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Aliun [14]
2 years ago
5

Calculate the amount of heat needed to melt 35.0 g of ice at 0 ºC.Express your answer in kilojoules

Physics
2 answers:
PolarNik [594]2 years ago
6 0
The amount of heat will be equal to Lm.

Where L is the latent heat of fusion and m is mass of the ice.

Latent heat of ice = 80cal/g.

So the amount of heat required here will be 35× 80cal

= 2,800 cal.
Strike441 [17]2 years ago
5 0
That would be 2,800 calories
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Answer:

A

Explanation:

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An 97 kg climber climbs to the top of Mount Everest, which has a peak height of 8850 m above sea level. What is the climber’s po
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Eg=(97)(9.8)(8850)
Eg=8412810J
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A student claims that any object in motion must experience a force that keeps it in motion. Do you agree or disagree? Explain yo
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Answer:

I disagree

Explanation:

I think the students claim is wrong because according to Newton's First Law an object that is in motion stays in motion unless acted upon by an unbalanced force. Which makes the students claim wrong because a object doesn't require another force to keep it moving.

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3 years ago
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Which of the following statements is TRUE about updating the exposure control plan?
iren2701 [21]

Statements that are true as regards exposure control plan and its updating are;

<em>Updates must have the  reflection of changes in tasks as well in procedures.</em>

<em>Updates must reflect changes in positions that affect occupational exposure.</em>

<em>Updates must have the cost of PPE that is needed and  necessary to reduce exposure</em>

An exposure control plan can be regarded as  the framework for compliance between the employer and the workers.

  • This framework give room for the employer to creates a written plan that will help in protecting their workers from bloodborne pathogens.

  • This plan gives hope to workers in term of protection when working with their Employer.

  • There are some elements that is associated with  Exposure Control Plan, and theses are;
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brainly.com/question/1203927?referrer=searchResults

3 0
2 years ago
Heptane and water do not mix, and heptane has a lower density (0.684 g/mL) than water (1.00 g/mL). A graduated cylinder contains
lakkis [162]

Given that the density of heptane is

d_h=\frac{0.684g}{mL}

The mass of heptane is

m_h=31\text{ g}

The density of water is

d_w=\frac{1g}{mL}

The mass of water is

m_w=37\text{ g}

The volume of heptane will be

\begin{gathered} V_h=\frac{m_h}{d_h} \\ =\frac{31}{0.684} \\ =45.32\text{ mL} \end{gathered}

The volume of water will be

\begin{gathered} V_w=\frac{m_w}{d_w} \\ =\frac{37}{1} \\ =37\text{ mL} \end{gathered}

Thus, the volume of heptane is 45.32 mL and the volume of water is 37 mL.

The total volume of liquid in the cylinder will be

\begin{gathered} V=V_h+V_w \\ =45.32+37 \\ =82.32\text{ mL} \end{gathered}

The total volume of liquid in the cylinder will be 82.32 mL.

7 0
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