Explanation:
Q1. Given:
v = 0 m/s
a = -5.5 m/s²
t = 3.5 s
Find: Δx
Δx = vt − ½ at²
Δx = (0 m/s) (3.5 s) − ½ (-5.5 m/s²) (3.5 s)²
Δx ≈ 33.7 m
Q2. Given:
Δx = 400 m
v₀ = 7.0 m/s
v = 35 m/s
Find: a
v² = v₀² + 2aΔx
(35 m/s)² = (7.0 m/s)² + 2a (400 m)
a = 1.47 m/s²
Answer:
Explanation:
moment of inertia of flywheel = 1/2 m R²
= .5 x 1500 x .6²
= 270 kg m²
If required angular velocity be ω
rotational kinetic energy = 1/2 I ω²
= .5 x 270 x ω² = 135 ω²
kinetic energy of bus when its velocity is 20 m/s
= 1/2 x 10000 x 20²
= 2000000 J
Given 90 % of rotational kinetic energy is converted into bus's kinetic energy
135 ω² x 0.9 = 2000000 J
ω²=16461
ω = 128.3 radian /s
b )
Let the height required be h .
Total energy of bus at the top of hill = mgh + 1/2 m v²
m ( gh + .5 v²)
= 10000 ( 9.8h + .5 x 3²)
From conservation of mechanical energy theorem
10000 ( 9.8h + .5 x 3²) = 2000000
9.8h + .5 x 3² = 200
9.8h = 195.5
h = 19.95 m .
The ramp does 480J of useful work with an efficiency of 80% .
<h3>What is efficiency of work done ?</h3>
- Efficiency is the ratio of the useful energy released by a system to the input energy .
- Mathematically, efficiency of energy = out put energy/ input energy
<h3>
What is the useful work done by the ramp having efficiency 80% and an input work done 600J?</h3>
- The efficiency =output work done/ input work done
- 80% =output work done/ 600J
- output work done =( 80×600)/100
=480J
Thus, we can conclude that the useful work done by the ramp is 480J.
Learn more about efficiency of energy here:
brainly.com/question/14280607
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Answer:
C. There is a net force on the box.
Explanation:
If the box was stationary, sitting on the roof, the correct answer choice would be A. In this scenario, the box is fallign downward, meaning that there is a net force pushing the box downward.