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larisa [96]
3 years ago
15

A scientist tested Charles’s law, which states that volume is directly related to temperature. Assume constant pressure. What wi

ll the scientist observe?
A) as volume decreases, temperature increases
B) as volume decreases, temperature decreases
C) as volume increases, temperature stays constant
D) as volume decreases, temperature can increase or decrease
Physics
2 answers:
liubo4ka [24]3 years ago
7 0

Answer:

B) as volume decreases, temperature decreases

Explanation:

As per ideal gas equation we know that

PV = nRT

now we know that

V = (\frac{nR}{P}) T

so here we know that since the pressure is constant so volume is directly proportional to the temperature

So here if we change the volume then the same change will occur to the temperature

so correct answer will be

B) as volume decreases, temperature decreases

sergeinik [125]3 years ago
5 0

Based on the options given, the most likely answer to this query is 

B) as volume decreases, temperature decreases.
Charles' law states that when volume increase or decrease, accordingly, temperature becomes similar.

Thank you for your question. Please don't hesitate to ask in Brainly your queries. 
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The cornea behaves as a thin lens of focal length approximately 1.80 {\rm cm}, although this varies a bit. The material of which
spayn [35]

Answer:

The height of the image will be "1.16 mm".

Explanation:

The given values are:

Object distance, u = 25 cm

Focal distance, f = 1.8 cm

On applying the lens formula, we get

⇒  \frac{1}{v} -\frac{1}{u} =\frac{1}{f}

On putting estimate values, we get

⇒  \frac{1}{v} -\frac{1}{(-25)} =\frac{1}{1.8}

⇒  \frac{1}{v} =\frac{1}{1.8} -\frac{1}{25}

⇒  v=1.94 \ cm

As a result, the image would be established mostly on right side and would be true even though v is positive.

By magnification,

m=\frac{v}{u} and m=\frac{h_{1}}{h_{0}}

⇒  \frac{v}{u} =\frac{h_{1}}{h_{0}}

⇒  \frac{1.94}{25}=\frac{{h_{1}}}{15}

⇒  {h_{1}}=1.16 \ mm

8 0
3 years ago
If 34.7 g of O2 reacts with iron to form 79.34 g of iron oxide, how much iron was used in the reaction?
zhuklara [117]

Answer: B. 44.64 g

Explanation:

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

Mass of reactants = mass of iron + mass of oxygen = mass of iron + 34.7 g

Mass of product = mass of iron oxide = 79.34 g

As Mass of reactants = Mass of product

mass of iron + 34.7 g = 79.34 g

mass of iron = 44.64 g

Thus 44.64 g of iron was used in the reaction

6 0
3 years ago
Two objects, with different sizes, masses, and temperatures, are placed in thermal contact. in which direction does the energy t
s344n2d4d5 [400]
<span>(c) energy travels from the object at higher temperature
to the object at lower temperature.

Size and mass have no effect.</span>
3 0
3 years ago
Since monsoons are storms that usually occur during a specific time of year in certain regions, you could not compare them to th
sweet [91]
In the given statement: "<span>Since monsoons are storms that usually occur during a specific time of year in certain regions, you could not compare them to thunderstorms. </span>" is false. Therefore, among the given choices, the correct answer is B. False.
4 0
3 years ago
Read 2 more answers
A 3.00 x 10^2-W electric immersion heater is
andre [41]

Answer

t = 367.77 s = 6.13 min

Explanation:

According to the law of conservation of energy:

Heat\ Supplied\ By \ Heater = Heat\ Absorbed\ by\ Glass + Heat\ Absorbed\ by\ Water\\Pt = m_gC_g\Delta T_g + m_wC_w\Delta T_w\\

where,

P = Electric Power of Heater =  300 W

t = time required = ?

m_g = mass of glass = 300 g = 0.3 kg

m_w = mass of water = 250 g = 0.25 kg

C_g = speicific heat of glass = 840 J/kg.°C

C_w =  specific heatof water = 4184 J/kg.°C

ΔT_g = ΔT_w = Change in Temperature of Glass and water = 100°C - 15°C

ΔT_g = ΔT_w = 85°C

Therefore,

(300\ W)(t) = (0.3\ kg)(840\ J/kg.^oC)(85^oC)+(0.25\ kg)(4184\ J/kg.^oC)(85^oC)\\

<u>t = 367.77 s = 6.13 min</u>

8 0
3 years ago
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