Lesson 2 History of Physical Science
Write an expository essay explaining how science builds on itself. Use at least two specific examples from this lesson.

The bus will travel a further 20 m before coming to rest.

The term acceleration has to do with a change in velocity with time. Now we have;

u = 20 m/s

v = 10 m/s

s =  60 m

Now;

v^2 = u^2 -2as

v^2 -  u^2 = -2as

(10)^2 - (20)^2 = - 2 * a * 60

a = (10)^2 - (20)^2/ - 2 * 60

a = 100 - 400/ - 2 * 60

a = 2.5 m/s^2

At that time;

v = 0 m/s

u = 20 m/s

a = 2.5 m/s^2

s = ?

Hence;

u^2 = 2as

s = u^2/2a

s = (20)^2/ 2 *  2.5

s = 400/5

s = 80 m

Hence, the bus will travel a further 20 m before coming to rest.

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The speed of the second ball after impact is 1.43 m/s at 60.8⁰.

Apply the principle of conservation of linear momentum;

0.6(2) = 0.6(1.5 sin30) + 0.6(vy)

1.2 = 0.45 + 0.6vy

0.6vy = 0.75

vy = 0.75/0.6

vy = 1.25 m/s

0.6(2) = 0.6(1.5 cos30) + 0.6(vx)

1.2 = 0.78 + 0.6vx

0.6vx = 0.42

vx = 0.42/0.6

vx = 0.7 m/s

v = √(vy² + vx²)

v = √(1.25² + 0.7²)

v = 1.43 m/s

θ = arc tan(vy/vx)

θ = arc tan (1.25/0.7)

θ = 60.8⁰

Thus, the speed of the second ball after impact is 1.43 m/s at 60.8⁰.

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a.

b.

Since k is the spring constant of both springs = 250 N/m. The equivalent spring constant in parallel is k' = k + k

= 2k

= 2 × 250 N/m

= 500 N/m

Now since A is the maximum distance the block is pulled from its equilibrium position, the total energy of the block is E = 1/2kA

Also, 1/2k'A² = 1/2k'x² + 1/2Mv² where

Making v subject of the formula, we have

v = √[k'(A² - x²)/M]

Substituting the values of the variables into the equation, we have

v = √[k'(A² - x²)/M]

v = √[500 N/m(A² - (0)²)/10]

v = √[50 N/m(A² - 0)]

v = [√50]A m/s

v = [5√2] A m/s

v = 7.07 A m/s

So, the speed of the block of mass when the springs are connected in parallel is 7.07 A m/s

Since velocity of spring v = ω√(A² - x²) where

Making ω subject of the formula, we have

ω = v/√(A² - x²)

Since v = 7.07 A m/s

Substituting the values of the other variables into the equation, we have

ω = v/√(A² - x²)

ω = 7.07 A m/s/√(A² - 0²)

ω = 7.07 A m/s/√(A² - 0)

ω = 7.07 A m/s/√A²

ω = 7.07 A m/s/A m

ω = 7.07 rad/s

So, the angular velocity when the two springs are in parallel is 7.07 rad/s

Since k is the spring constant of both springs = 250 N/m. The equivalent spring constant in parallel is 1/k" = 1/k + 1/k

= 2/k

⇒ k" = k/2

k" = 250 N/m ÷ 2

= 125 N/m

Now since A is the maximum distance the block is pulled from its equilibrium position, the total energy of the block is E = 1/2kA

Also, 1/2k"A² = 1/2k"x² + 1/2Mv'² where

Making v subject of the formula, we have

v = √[k"(A² - x²)/M]

Substituting the values of the variables into the equation, we have

v = √[k"(A² - x²)/M]

v = √[125 N/m(A² - (0)²)/10]

v = √[125 N/m(A² - 0)]

v = [√125]A m/s

v = [5√5] A m/s

v = 11.2 A m/s

So, the speed of the block of mass when the springs are connected in series is 11.2 A m/s

Since velocity of spring v = ω√(A² - x²) where

Making ω subject of the formula, we have

ω = v/√(A² - x²)

Since v = 11.2 A m/s

Substituting the values of the other variables into the equation, we have

ω = v/√(A² - x²)

ω = 11.2 A m/s/√(A² - 0²)

ω = 11.2 A m/s/√(A² - 0)

ω = 11.2 A m/s/√A²

ω = 11.2 A m/s/A m

ω = 11.2 rad/s

So, the angular velocity when the two springs are in series is 11.2 rad/s

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We obtain the magnitude of the acceleration as   1.55 m/s^2 while its component parallel to the ground is 0.775  m/s^2.

