Complete Question
The pKb values for the dibasic base B are pKb1=2.10 and pKb2=7.54. Calculate the pH at each of the points in the titration of 50.0 mL of a 0.60 M B(aq) solution with 0.60 M HCl(aq).
(a) before addition of any HCl (b) after addition of 25.0 mL of HCl
Answer:
a The value is [tex]pH =12.81[/tex]
b [tex]pH = 11.9[/tex]
Explanation:
From the question we are told that
The first pKb value for B is [tex]pK_b_1 = 2.10[/tex]
The second pKb value for B is [tex]pK_b_2 = 7.54[/tex]
The volume is [tex]V = 50.0 mL =[/tex]
The concentration of B is [tex][B] = 0.60 M[/tex]
The concentration of [tex]C_A = 0.60 M[/tex]
Generally the reaction equation showing the first dissociation of B is
[tex]\ce{B_{(aq) } + H_2O _{(l)} <=> BH^+ _{(aq)} + OH^- _{(aq)} }[/tex]
Here the ionic constant for B is mathematically represented as
[tex]K_i = \frac{[BH^+] [OH^-]}{[B]}[/tex]
Let denot the concentration of [BH^+] as z and since [tex][BH^+] = [OH^-][/tex] then [tex][OH^-][/tex] is also z
So [B] = 0.60 - z
Here [tex]K_i[/tex] is ionic constant for the first reaction of a dibasic base B and the value is
[tex]K_i = 7.94 *10^{-3}[/tex]
So
[tex] 7.94 *10^{-3}= \frac{z^2}{ 0.60 - z}[/tex]
=> [tex]z^ 2 + 0.00794 z - 0.00476[/tex]
using quadratic formula to solve this equation
[tex]z = 0.0651[/tex]
Hence the concentration of [tex]OH^{-}[/tex] is [tex][OH^-] =0.0651[/tex]
Generally [tex]pOH = -log [OH^-][/tex]
=> [tex]pOH = -log (0.065)[/tex]
=> [tex]pOH = 1.187 [/tex]
Generally the pH is mathematically represented as
[tex]pH = 14 - 1.187[/tex]
[tex]pH =12.81[/tex]
Generally the volume of [tex]HCl[/tex] at the second dissociation of the base B is [tex] 50 mL [/tex]
The volume of the [tex]HCl[/tex] half way to the first dissociation of the base is 25mL
Now the pOH at half way to the first dissociation of the base is
[tex]pOH = -log(K_i)[/tex]
=> [tex]pOH = -log(0.00794)[/tex]
=> [tex]pOH = 2.100[/tex]
Generally the pH after addition of 25.0 mL of HCl is
[tex]pH = 14 - 2.100[/tex]\
=> [tex]pH = 11.9[/tex]
The first dissociation's equation is as follows:
[tex]B(aq) + H_2O(l) \leftrightharpoons BH^{+} (aq) + OH^{-}(aq) \\\\[/tex]
Constant of base ionization
[tex]\to K_{bl}=\frac{[BH^{+}][OH^{-}]}{[B]}\\\\ \to 7.94\times 10^{-3} = \frac{x\times x}{(0.95- x)} \\\\\to 7.94\times 10^{-3} = \frac{x^2}{(0.95- x)} \\\\\to x^2=7.94\times 10^{-3} (0.95-x) \\\\\to x^2=7.543\times 10^{-3} - 7.94\times 10^{-3} x) \\\\\to x^2=7.543\times 10^{-3} - 7.94\times 10^{-3} x) \\\\\to x = 0.0830\ M\\\\[/tex]
So,
[tex]\to [OH^{-}] = 0.0830\ M\\\\[/tex]
The second dissociation of the base equation is
[tex]BH^{+}\ (aq) + H_20\ (l) \leftrightharpoons BH_2^{2+}\ (aq) + OH^{-}\ (aq) \\\\[/tex]
Constant of base ionization
[tex]\to K_{bl}=\frac{[BH^{+}][OH^{-}]}{[B]}\\\\ \to 3.2 \times 10^{-8} =\frac{y \times (0.0830+y)}{(0.0830- y)}\\\\[/tex]
[tex]\to y= \frac{ 3.2 \times 10^{-8} \times (0.0830- y)}{ (0.0830+y)} \\\\ \to y= \frac{ 3.2 \times 10^{-8} \times (0.0830- y)}{ (0.0830+y)} \\\\ \to y = 3.2\times 10^{-8}[/tex]
So,
[tex]\to [OH^{-}] = 0.0830\ M \\\\\to pOH = 1.08 \\\\\to pH = 14.00 - pOH = 12.92\\\\[/tex]
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What is the volume of 11.2 g of O2 at 7.78 atm and 415 K?
Answer:
1.53 L
Explanation:
Step 1: Given data
Mass of oxygen (m): 11.2 gPressure (P): 7.78 atmTemperature (T): 415 KIdeal gas constant (R): 0.0821 atm.L/mol.KStep 2: Calculate the moles (n) corresponding to 11.2 g of oxygen
The molar mass of oxygen is 32.00 g/mol.
11.2 g × (1 mol/32.00 g) = 0.350 mol
Step 3: Calculate the volume of oxygen
We will use the ideal gas equation.
P × V = n × R × T
V = n × R × T / P
V = 0.350 mol × (0.0821 atm.L/mol.K) × 415 K / 7.78 atm
V = 1.53 L
What element has 6 protons,7 neutrons and 6 electrons
Answer:
corbon 12
Explanation:
carbon atoms with the usual 6 nutrons have a mass number of 12
6 protons plus 7 neutrons equals 13 for carbon nuclei with 7 neutrons, these atoms make up the carbon-13 element.
What is carbon-13?
A naturally occurring stable isotope of carbon with a nucleus comprising six protons and seven neutrons is carbon-13 (13C). It makes up roughly 1.1% of all the native carbon on Earth and is one of the environmental isotopes.
Since they are carbon atoms with extra neutrons, carbon 12, 13, and 14: You could also say that the difference in mass between these isotopes is molecular mass.
Therefore, molecular mass is determined by adding the number of protons and neutrons (if you have 1 mole of each isotope, carbon 14 would have the greatest mass).
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Examples 36
A cylinder is 350cm³ long and two gases A and B have relative molecular weights of 64 and 16 respectively. If the two gases, at the same temperature and pressure, are released simultaneously at both ends determine the distance from one end at which the gases meet.
2x = 350- x
3x = 350
x = 350/x
therefore,
x = 116.67cm.
so, (350 - x)cm³ = (350- 166.67)cm³ = 233.33cm³
Hence distance covered by gas A is 116.67cm that by B is 233.33cm
Calculate the final volume of a solution prepared by diluting a 4.0 M solution with a volume of 7.0 mL to 0.80 M in milliliters.
Answer:
Explanation:
Let the final volume be v .
from the formula
S₁ V₁ = S₂V₂
4 x 7 = .8 x V₂
V₂ = 35 mL
Final volume of the solution will be 35 mL .
What is the process of cell eating called
Answer:
Phagocytosis
Explanation:
A 13.0-L scuba diving tank contains a helium-oxygen (heliox) mixture made up of 23.6 g of He and 4.85 g of O2 at 298 K. Calculate the mole fraction of each component in the mixture.
Answer:
[tex]x_{He}=0.975\\x_{O_2}=0.025[/tex]
Explanation:
Hello.
In this case, since we know the mass of both helium and oxygen, in order to obtain the mole fractions we first need the compute the moles by using their atomic masses, 4.00 g/mol and 32.00 g/mol respectively as shown below:
[tex]n_{He}=23.6gHe*\frac{1molHe}{4.00gHe}=5.90molHe\\ \\n_{O_2}=4.85gO_2*\frac{1molO_2}{32.00gO_2}=0.152molO_2\\[/tex]
Therefore, the mole fractions are:
[tex]x_{He}=\frac{n_{He}}{n_{He}+n_{O_2}}=\frac{5.90}{5.90+0.152} \\\\x_{He}=0.975\\\\x_{O_2}=\frac{n_{O_2}}{n_{He}+n_{O_2}} =\frac{0.152}{5.90+0.152} \\\\x_{O_2}=0.025[/tex]
Best regards!
A graduated cylinder is filled to an initial volume of 25mL. A solid object is dropped into the graduated cylinder. The final volume of the graduated cylinder is 60mL. The mass of the object is 140g. What is the density of the solid object?
Answer:
The answer is 4 g/mLExplanation:
The density of a substance can be found by using the formula
[tex]density = \frac{mass}{volume} \\ [/tex]
volume = final volume of water - initial volume of water
volume = 60 - 25 = 35 mL
From the question we have
[tex]density = \frac{140}{35} \\ [/tex]
We have the final answer as
4 g/mLHope this helps you
Which of these four elements is the most reactive metal?
Answer:
Rubidium
Answer: Rubidium is the most reactive metal. Explanation: Metals are the elements that looses electrons and thus, their chemical reactivity will be the tendency to loose electrons.
Explanation:
the de broglie equation predicts that the wavelength (in m) of a proton moving at 1000 m/s is what?
(h=6.63 x 10^-34J•s; mass od a photon = 1.673 x 10^-24
Explanation:
from
de Broglie wavelength= h/ mv
substitute the values
joules= mass(kg) × acceleration (m/s^2) × distance(m)= kgm^2/s^2.
sorry I don't have my calc with me
According to De Broglie's Hypothesis, matter has both particle and wave properties. The wavelength of De Broglie is given as λ = h p,
wavelength (in m) of a proton = 11.09 x[tex]10^ {-55[/tex] m.
What is de Broglie statement?According to De Broglie's Hypothesis, matter has both particle and wave properties. The wavelength of De Broglie is given as λ = h p, where represents the particle momentum, and can be written as: λ= h m v.The de Broglie hypothesis demonstrated that wave-particle duality was a fundamental principle shared by both radiation and matter, rather than an aberrant behavior of light.De Broglie's equation does not apply to larger particles; rather, it only applies to microscopic particles such as electrons, protons, and neutrons. De Broglie's equation is only valid for moving microscopic particles.λ= h m v.
= 6.63 x [tex]10^{-34}[/tex] x 1.673 x [tex]10^{-24[/tex]x 1000
= 11.09 x[tex]10^ {-55[/tex] m
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Given the following balanced equation:
9 Fe2O3 + 2 NH3 → 6 Fe3O4 + N2 + 3 H2O
A) how many grams of NH3 are needed to react with 125 grams of Fe2O3?
B) how many grams of Fe3O4 will be produced?
Answer:
What mass of sodium hydroxide is needed to react completely with 10.0 g of iron( III) chloride? The conversion sequence is: A. B.
Explanation:
make me as brain liest
which of the two pure substance consists of different types of Element
Answer:Two Main Types of Pure Substances
Elements and compounds are the two types of pure substances. Examples of common elements include carbon, nitrogen and hydrogen. They consist of one type of atom and cannot break down into something else. Every pure carbon substance, for example, has the same particles in it.
Explanation:I think
A hydrocarbon molecule contains carbon and hydrogen atoms in equal numbers. Its molar mass is 130.18 g/mol. What is the molecular formula for the hydrocarbon
Answer:
The molecular formula of the hydrocarbon is C10H10
Explanation:
Here, we are interested in finding the molecular formula for the hydrocarbon.
Since there are equal number of moles of carbon and hydrogen, then we have the molecular formula looking like;
CnHn
Kindly recall that the atomic mass of carbon is 12 a.m.u while that of hydrogen is 1 amu
so calculating the atomic mass of the compound, we have;
12(n) + 1(n) = 130.18
13n = 130.18
n = 130.18/13
n = 10.01
So the molecular formula will be C10H10
Answer:
C10H10
Explanation:
A hydrocarbon is a binary compound of carbon and hydrogen. Hence a hydrocarbon is a compound of the general formula (CH)n
Thus;
(12 + 1) n = 130.18
n= 130.18/13
n = 10
Hence the molecular formula of the compound is C10H10
In which atmospheric layer is the ozone layer?
A.troposphere
B.mesosphere
C.stratosphere
D.thermosphere
Answer:
stratosphere
Explanation:
contains a high concentration of ozone in relation to other parts of the atmosphere, although still small in relation to other gases in the stratosphere.
Answer:
stratosphere
Explanation: Most atmospheric ozone is concentrated in a layer in the stratosphere, about 9 to 18 miles (15 to 30 km) above the Earth's surface. Ozone is a molecule that contains three oxygen atoms.
A chemical engineer must calculate the maximum safe operating temperature of a high-pressure gas reaction vessel. The vessel is a stainless-steel cylinder that measures 41.0cm wide and 49.2cm high. The maximum safe pressure inside the vessel has been measured to be 3.70MPa. For a certain reaction the vessel may contain up to 2.50kg of dinitrogen difluoride gas.
Required:
Calculate the maximum safe operating temperature the engineer should recommend for this reaction. Write your answer in degrees Celsius. Be sure your answer has the correct number of significant digits.
Answer:
[tex]T=2.78x10^3 \°C[/tex]
Explanation:
Hello,
In this case, considering that the safe temperature may be computed via the ideal gas law as we now the pressure, mass and volume via the dimensions:
[tex]V=\pi r^2 h=\pi *(41.0cm)^2*49.2cm=2.60x10^5cm^3*\frac{1L}{1000cm^3} =260L[/tex]
The pressure in atm is:
[tex]P=3.70MPa*\frac{1x10^6Pa}{1MPa} \frac{1atm}{101325Pa} =36.5atm[/tex]
And the moles considering the mass and molar mass (66 g/mol) of dinitrogen difluoride (N₂F₂):
[tex]n_{N_2F_2}=2.50kg*\frac{1000g}{1kg}*\frac{1mol}{66g} =37.9mol[/tex]
In sich a way, by applying the ideal gas equation, which is not the best assumption but could work as an approximation due to the high temperature, the temperature, with three significant figures, will be:
[tex]T=\frac{PV}{nR}=\frac{36.5Pa*260L}{37.9mol*0.082\frac{atm*L}{mol*K} }\\ \\T=3053.6K-273.15\\\\T=2.78x10^3 \°C[/tex]
Best regards.
calculate the mol fraction of ethanol and water in a sample of rectified spirit which contains 95% of ethanol by mass
Answer:
We are given that there is 95% ethanol by mass in rectified spirit
so, we can say that in a 100g sample, we have 95 grams of ethanol and 5 grams of water
we will find the number of moles of ethanol and water in 100g solution of rectified spirit and use that to calculate the mole fraction
Moles of Ethanol:
Molar mass of ethanol = 46 grams / mol
Number of moles = Given mass / molar mass
Number of moles = 95 / 46
Moles of Ethanol = 2 moles (approx)
Moles of Water:
Molar mass of water = 18 grams per mol
Number of moles = Given mass / molar mass
Moles of water = 5 / 18
Moles of water = 0.28 moles (approx)
Mole Fractions:
Mole fraction of a specific compound is the number of moles of that compound divided by the total number of moles in the solution
Mole fraction of Ethanol:
Moles of ethanol / (moles of ethanol + moles of water)
2 / (2 + 0.28)
2 / (2.28) = 0.9 (approx)
Mole fraction of Water:
Moles of water / (Moles of ethanol + moles of water)
0.28 / (2 + 0.28)
0.28 / (2.28) = 0.1 (approx)
Facilitated diffusion is a form of passive transport and involves the movement of molecules (like glucose) that are too large to just go through the cell membrane without help. These larger molecules move through the cell membrane from High concentration to Low concentration with the help of _____________________________.
Answer: Membrane Protiens
Explanation:
In facilitated diffusion, molecules diffuse across the plasma membrane with assistance from membrane proteins, such as channels and carriers. A concentration gradient exists for these molecules, so they have the potential to diffuse into (or out of) the cell by moving down it.
What is the heat gjoules transferred by a chemical reaction to the reservoir of a calorimeter containing 135g of dilute aqueous solution (c=4.184 J/g^ C) the reaction causes the temperature of the reservoir to rise from 23.0 27.0 degrees * C
Answer:
Required heat = 2,259.36 N (approx)
Explanation:
Given:
Mass = 135 g
Specific heat (c) =4.184 J/gC
Change in temperature ΔT = 27-23 = 4 c
Find:
Required heat
Computation:
Q = mcΔT
Q = (135)(4.184)(4)
Required heat = 2,259.36 N (approx)
The heat transferred by the chemical reaction to the reservoir of a calorimeter containing 135 g of dilute aqueous solution is 2259.36 J
To solve this question, we'll begin by calculating the change in the temperature of the reservoir. This can be obtained as follow:
Initial temperature (T₁) = 23 °C
Final temperature (T₂) = 27 °C
Change in temperature (ΔT) =?ΔT = T₂ – T₁
ΔT = 27 – 23
ΔT = 4 °CFinally, we shall determine the heat transferred to the reservoir.
Change in temperature (ΔT) = 4 °C
Mass (M) = 135 g
Specific heat capacity (C) = 4.184 J/gºC
Heat (Q) =?Q = MCΔT
Q = 135 × 4.184 × 4
Q = 2259.36 JTherefore, the heat transferred by the chemical reaction to the reservoir of a calorimeter containing 135 g of dilute aqueous solution is 2259.36 J
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Which of the following are decomposers?
O A. mice
O B. snakes
O C. fungi
OD. spiders
ASAP
Is lead a representative metal or transitional metal?
Answer:
It's a representative metal
Explanation: Transitional metals are metals of various chemical elements and have valence electrons—i.e., electrons that can participate in the formation of chemical bonds.
List the steps in the water cycle, in the correct order?
pls help! I'll give u 10 points and the brainlest answer!!!
Answer:
371km
Explanation:
the lower the more fuel there is
a change of matter is a physical change
True or False
Answer:
true
Explanation:
hope it helps
Answer:
I'm pretty sure it's true
Explanation:
20 characters
A block of aluminum weighing 100g is cooled from 108.4°C to 68.2°C
with the release of 1080 joules of heat. From this data, calculate the
specific heat of aluminum
Answer:
Specific heat of aluminum = 0.27 j/g.°C
Explanation:
Given data:
Mass of Al = 100 g
Initial temperature = 108.4°C
Final temperature = 68.2°C
Heat released = -1080 j
Specific heat of aluminum = ?
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = 68.2°C - 108.4°C
ΔT = - 40.2°C
Q = m.c. ΔT
-1080 j = 100 g ×c ×- 40.2°C
-1080 j = -4020 g.°C ×c
c = -1080 j/-4020 g.°C
c = 0.27 j/g.°C
If 1.4434 moles of H2O are produced, how many moles of N2 will also be produced?
Answer:
1 gram of N2 is equal to 0.035697202053303 mole. i dont know the rest but intried to help
. hopefully this will give you someone else a starting point. goodluck
For each of the following molecules draw the Lewis structure on a separate sheet of paper. MAKE SURE TO FOLLOW THE RULES FROM CLASS (ie do not break the octet rule unless necessary to connect all the atoms). Then based on your structure indicate:
the total number of valence electrons.
the electronic and molecular shapes (choose from: linear, trigonal planar, bent, tetrahedral, trigonal pyramidal, trigonal bipyramidal, seesaw, T-shaped, octahedral, square pyramidal, or square planar).
whether or not the molecule is polar (Y/N).
Note: The central atom is the first atom listed, except for HCN, H2CO, and OCN-, where carbon is the central atom (underlined).
Formula Valence electrons Electronic Shape Molecular Shape Polar (Y/N)
HCN
PH3
CHCl3
NH4+
H2CO
SO42-
SeF2
CO2
O2
ClO4-
HBr
PF5
BeH2
PO43-
BH3
Br3-
Answer:
Kindly check the explanation section.
Explanation:
Without mincing words let us dive right into the solution to the question above, taking each compound at a time.
NB: Kindly Check attachment for the Lewis Structure of each of the chemical compounds.
Therefore, the number of valence electrons, electronic shape, molecular shape and whether the molecules are polar(Polarity) is given below for each chemical compound.
(1). Compound: HCN
(a). number of valence electrons = 10.
(b). electronic shape =linear.
(c). molecular shape = linear.
(d). Polarity = Y.
(2). Compound: PH3
(a). number of valence electrons = 8.
(b). electronic shape = Tetrahedral.
(c). molecular shape = Trigonal Pyramidal.
(d). Polarity = Y.
(3). Compound: CHCl3.
(a). number of valence electrons = 26.
(b). electronic shape = tetrahedral.
(c). molecular shape = tetrahedral.
(d). Polarity = Y.
(4). Compound: NH4^+
(a). number of valence electrons = 8
(b). electronic shape = tetrahedral
(c). molecular shape = tetrahedral
(d). Polarity = Y.
(5). Compound: H2CO
(a). number of valence electrons = 12.
(b). electronic shape = Trigonal planar.
(c). molecular shape = Trigonal planar
(d). Polarity = Y.
(6). Compound: SO4^2-
(a). number of valence electrons = 32.
(b). electronic shape = Tetrahedral.
(c). molecular shape = Tetrahedral.
(d). Polarity = N.
(7). Compound: SeF2.
(a). number of valence electrons = 20.
(b). electronic shape = Tetrahedral.
(c). molecular shape = bent.
(d). Polarity = Y.
(8). Compound: CO2.
(a). number of valence electrons = 16.
(b). electronic shape = linear.
(c). molecular shape = linear.
(d). Polarity = N.
(9). Compound: O2
(a). number of valence electrons = 32.
(b). electronic shape = Trigonal planar.
(c). molecular shape = Linear.
(d). Polarity = N.
(10). Compound: ClO4-.
(a). number of valence electrons = 32.
(b). electronic shape = Tetrahedral.
(c). molecular shape = Tetrahedral.
(d). Polarity = N.
(11). Compound: HBr.
(a). number of valence electrons = 8.
(b). electronic shape = Linear.
(c). molecular shape = Linear.
(d). Polarity = Y.
(12). Compound: PF5.
(a). number of valence electrons = 40.
(b). electronic shape = Trigonal Bipyramidal.
(c). molecular shape = Trigonal Bipyramidal.
(d). Polarity = N.
(13). Compound: BeH2.
(a). number of valence electrons = 4.
(b). electronic shape = Linear.
(c). molecular shape = Linear.
(d). Polarity = N.
(14). Compound: PO4^3-.
(a). number of valence electrons = 32.
(b). electronic shape = Tetrahedral.
(c). molecular shape = Tetrahedral.
(d). Polarity = N.
(15). Compound: BH3.
(a). number of valence electrons = 6.
(b). electronic shape = Trigonal planar.
(c). molecular shape = Trigonal planar.
(d). Polarity = N
(16). Compound: Br3-.
(a). number of valence electrons = 32.
(b). electronic shape = Trigonal Bipyramidal.
(c). molecular shape = Linear.
(d). Polarity = N.
What is the density of a liquid with the same mass as in problem d, if it has a volume of 43.2 mL?
(part d: A liquid has a volume of 62.7 mL and a density of 2.59 g/mL. What is its mass?)
Answer:
3.76
Explanation:
You start by finding the mass of part d. Mass= Density times Volume.In this case you would multiply 62.7 and 2.59 which is 162.393 rounded is 162.4. Then density= mass/ volume which is 162.4/ 43.2. density equals 3.76
Which of the following substances would have the greatest ductility?
A. Fe(s)
B. SiO2(s)
C. C(s)
D. NaCl(s)
Fe(s) would have the greatest ductility.
What is ductility?Ductility is the capability of a fabric to be drawn or plastically deformed without fracture. it's far therefore a demonstration of how 'gentle' or malleable the fabric is. The ductility of steel varies relying on the sorts and levels of alloying factors gift.
What are malleability and ductility?Ductility is the property of metallic associated with the capability to be stretched into twine without breaking. Malleability is the assets of metallic associated with the ability to be hammered into thin sheets without breaking. The outside force or strain is tensile pressure.
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what are molecular compounds
Answer:
inorganic compounds that take the form of discrete molecules
Answer:
Compounds that take the form of discrete molecules.
Explanation:
Rather than forming ions, the atoms of a molecule share their electrons in such a way that a bond forms between pairs of atoms.
A compound is 49 and 6.7 hydrogen and 53 oxygen what’s the empirical formula
Answer:
dangerous woman by ariana grande
Explanation:
Answer:
C6H10O6
Explanation:
Convert % and make them to grams. Convert grams to mols. Use mole ratios to find empirical formula. I will calculate it in a second. Hope this helps for now. This is my quick answer imma check my work one sec.
1) A hanglider flies with a horizontal velocity of 13 m/s when his wallet falls out. If the hanglider is 1450 m above the earth when the wallet falls out, find the horizontal distance the wallet travels before hitting the ground.
Answer:
The wallet is falling from a height of 1450 m
We are given that:
downward acceleration (a) = 10 m/s/s
downward displacement (s) = 1450 m
downward initial velocity (u) = 0 m/s (Since the wallet was dropped, it had no initial velocity)
horizontal initial velocity (v) = 13 m/s
Since there is no external force being applied on the wallet except the force of gravity, the wallet will keep moving at a constant velocity of 13 m/s
To calculate the horizontal distance travelled by the wallet, we need to know how long the wallet was airborne
To calculate how long the wallet was airborne, we have to find out that how long it took the wallet to hit the ground
Time taken for the wallet to hit the ground:
From the second equation of motion:
s = ut + 1/2 at² ------------(for vertical motion of the wallet)
1450 = (0)(t) + 1/2 (10)(t)²
1450 = 5t²
t² = 290
t = 17 (approx)
Time taken for the wallet to hit the ground = 17 seconds
Horizontal distance travelled by the wallet:
Since the wallet hits the ground after 17 seconds, it will move horizontally at a constant velocity of 13 m/s for 17 seconds
horizontal acceleration = 0 m/s/s
From the second equation of motion:
s = ut + 1/2at²
s = ut + 1/2 (0)t²
s = ut
here, u is the horizontal initial velocity of the wallet and the time taken by the wallet to hit the ground is 't'
s = (13)(17)
s = 221 m
Hence, the horizontal distance travelled by the wallet while falling from a height of 1450 m is 221 m