How much work is done in lifting a 2kg objext to a height of 1m​

Answers

Answer 1
Answer:

Answer:

19.6 J

Explanation:

The following data were obtained from the question:

Mass (m) of object = 2 Kg

Height (h) = 1 m

Workdone =?

NOTE: Acceleration due to gravity (g) = 9.8 m/s²

Thus, we can obtain the workdone in lifting the object by using the following formula:

Workdone = mgh

Workdone = 2 × 9.8 × 1

Workdone = 19.6 J

Therefore, the workdone in lifting the object to height of 1 m is 19.6 J


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Answer the in under 30 minutes and I will give brainliestAfter which event in Indonesia did one of the worst tsunamis form?


a hurricane


a volcanic eruption


a landslide


an earthquake

Answers

Answer: I believe the answer is an earthquake.

Explanation: Sorry If I am wrong!

Answer:

earthquake

Explanation:

I took the k12 test

Pls help ASAP I will give brainliest

Answers

Answer:

Lemon

HCI

Blood

Saliva

Bleach

NaOH

Explanation:

Blood 7.35-7.45

Bleach 12.6

Saliva 6.2-7.6

Lemon 2-3

HCI 3.01

NaOH 13

If 4.0 mol aluminum and 7.0 mol hydrogen bromide react according to the following equation, how many moles of hydrogen are formed and what is the limiting reactant?

Answers

Answer:

Moles of hydrogen formed = 3.5 moles

Explanation:

Given that:-

Moles of aluminium= 4.0 mol

Moles of hydrogen bromide = 7.0 mol

According to the reaction:-

2Al_((s))+6HBr_((aq))\rightarrow 2AlBr_3_((aq))+3H_2_((g))

2 moles of aluminum react with 6 moles of hydrogen bromide

1 mole of aluminum react with 6/2 moles of hydrogen bromide

4 moles of aluminum react with (6/2)*4 moles of hydrogen bromide

Moles of hydrogen bromide = 12 moles

Available moles of hydrogen bromide = 7.0 moles

Limiting reagent is the one which is present in small amount. Thus, hydrogen bromide is limiting reagent. (7.0 < 12)

The formation of the product is governed by the limiting reagent. So,

6 moles of hydrogen bromide on reaction forms 3 moles of hydrogen

1 moles of hydrogen bromide on reaction forms 3/6 moles of hydrogen

7 moles of hydrogen bromide on reaction forms (3/6)*7 moles of hydrogen

Moles of hydrogen formed = 3.5 moles

Answer:

3.5 mol H2, HBr (limiting reactant)

Explanation:

4.0 mol Al × 3 mol H2/ 2 mol Al = 6.0 mol H2

7.0 mol HB ×3 mol H2/ 6mol HBr = 3.5 mol H2

Since 7.0mol of HBr will produce less H2 than 4.0mol of Al, HBr will be the limiting reactant, and the reaction will produce 3.5mol of H2.

Predict whether the compounds are soluble or insoluble in water

Answers

Answer:

The polar compounds are soluble in water while non polar are insoluble in water.

Explanation:

Solvent is the that part of solution which is present in large proportion and have ability to dissolve the solute. In simplest form it is something in which other substance get dissolve. The most widely used solvent is water, other examples are toluene, acetone, ethanol, chloroform etc.

Water is called universal solvent because of high polarity all polar substance are dissolve in it. Hydrogen is less electronegative while oxygen is more electronegative and because of difference in electronegativity hydrogen carry the partial positive charge while oxygen carry partial negative charge.

Water create electrostatic interaction with other polar molecules. The negative end of water attract the positive end of polar molecules and positive end of water attract negative end of polar substance and in this way polar substance get dissolve in it.

Example:

when we stir the sodium chloride into water the cation Na⁺ ions are surrounded by the negative end of water i.e oxygen and anion Cl⁻ is surrounded by the positive end of water i.e hydrogen and in this way all salt is get dissolved.

