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3 years ago

If a substance conducts heat easily, it is considered to be agood insulator.bad substance.good conductor.poor conductor.

2 answers:

7 0

If a Substance conducts heat easily then it is considered a good conductor, since the electrons can move freely within the substance.

Yuliya22 [10]3 years ago

3 0

I think the answer is good conductor!

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A 4.0-n puck is traveling at 3.0 m/s. it strikes an 8.0-n puck, which is stationary. the two pucks stick together. their common

Natalka [10]

<span>1.0 m/s Momentum = mass x velocity Total Momentum before any collision = total momentum afterwards 4.0 x 3.0= 12 :g x momentum before (x g because using weight) Afterwards, if the velocity of the two joined is v then we get: 'momentum x g'=12v so 12v=12 so v=1m/s</span>

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2 years ago

hich muscle fibers are best suited for activities that involve lifting large, heavy objects for a short period of time? cardiac

Temka [501]

Answer:

Dead lifting uses tho muscle fundamentals

Explanation:

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3 years ago

HURRY Which change is an example of transforming potential energy to kinetic energy

Jet001 [13]

Answer:

C. changing nuclear energy to radiant energy

Explanation:

Nuclear energy takes atoms in their potential state, split them (fission) or fuse them (fusion) creating chain reactions of radiant energy. Most nuclear electrical power plants use fission, radiant energy heats water making steam to spin turbines.

Or think of the atom bomb. Definitely potential energy until the fuse starts detonation and chain reactions. The radiant kinetic energy and shock waves were horrendous.

3 0

2 years ago

Read 2 more answers

How many atoms are in 3.5 moles of sulfur

irina [24]

There are 2.1077x10^24 atoms of sulfur in 3.50 mols of sulfur.

5 0

2 years ago

Which of the following expressions will have units of kg⋅m/s2? Select all that apply, where x is position, v is velocity, m is m

netineya [11]

Answer: $m \frac{d}{dt}v_{(t)}$

Explanation:

In the image attached with this answer are shown the given options from which only one is correct.

The correct expression is:

$m \frac{d}{dt}v_{(t)}$

Because, if we derive velocity $v_{t}$ with respect to time $t$ we will have acceleration $a$ , hence:

$m \frac{d}{dt}v_{(t)}=m.a$

Where $m$ is the mass with units of kilograms ( $kg$ ) and $a$ with units of meter per square seconds $\frac{m}{s}^{2}$ , having as a result $kg\frac{m}{s}^{2}$

The other expressions are incorrect, let’s prove it:

$\frac{m}{2} \frac{d}{dx}{(v_{(x)})}^{2}=\frac{m}{2} 2v_{(x)}^{2-1}=mv_{(x)}$ This result has units of $kg\frac{m}{s}$

$m\frac{d}{dt}a_{(t)}=ma_{(t)}^{1-1}=m$ This result has units of $kg$

$m\int x_{(t)} dt= m \frac{{(x_{(t)})}^{1+1}}{1+1}+C=m\frac{{(x_{(t)})}^{2}}{2}+C$ This result has units of $kgm^{2}$ and $C$ is a constant

$m\frac{d}{dt}x_{(t)}=mx_{(t)}^{1-1}=m$ This result has units of $kg$

$m\frac{d}{dt}v_{(t)}=mv_{(t)}^{1-1}=m$ This result has units of $kg$

$\frac{m}{2}\int {(v_{(t)})}^{2} dt= \frac{m}{2} \frac{{(v_{(t)})}^{2+1}}{2+1}+C=\frac{m}{6} {(v_{(t)})}^{3}+C$ This result has units of $kg \frac{m^{3}}{s^{3}}$ and $C$ is a constant

$m\int a_{(t)} dt= \frac{m {a_{(t)}}^{2}}{2}+C$ This result has units of $kg \frac{m^{2}}{s^{4}}$ and $C$ is a constant

$\frac{m}{2} \frac{d}{dt}{(v_{(x)})}^{2}=0$ because $v_{(x)}$ is a constant in this derivation respect to $t$

$m\int v_{(t)} dt= \frac{m {v_{(t)}}^{2}}{2}+C$ This result has units of $kg \frac{m^{2}}{s^{2}}$ and $C$ is a constant

6 0

2 years ago