The acceleration refers to change of speed with time. From the question;

It the follows that;

v^2 = u^2 + 2as

And;

u = 0m/s  owing to the fact that motion started from rest

v^2 = 2as

a = v^2/2s

a = (10.43 m/s)^2/2 * 35.0 m

a = 1.55 m/s^2

For the acceleration component which is parallel to the ground;

ax = a sin θ

ax =  1.55 m/s^2 cos 60

ax = 0.775  m/s^2

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The upward force exerted on the board by the support is mathematically given as

Fu= 764.8 N

Generally, the equation for is  mathematically given as

Considering that the Net Force on the system is null

The weight of the children plus the weight of the board equals the upward force imposed on the support.

The upward force

Fu= 440 + 272 + 52.8 N

Fu= 764.8 N

In conclusion, he upward force exerted on the board by the support

Fu= 764.8 N

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(a) The peak voltage across the resistor is 339.41 V.

(b) The maximum current through the resistor is 0.23 A.

(c) The rms current through the resistor is 0.16 A.

(d)  The average power dissipated by the resistor is 38.4 W.

Vrms = 0.7071V₀

where;

V₀ = Vrms/0.7071

V₀ = 240/0.7071

V₀ = 339.41 V

I(rms) = V(rms)/R

I(rms) = (240)/(1,540)

I(rms) = 0.16 A

I₀ = I(rms)/0.7071

I₀ = (0.16)/0.7071

I₀ = 0.23 A

P = I(rms) x V(rms)

P = 0.16 x 240

P = 38.4 W

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Heat transfer in a closed system is the addition of changes in internal energy and the total amount of work done by it.  The final energy of the system is 35.5kJ.  

Heat transfer is the transfer of heat energy due to temperature differences.

The paddle-wheel paintings are quantities of workdone, 500 N.m or 0.5kJ.

The preliminary (initial) power of the device is 10 kJ.

Total warmness transferred in the course of the method is 30 kJ

Total warmness misplaced in the course of the method to the encompassing air is 5 kJ.

The energy of the system is given as:

The energy of the system = Energy in - Energy out

The energy of the system = Initial energy + Energy transferred + Work done - Energy lost

Energy of the system = 10 + 30 + 0.5 - 5 kJ

Energy of the system = 35.5 kJ

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Answer:

Behaviorists give prominence to the external behavior of individuals and believe that behavior is a response to external stimuli. On the other hand, psychoanalysis emphasizes the centrality of the human mind. They believe that the unconscious has the potential to motivate behavior.

Explanation:

Answer: Behaviorists give prominence to the external behavior of individuals and believe that behavior is a response to external stimuli. On the other hand, psychoanalysis emphasizes the centrality of the human mind. They believe that the unconscious has the potential to motivate behavior.

Explanation:

The horizontal force required to move the crate at a steady speed across the floor will be 58.86 N.

It is a type of opposition force acting on the surface of the body that tries to oppose the motion of the body. its unit is Newton (N).

The given data in the problem is;

Mass,m = 24 kg

The coefficient of kinetic friction, μ=0.25

The friction force is defined as the product of the coefficient of friction and normal reaction.

f=μN

If there is no other force, the horizontal force is equal to the friction force;

F=f

The horizontal force required to move the crate at a steady speed across the floor is found as;

F = μN

The normal reaction force is;

N=W

N=mg

F = μmg

F=0.25 ×  24-kg × 9.81 m/s²

F = 58.86 N

Hence horizontal force is required to move the crate at a steady speed across the floor will be 58.86 N.

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Initial speed of ball will be 2.7m/s.

When a body is rotating about an axis, then it has kinetic energy.

And this energy is called rotational kinetic energy.

It is given as -  R.K.E. = 1/2 Iω²

And if a ball is rolling without slipping.

Then the moment of inertia of the solid ball is written as -

I = 25MR²

Vi = Rω

Here it is given in the problem that-

height(h) = 0.53m

Now by the conservation of energy we can write the equation as -

1/2MVi² + 1/2Iω² = Mgh

so that -

(1/2)MVi² + (1/2)×(2/5MR²) ×(Vi/R)² = Mgh(1/2)Vi² + (1/5)Vi²

= gh(7/10)Vi² = 9.8 × 0.53

Vi = 2.7 m/s

So that the initial velocity of ball came out to be 2.7m/s after applying all concepts or rotational motion.

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Considering the Coulomb's Law, the force between the two ballons is 9.216×10⁻⁸ N.

Charged bodies experience a force of attraction or repulsion on approach.

From Coulomb's Law it is possible to predict what the electrostatic force of attraction or repulsion between two particles will be according to their electric charge and the distance between them.

From Coulomb's Law, the electric force with which two point charges at rest attract or repel each other is directly proportional to the product of the magnitude of both charges and inversely proportional to the square of the distance that separates them:

[tex]F=k\frac{Qq}{d^{2} }[/tex]

where:

The force is attractive if the charges are of opposite sign and repulsive if they are of the same sign.

In this case, you know that two balloons have a negative charge of 1.6×10⁻¹⁰ C and the balloons are 0.05 m apart.

