u₁ sinθ₁ =u₂ sinθ₂=u₃ sinθ₃
For maximum range we can write
D = u₁² sin2θ₁ /g
1.5 D = u₂² sin2θ₂ / g
2 D =u₃² sin2θ₃ / g
1.5 D / D = u₂² sin2θ₂ /u₁² sin2θ₁
1.5 = u₂ cosθ₂ /u₁ cosθ₁ ( since , u₁ sinθ₁ =u₂ sinθ₂ )
u₂ cosθ₂ >u₁ cosθ₁
u₂ sinθ₂ < u₁ sinθ₁
2u₂ sinθ₂ / g < 2u₁ sinθ₁ /g
Time of flight B < Time of flight A
Similarly we can prove
Time of flight C < Time of flight B
Hence Time of flight A is greatest .