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3 years ago

8

Which of the following is a physical change?

2 answers:

yanalaym [24]3 years ago

4 0

A physical change is when the natural state of an object has or will be changed. Out of the following choices, sawing a piece of wood in half will be the physical change.

Credit for the answer explanation: oopsydaisy

NOTE: THIS ANSWER IS NOT MY ANSWER, BUT I HAVE CONFIRMED THAT IT IS CORRECT.

gladu [14]3 years ago

3 0

The answer would be A

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25% part (c) assume that d is the distance the cheetah is away from the gazelle when it reaches full speed. Derive an expression

levacccp [35]

maximum speed of cheetah is

$v_1 = v_{max}$

speed of gazelle is given as

$v_2 = v_{g}$

Now the relative speed of Cheetah with respect to Gazelle

$v_{12} = v_1 - v_2$

$v_{12} = v_{max} - v_g$

now the relative distance between Cheetah and Gazelle is given initially as "d"

now the time taken by Cheetah to catch the Gazelle is given as

$d = v_{12}* t$

so by rearranging the terms we can say

$t = \frac{d}{v_{12}}$

$t = \frac{d}{v_{max} - v_g}$

so above is the relation between all given variable

6 0

3 years ago

328,000 mm + 137 m = ______________m?

svp [43]

Answer:

465m.

Explanation:

Convert all units to meters. So,

328 + 137 = 465m.

5 0

3 years ago

scientific endeavor should only be pursued after known tradeoffs associated with the research have been considered. True or Fals

AfilCa [17]

This is false. There were many times in history when people discovered something that they didn't even know was possible or didn't even plan to discover it. Knowing tradeoffs doesn't mean that something won't surprise you or that all will go according to plan.

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3 years ago

the sound sorce has a frequency of 260Hz , how much will the frequency be registered by the reciver. plss help me

lina2011 [118]

130 Hz will be reached

4 0

3 years ago

An object of mass, m1 with a velocity, v1 collides with another object at rest (v2 = 0) with a mass, m2. After the collision, m1

goblinko [34]

Answer:

$v"_{1} = v_{1} tan$ Θ

$v^{"} _{2} = \frac{m_{1}v_{1}}{m_{2}cos}$ Θ

Θ = $tan^{-1}(\frac{v^{"} _{1} }{v_{1} } )$

Explanation:

Applying the law of conservation of momentum, we have:

Δ $p_{x = 0}$

$p_{x} = p"_{x}$

$m_{1}v_{1} = m_{2}v"_{2} cos$ Θ (Equation 1)

Δ $p_{y} = 0$

$p_{y} = p"_{y}$

$0 = m_{1} v"_{1} - m_{2} v"_{2} sin$ Θ (Equation 2)

From Equation 1:

$v"_{2} = \frac{m_{1}v_{1}}{m_{2}cos}$ Θ

From Equation 2:

$m_{2} v"_{2}$ sinΘ = $m_{1} v_{1}$

$v"_{1} = \frac{m_{2} v"_{2}sinΘ}{m_{1} }$

Replacing Equation 3 in Equation 4:

$v"_{1}=\frac{m_{2}\frac{m_{1}v_{1}}{m_{2}cosΘ}sinΘ}{m_{1}}$

$v"_{1}=v_{1}\frac{sinΘ}{cosΘ}$

$v"_{1}=v_{1}tan$ Θ (Equation 5)

And we found Θ from the Equation 5:

tanΘ= $\frac{v"_{1}}{v_{1}}$

Θ= $tan^{-1}(\frac{v"_{1}}{v_{1}})$

7 0

3 years ago