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2 years ago

5

How long does it take light to travel from jupiter to earth?.

1 answer:

valentina_108 [34]2 years ago

7 0

The answer is : 35 to 52 minutes.
Hope this helps!
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A 10.0g piece of copper wire, sitting in the sun reaches a temperature of 80.0 C. how many Joules are released when the copper c

Zolol [24]

Answer:

150.8 J

Explanation:

The heat released by the copper wire is given by:

$Q=mC_s \Delta T$

where:

m = 10.0 g is the mass of the wire

Cs = 0.377 j/(g.C) is the specific heat capacity of copper

$\Delta T=40.0 C - 80.0 C=-40.0 C$ is the change in temperature of the wire

Substituting into the equation, we find

$Q=(10.0 g)(0.377 J/gC)(-40.0^{\circ})=-150.8 J$

And the sign is negative because the heat is released by the wire.

6 0

3 years ago

A train Traveled from hong kong to beijing. It traveled at an average speed of 160 km/h in the first four hours. After that, it

4vir4ik [10]

The answer is 7 hours

3 0

3 years ago

Difinition of gravity

erastovalidia [21]

Answer:

it's all around you and it can't be destroyed

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3 years ago

Read 2 more answers

Which quantity is measured in newton seconds (Ns)?

pshichka [43]

Answer:

Impulse

Explanation:

Impulse is force times time

7 0

2 years ago

A 23.5 g piece of aluminum metal is initially at 100.0°C. It is dropped into a coffee cup-calorimeter containing 130.0 g of wate

vivado [14]

Answer: The molar heat capacity of aluminum is $25.3J/mol^0C$

Explanation:

$heat_{absorbed}=heat_{released}$

As we know that,

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$ .................(1)

where,

q = heat absorbed or released

$m_1$ = mass of water = 130.0 g

$m_2$ = mass of aluminiunm = 23.5 g

$T_{final}$ = final temperature = $26.0^oC=(273+26)K=299K$

$T_1$ = temperature of water = $23^oC=(273+23)K=296K$

$T_2$ = temperature of aluminium = $100^oC=273+100=373K$

$c_1$ = specific heat of water= $4.184J/g^0C$

$c_2$ = specific heat of aluminium= ?

Now put all the given values in equation (1), we get

$130.0\times 4.184\times (299-296)=-[23.5\times c_2\times (299-373)]$

$c_2=0.938J/g^0C$

Molar mass of Aluminium = 27 g/mol

Thus molar heat capacity = $0.938J/g^0C\times 27g/mol=25.3J/mol^0C$

5 0

3 years ago