A non reactive metal can be identified from other metals because it's surface is?

An unknown substance has a mass of 11.9 g . When the substance absorbs 1.071×102 J of heat, the temperature of the substance is

Mashcka [7]

Answer: The most likely identity of the substance is iron.

Explanation:

The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.

$Q=m\times c\times \Delta T$

Q = Heat absorbed= $1.071\times 10^2$ Joules

m= mass of substance = 11.9 g

c = specific heat capacity = ?

Initial temperature of the water = $T_i$ = 25.0°C

Final temperature of the water = $T_f$ = 45.0°CChange in temperature , $\Delta T=T_f-T_i=(45-25)^0C=20^0C$

Putting in the values, we get:

$1.071\times 10^2=11.9\times c\times 20^0C$

$c=0.45J/g^0C$

The specific heat of 0.45 is for iron and thus the substance is iron.

3 0

3 years ago

If the tension in the rope is 160 n, how much work does the rope do on the skier during a forward displacement of 270 m?

Lunna [17]

If the tension in the rope is 160 n, - 43200 J work doen by the rope on the skier during a forward displacement of 270 m.

Given,

Tension force in the rope is (T) = 160 N

Displacement of the skier (S) = 270 m

The displacement takes place in forward direction while the direction of the tension in the rope is opposite to it.

Therefore, work done by the rope on the skier is,

$W=T.S$

⇒ $W=270*160*cos\pi \\W=-43200 J$

Hence work done by the rope is - 43200 J.

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8 0

2 years ago

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Andrea's near point is 20.0 cm and her far point is 2.0 m. Her contact lenses are designed so that she can see objects that are

Shalnov [3]

Answer:

the closest distance that she can see an object clearly when she wears her contacts is 22.2 cm

Explanation:

Given the data in the question,

near point = 20 cm

far point = 2 m = 200 cm

Now, for an object that is infinitely far away, the image is at is its far point.

so using the following expression, we can determine the focal length

1/f = 1/i + 1/o

where f is the focal length, i is the image distance and o is the object distance.

here, far point i = 2 m = 200 cm and v is ∞

so we substitute

1/f = 1/(-200 cm) + 1/∞

f = -200 cm

Also, for object at its closest point, the image appear at near point,

so

1/f = 1/i + 1/o

we make o the subject of formula

o = ( i × f ) / ( i - f )

given that near point i = 20 cm

we substitute

o = ( -20 × -200 ) / ( -20 - (-200) )

o = 4000 / 180

o = 22.2 cm

Therefore, the closest distance that she can see an object clearly when she wears her contacts is 22.2 cm

4 0

3 years ago

What question can a student BEST answer when comparing and contrasting the models?

OverLord2011 [107]

Answer:

A

Explanation:

4 0

3 years ago

Birds sitting on a single power line don't get shocked. But if they were to place one foot on each of two lines, ___________ wou

Natalka [10]

Current would flow between them and they would receive a terrible shock.

5 0

3 years ago

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