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Vadim26 [7]
4 years ago
12

What is the difference between the energy of spring a, stretched 0.6 meters, and spring b, stretched 0.3 meters, if they have th

e same spring constant?

Physics
1 answer:
butalik [34]4 years ago
7 0

delE = 1/2 k (dX^2 - dx^2); where dX = .6 m and dx = .3 m.  

So we need k, the constant, to get the "difference."  

But I'm guessing you really want E/e = 1/2 kdX^2/1/2 kdx^2 = (dX/dx)^2 = 4 so E = 4e, which is to say spring A has 4 times more energy than B. ANS.

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The 18th hole at Pebble Beach Golf Course is a dogleg to the left of length 496.0 m. The fairway off the tee is taken to be the
dimulka [17.4K]

Answer:

(472i + 80.3j) m

Explanation:

Given the following :

Distance of tee shot = 300m

Distance of second shot = 189.0 m

Displacement r1 of tee shot :

r1 = 300 mi

Displacement r2 of second shot :

r2 = 172.0 mi + 80.3 mj

The final displacement of the golf ball from the tee:

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r_final = (300i)m + (172.0i + 80.3j) m

r_final = (300 + 172)i m + 80.3j m

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7 0
3 years ago
4 The time to failure of an electrical component has a Weibull distribution with parameters λ = 0.056 and a = 2.5. A random coll
vivado [14]

Answer:

the probability that at least 125 of the 500 components will have failure times larger than 20 is 0.7939

Explanation:

see the attached file

5 0
4 years ago
Fill in the blank.
BabaBlast [244]

Answer:B

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4 0
4 years ago
A book that weighs 16N is on top of a shelf that is 3m high. What is the work required to
DENIUS [597]

48J

Explanation:

Given parameters:

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Unknown:

Work done to raise the book = ?

Solution:

Work done is defined as the product of force and distance. The force here is the weight of the book which signifies that it was carried in the vicinity of gravitational force.

   Work done = Force x  distance = weight x distance

Weight is a type of force;

  Input the parameters;

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The unit of workdone is in Joules

learn more:

Work done brainly.com/question/9100769

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8 0
3 years ago
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vesna_86 [32]

Answer:

12.974 m/s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

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a = Acceleration = \mu g

v^2-u^2=2as\\\Rightarrow -u^2=2\mu gs-v^2\\\Rightarrow u=\sqrt{v^2-2\mu gs}\\\Rightarrow u=\sqrt{0^2-2\times -(9.81\times 0.66)\times 13}\\\Rightarrow u=12.974\ m/s

Car's original speed before braking was 12.974 m/s

6 0
4 years ago
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