The chemicals that can dissolve in a certain solvent to create a homogenous mixture known as a solution are said to be soluble chemicals. The compounds that are soluble are: KNO_3, AgNO_3, and CuBr_2.

As per this,

  • Nitrate (NO^{3-) salts are often soluble in water. KNO_3 is potassium nitrate.
  • The majority of nitrate (NO^{3-) salts, including silvernitrate, are soluble in water, including AgNO_3.
  • The majority of bromide (Br^-) salts, including copper(II) bromide, are water soluble.

Insoluble:

  • Lead(II) chloride, or PbCl_2, is an exception and is regarded as being insoluble in water among the chloride (Cl^-) salts.
  • Barium sulphate, also known as BaSO4, is an exception to the rule of most sulphate (SO^{4-) compounds being soluble in water.

Thus, these are the classification of the compounds as per their solubility.

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Your question seems incomplete, the prpbable complete question is:

Predict whether the following compounds are soluble or insoluble in water. Soluble Insoluble PbCl2, BaSO4, KNO3, AgNO3, and CuBr2.

1. Potassium (K) has an atomic mass of 39.0983 amu and only two naturally-occurring isotopes. The K-41 isotope (40.9618 amu) has a natural abundance of 6.7302%. What is the mass (in amu) of the other isotope

Answers

Answer:

38.96383282 amu

Explanation:

39.0983 = (40.9618 * 0.067302) + ( ? * (1-0.067302)

39.0983 = 2.756811064 + ( ? * 0.932698)

subtract 2.756811064 from both sides

36.34148894 = ( ? * 0.932698)

divide both sides by 0.932698

? = 38.96383282 amu

Answer:

38.96383282 amu

Explanation:

39.0983 = (40.9618  0.067302) + ( ?  (1-0.067302)

39.0983 = 2.756811064 + ( ?  0.932698)

subtract 2.756811064 from both sides

36.34148894 = ( ?  0.932698)

divide both sides by 0.932698

? = 38.96383282 amu

A solution contains 1.694 mg CoSO4 (155.0 g/mol) per milliliter. Calculate (a) the volume of 0.08640 M EDTA needed to titrate a 25.00-mL aliquot of this solution. (b) the volume of 0.009450 M Zn2 needed to titrate the excess reagent after addition of 50.00 mL of 0.08640 M EDTA to a 25.00-mL aliquot of this solution.

Answers

Moles of any substance is given by the molar mass and the mass. The volume of EDTA required to titrate is 3.16 mL and the volume of zinc required to titrate is 22.8 mL.

What is molarity?

Molarity is the property of the solution that gives the concentration of the solute present in the solution.

Given,

Mass of cobalt sulfate = 1.697 gm

The molar mass of cobalt sulfate = 155 g/mol

In the first part, the volume of the aliquot is 25 mL and the molarity is 0.08640 M.

The reaction is shown as:

Co²⁺ + H₄Y → CoH₂Y + 2H⁺

Moles of cobalt: n = 0.001694 ÷ 155 = 0.0000109 moles

In 25 ml aliquot moles of cobalt are, 0.000273 moles.

The volume of EDTA solution is calculated as:

V = moles ÷ Molarity

= 0.000273 mole ÷ 0.0864 mol/ L

= 3.16 mL

Hence, 3.16 mL of EDTA is required.

For the second part, moles of EDTA are calculated as:

n = 0.008640 × 0.050 = 4.32 ×10⁻⁴

In a 25 mL, sample moles of EDTA are 2.16 × 10⁻⁴ moles.

Excess moles of EDTA in the solution:

0.000432 - 0.000216 = 0.000216 moles

The volume of EDTA is calculated as:

V = 0.000216 ÷ 0.009450

= 0.0228 L

Hence, 22.8 mL of zinc is required.

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Answer:

the answer is in the screenshot

Explanation:

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