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kolbaska11 [484]
3 years ago
11

Which is not an example of how an object gains elastic potential energy by stretching?

Physics
1 answer:
lora16 [44]3 years ago
4 0
C because you don’t put a force of which you choose and the force wouldn’t be strong enough
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Please help me......
snow_tiger [21]
A. The statement is true
6 0
3 years ago
Engineering solutions use scientific laws to predict behaviors of objects in certain situations. Will used Newton's Laws of Moti
Alinara [238K]

D) Scientific laws do not account for unseen variations, like wind

Explanation:

Will model in predicting the path of an arrow he was about to shoot failed because scientific laws most times do not account for unseen variations like wind.

Scientific laws are the description of an observed phenomenon in nature.

  • Most scientific laws have exceptions.
  • Exceptions in scientific laws are conditions in which the law will not hold true.
  • There are exceptions to newton's law of motion which Will did not take into account.

learn more:

Newton's law brainly.com/question/11411375

#learnwithBrainly

6 0
3 years ago
Compare and contrast the average kinetic energy of 0.5 L of coffee at 34ÁC,
Andre45 [30]
Compared to coffee at room temperature, the molecules of the coffee at 34°C will be moving faster and colliding with one another more frequently.
6 0
3 years ago
g In 1956, Frank Lloyd Wright proposed the construction of a mile-high building in Chicago. Suppose the building had been constr
Lorico [155]

To solve this problem it is necessary to apply the concepts related to acceleration due to gravity, as well as Newton's second law that describes the weight based on its mass and the acceleration of the celestial body on which it depends.

In other words the acceleration can be described as

a = \frac{GM}{r^2}

Where

G = Gravitational Universal Constant

M = Mass of Earth

r = Radius of Earth

This equation can be differentiated with respect to the radius of change, that is

\frac{da}{dr} = -2\frac{GM}{r^3}

da = -2\frac{GM}{r^3}dr

At the same time since Newton's second law we know that:

F_w = ma

Where,

m = mass

a =Acceleration

From the previous value given for acceleration we have to

F_W = m (\frac{GM}{r^2} ) = 600N

Finally to find the change in weight it is necessary to differentiate the Force with respect to the acceleration, then:

dF_W = mda

dF_W = m(-2\frac{GM}{r^3}dr)

dF_W = -2(m\frac{GM}{r^2})(\frac{dr}{r})

dF_W = -2F_W(\frac{dr}{r})

But we know that the total weight (F_W) is equivalent to 600N, and that the change during each mile in kilometers is 1.6km or 1600m therefore:

dF_W = -2(600)(\frac{1.6*10^3}{6.37*10^6})

dF_W = -0.3N

Therefore there is a weight loss of 0.3N every kilometer.

4 0
3 years ago
Calculate the mass of an object with a density of 102.5 g/mL and volume of 375 mL.
AveGali [126]

Answer:

38,437.5

Explanation:

Density(d)= 102.5g/ml

Volume (v)=375ml

Mass(m) = ?

D =m/v

102.5= m/375

102.5*375=m

38,437.5=m

therefore Mass = 38,437.5g/ml.

6 0
3 years ago
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