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3 years ago

Which of these is an example of a physical change?

Physics

1 answer:

azamat3 years ago

3 0

Answer:

A) Metal denting.

Explanation:

Physical appearance is when the "appearance" of the substance changes. In this case, a presumably smooth surface metal sheet was dented.

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A 750 g air-track glider attached to a spring with spring constant 14.0 N/m is sitting at rest on a frictionless air track. A 20

alexandr402 [8]

Answer:

the amplitude of the subsequent oscillations is 0.11 m

the period of the subsequent oscillations is 1.94 s

Explanation:

given Information:

the mass of air-track glider, $m_{1}$ = 750 g = 0.75 kg

spring constant, k = 13.0 N/m

the mass of glider, $m_{2}$ = 200 g = 0.2 kg

the speed of glider, $v_{2}$ = 170 cm/s = 1.7 m/s

the amplitude of the subsequent oscillations is A = 0.11 m

according to mechanical enery equation, we have

$A = \sqrt{\frac{m_{1} +m_{2} }{k} }v_{f}$

where

A is the amplitude and $v_{f}$ is the final speed.

to find $v_{f}$ , we can use momentum conservation lwa, where the initial momentum is equal to the final momentum.

$P_{f} = P_{i}$

$(m_{1} +m_{2} )v_{f} = m_{1} v_{1} +m_{2}v_{2}$

$v_{1}$ = 0, thus

(0.75+0.2) $v_{f}$ = (0.75)(0)+(0.2)(1.7)

0.95 $v_{f}$ = 0.34

$v_{f}$ = 0.36 m/s

Now we can calculate the amplitude

$A = \sqrt{\frac{0.75 +0.2 }{10} }0.36$

A = 0.11 m

the period of the subsequent oscillations is T = 1.94 s

the equation for period is

T = 2π $\sqrt{\frac{m_{1}+m_{2} }{k} }$

T = 2π $\sqrt{\frac{0.75+0.2 }{10} }$

T = 1.94 s

7 0

3 years ago

A man slides on snow without friction starting at 8.96m/s at the top of an inclined plane with height 8.21m. What is his speed a

hammer [34]

Answer:

V2 = 15.53 [m/s]]

Explanation:

In order to solve this problem we must use the principle of energy conservation, where potential energy is transformed into kinetic energy. At the bottom is taken as a reference level of potential energy, where the value of this energy is equal to zero.

Above the inclined plane we have two energies, kinetics and potential. While when the sled is at the reference level all this energy will have been transformed into kinetic energy.

$E_{1}=E_{2}\\ m*g*h+(\frac{1}{2} )*m*v_{1} ^{2}=\frac{1}{2}*m*v_{2} ^{2} \\(9.81*8.21)+(0.5*8.96^{2} )=(0.5*v_{2}^{2} )\\(0.5*v_{2}^{2} )=120.68\\v_{2} ^{2}=241.36\\v_{2} =\sqrt{241.36}\\ v_{2} =15.53[m/s]$