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tankabanditka [31]
4 years ago
15

In which situation is no work considered to be done by a force?

Chemistry
2 answers:
Masja [62]4 years ago
7 0

If the angle is either 0 or 180, that means that there is either negative or positive work, so A and D are not correct.

If the angle is 45, then there is still some work involved.

The only option where there is no work done by a force is B. when the angle is between the force and displacement is 90. 

san4es73 [151]4 years ago
6 0

Answer:

The correct answer is option B.

Explanation:

Work done is defined as product of applied force to an object and displacement that is distance covered by an object when force applied.

W=Force\times distance\times \cos\theta

  • When \theta =0^o or 180^o. positive or negative work will be done.
  • When \theta =45^o s, some amount wok will be done
  • When \theta =90^o no work will be done.
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Kendrick sees a display of rock samples at a museum. The information included in the display is as follows:
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Because it's in a different place

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Match the scientist to his contribution to the atomic theory - Thomson - Rutherford - Bohr A. Electron energy levels B. Nucleus
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What is the concentration of kl solution if 20.68g of solute was added to enough water to form 100ml solution?
rewona [7]

Answer:

1.25 M

Explanation:

Step 1: Given data

Mass of KI (solute): 20.68 g

Volume of the solution: 100 mL (0.100 L)

Step 2: Calculate the moles of solute

The molar mass of KI is 166.00 g/mol.

20.68 g × 1 mol/166.00 g = 0.1246 mol

Step 3: Calculate the molar concentration of KI

Molarity is equal to the moles of solute divided by the liters of solution.

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5 0
3 years ago
Calculate the final temperature of the system: A 50.0 gram sample of water initially at 100 °C and a 100 gram sample initially a
katrin2010 [14]

Answer:

The final temperature of the system is 42.46°C.

Explanation:

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1\times c\times (T_f-T_1)=-(m_2\times c\times (T_f-T_2))

where,

c = specific heat of water= 4.18J/g^oC

m_1 = mass of water sample with 100 °C= 50.0 g

m_2 = mass of water sample with 13.7 °C= 100.0 g

T_f = final temperature of system

T_1 = initial temperature of 50 g of water sample= 100^oC

T_2 = initial temperature of 100 g of water =13.7^oC

Now put all the given values in the given formula, we get

50.0 g\times 4.184 J/g^oC\times (T_f-100^oC)=-(100 g\times 4.184 J/g^oC\times (T_f-13.7^oC))

T_f=42.46^oC

The final temperature of the system is 42.46°C.

5 0
4 years ago
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