Answer:
his acceleration rate is -0.00186 m/s²
Explanation:
Given;
initial position of the car, x₀ = 100 miles = 160, 900 m ( 1 mile = 1609 m)
time of motion, t₀ = 60 minutes = 60 mins x 60 s = 3,600 s
final position of the car, x₁ = 150 miles = 241,350 m
time of motion, t₁ = 100 minutes = 100 mins x 60 s = 6,000 s
The initial velocity is calculated as;
u = 160, 900 m / 3,600 s
u = 44.694 m/s
The final velocity is calculated as;
v = 241,350 m / 6,000 s
v = 40.225 m/s
The acceleration is calculated as;

Therefore, his acceleration rate is -0.00186 m/s²

Actually Welcome to the concept of Efficiency.
Here we can see that, the Input work is given as 2.2 x 10^7 J and the efficiency is given as 22%
The efficiency is => 22% => 22/100.
so we get as,
E = W(output) /W(input)
hence, W(output) = E x W(input)
so we get as,
W(output) = (22/100) x 2.2 x 10^7
=> W(output) = 0.22 x 2.2 x 10^7 => 0.484 x 10^7
hence, W(output) = 4.84 x 10^6 J
The useful work done on the mass is 4.84 x 10^6 J
Answer:
Explanation:
1.2(0) + 3(0.8) + 1.4(0.8/2) / (1.2 + 3 + 1.4) = 0.5285714... ≈ 0.53 m
The answer is true
Step by step explanation: