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Answer:
most commonly occurs because of the frequent pumping of water from the ground.
Explanation:
Answer:
a) Ws = 2.548 J
b) Wf = 1.153 J
c) v = 1.923 m / s
Explanation:
a) The work done by the spring force
Ws = ½ * k * x²
Ws = ½ * 260 N/m *0.14² m
Ws = 2.548J
b) The increase in thermal energy can by find using
Et = Wf
Wf = µ * m *g * x
Wf = 0.42 * 2.0 kg *9.8 m/s² * 0.14m
Wf = 1.153 J
c) The speed just as the block reaches can by fin using
EK = Ws + Et
Ek = ( 2.548 + 1.153 ) J = 3.7 J
Ek = ½ * m * v²
v² = 2* Ek / m
v = √[2 * 3.7 J / 2.0 kg]
v = 1.923 m / s
Answer:


Explanation:
d = Diameter of column = 0.5 inch
= Area of concrete = 
The strain in the system is conserved

Now


Stress is given by

The stress in the steel is 

The stress in the steel is 
Answer:
1.) 4m
2.) 37 m
3.) 62m
4.) 2.5 s
Explanation:
1.) Given that the
Thinking distance = 1m
Breaking distance = 3m
Stopping distance = breaking distance + thinking distance
Stopping distance = 1 + 3 = 4m
2.) Given that the
Stopping distance = 52 m
Thinking distance = 15m
Breaking distance = 52 - 15 = 37m
3.) The stopping distance = 76m
Thinking distance = 14m
Breaking distance = 76 - 14 = 62m
It take the brakes 62m to slow the car down to a stop.
4.) Given that a lorry travels 28m when stopping from a speed of 4m/s. If its braking distance was 18m, what was the driver’s reaction time?
Thinking = stopping distance - braking distance
Thinking distance = 28 - 18 = 10m
Speed = distance/time
4 = 10/reaction time
Reaction time = 10/4
Reaction time = 2.5 s
5.) Question incomplete