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KonstantinChe [14]
3 years ago
10

Why is it difficult to use the law of conservation of energy to calculate the

Physics
2 answers:
Liula [17]3 years ago
7 0

Answer:

Inelastic collisions actually do conserve energy, but the loss of energy to heat and mechanical vibration is hard to calculate so the math equating energy before and energy after is hard to balance.

Dmitriy789 [7]3 years ago
4 0

<u>Answer:</u>

The law of conservation of energy is difficult to use for calculate the

effects of a collision because of  energy loss

<u>Explanation:</u>

We know that Collisions are of two types which is elastic and inelastic. The Law of Conservation of energy can be defined as in isolated system the total energy is said to remain constant and stays conserved over time can be applied on elastic collisions only as they don't lose energy and can not be applied on inelastic collisions due to energy  loss  in them due to heat and mechanical vibration. Hence, law of conservation of energy can not be used to calculate effects of collision

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P6: An object of mass m sits on a spring of constant k in an elevator that is accelerating upwards with acceleration a. a) In te
tankabanditka [31]

Answer:

(a). The spring compressed is \dfrac{ma+mg}{k}.

(b). The acceleration is 1.5 g.

Explanation:

Given that,

Acceleration = a

mass = m

spring constant = k

(a). We need to calculate the spring compressed

Using balance equation

kx-mg=ma

x=\dfrac{ma+mg}{k}....(I)

The spring compressed is \dfrac{ma+mg}{k}.

(b). If the compression is 2.5 times larger than it is when the mass sits in a still elevator,

The compression is given by

x=2.5\times x_{0}

Here, acceleration is zero

So, x=2.5\times\dfrac{mg}{k}

We need to calculate the acceleration

Put the value of x in equation (I)

2.5\times \dfrac{mg}{k}=\dfrac{ma+mg}{k}

2.5\times\dfrac{mg}{k}=\dfrac{m}{k}(a+g)

a=2.5g-g

a=1.5g

Hence, (a). The spring compressed is \dfrac{ma+mg}{k}.

(b). The acceleration is 1.5 g.

8 0
3 years ago
Spiderman, whose mass is 70.0 kg, is dangling on the free end of a 12.2-m-long rope, the other end of which is fixed to a tree l
Lana71 [14]

Answer:

U = -3978.8 J

Explanation:

The work of the gravitational force U just depends of the heigth and is calculated as:

U = -mgh

Where m is the mass, g is the gravitational acceleration and h the alture.

for calculate the alture we will use the following equation:

h = L-Lcos(θ)

Where L is the large of the rope and θ is the angle.

Replacing data:

h = 12.2-12.2cos(58.4)

h = 5.8 m

Finally U is equal to:

U = -70(9.8)(5.8)

U = -3,978.8 J

5 0
3 years ago
Nikolas had an idea that he could use the compressed carbon dioxide in a fire extinguisher to propel him on his skateboard.
Vikentia [17]
The Newton’s law Nikolas would use to come up with this idea is the <span>Third law that states:

</span><span>When one body exerts a force on a second body, the second body simultaneously exerts a force equal in magnitude and opposite in direction on the first body.
</span>
So, in this case, let's name the first Body A which is the skateboard and the second body B which is <span>the compressed carbon dioxide in a fire extinguisher. Then, as shown in the figure below, according to the Third law:

</span>FA = -FB<span>

</span>

8 0
3 years ago
Read 2 more answers
Monochromatic light with wavelength 588 nm is incident on a slit with width 0.0351 mm. The distance from the slit to a screen is
Norma-Jean [14]

Answer:

Explanation:

A. Using

Sinစ= y/ L = 0.013/2.7= 0.00481

စ=0.28°

B.here we use

Alpha= πsinစa/lambda

= π x (0.0351)sin(0.28)/588E-9m

= 9.1*10^-2rad

C.we use

I(စ)/Im= (sin alpha/alpha) ²

So

{= (sin0.091/0.091)²

= 3*10^-4

6 0
3 years ago
An Olympic swimmer swims 50.0 meters in 23.1 seconds . What is his average speed
Aneli [31]
If you take 50 meters and divide by 23.1 seconds, you will get 2.16 meters per second.

So his average speed is 2.16 m/s.
3 0
3 years ago
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