Replacing in the Coulomb's Law, you get:

[tex]F=9x10^{9} \frac{Nm^{2} }{C^{2} } \frac{(-1.6x10^{-10} C)x(-1.6x10^{-10} C)}{(0.05 m)^{2} }[/tex]

Solving:

[tex]F=9x10^{9} \frac{Nm^{2} }{C^{2} } \frac{2.56x10^{-20} C^{2} }{(0.05 m)^{2} }[/tex]

[tex]F=9x10^{9} \frac{Nm^{2} }{C^{2} } 1.024x10^{-17}\frac{ C^{2} }{m^{2} }[/tex]

F= 9.216×10⁻⁸ N

Finally, the force between the two ballons is 9.216×10⁻⁸ N.

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(A)The total resistance of the parallel resistors is: [tex]R_{T} = \frac{R_{1} \times R_{2} \times R_{3}}{R_{2}R_{3}+R_{1}R_{3}+R_{1}R_{2}}[/tex]

(B) The current in the overall circuit is: [tex]I_{T} = \frac{V}{R_{1}} + \frac{V}{R_{2}} + \frac{V}{R_{3}}[/tex]

(C) The current through each resistance is as follows:

For the three resistances connected in parallel, R1, R2, R3 , the total resistance, [tex]R_{T}[/tex] is calculated as follows:

[tex]R_{T} = \frac{R_{1} \times R_{2} \times R_{3}}{R_{2}R_{3}+R_{1}R_{3}+R_{1}R_{2}}[/tex]

The current in the overall circuit is calculated using the formula:

[tex]I_{T} = \frac{V}{R_{1}} + \frac{V}{R_{2}} + \frac{V}{R_{3}}[/tex]

The current through each resistance is given as follows;

Current through R1, [tex]I_{1} = \frac{V}{R_{1}}[/tex]

Current through R2; [tex]I_{2} = \frac{V}{R_{2}}[/tex]

Current through R3; [tex]I_{3} = \frac{V}{R_{3}}[/tex]

In conclusion, the voltage across resistances in parallel is the same but the current varies.

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If the metal ion has a 3+ charge and the nonmetal ion has a 4- charge, the formula of the compound is X4Z3.

The formula of an ionic compound is determined by the charges of the ions that make up the compound. Recall that when the compound is  formed, the ions exchange valences.

Hence, is the metal ion has a 3+ charge and the nonmetal ion has a 4- charge, the formula of the compound is X4Z3.

Missing parts:

A metal ion (X) with a charge of 3+ is attracted to a nonmetal ion (Z) with a charge of 4-. Which of these formulas represents the resulting compound.

A. 7XZ

B. X3Z4

C. X4Z3

D. 4X3Z

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The strength of the electric field at that point and the force would this charge experiences at that point will be 4.587 N/C and  12.38 N.

The electric field strength is defined as the ratio of electric force to charge.

Given data;

q₁ = 5.4 C

F₁ is the electric force in case1

E is the electric field =?

F₂ is the electric force in case 2

q₂ is the charge 2

The strength of the electric field at that point is;

F₁=Eq₁

E₁=F/q₁

E₁=25.0 N / 5.4 C

E₁=4.587 N/C

The force would this charge experience at that point when the charge is 2.7 C;

F₂=Eq₂

F₂=4.587 N/C × 2.7 C

F₂ = 12.38 N

Hence the strength of the electric field at that point and the force would this charge experiences at that point will be 4.587 N/C and  12.38 N.

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The average speed of the person over the entire trip is 3.75 m/s.

Average speed = total distance/total time

Average speed = (d + d)/(d/Va + d/Vb)

where;

Average speed = (2Vavb)/(Va + Vb)

Average speed = (2 x 5 x 3)/(5 + 3)

Average speed = 3.75 m/s

Thus, the average speed of the person over the entire trip is 3.75 m/s.

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The coasts of Africa and South America were touched  125 million years ago due to continental drift.

The time can be computed from the ratio of distance and speed.

Given:

Distance = 5, 000,000

Speed = 4 cm/year = 0.04 m/year

The time is computed as:

Time = Distance / Rate

Time = 5,000,000 meters / 0.04 meters per year

Time = 125,000,000 years

Hence, the coasts of Africa and South America were touched  125 million years ago due to continental drift.

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The frequency f of the AC source is determined as 0.446 Xc.

The frequency of the AC source is calculated as follows;

Xc ≡ 1/2fC

where;

fC = 2Xc

f = 2Xc / C

f = (2Xc)/4.48

f = 0.446 Xc

Thus, the frequency f of the AC source is determined as 0.446 Xc.

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Hello!

1.

To solve, we can use the work-energy theorem.

Initially, the rod is at rest. Its center of mass is some height above the surface, so it contains gravitational potential energy.

Finally, the rod has rotational kinetic energy.

To find the initial energy, we have to find where the center of mass is. It is located at the midpoint of the rod, or l/2. To find its height above, though, we have to find its sine component with respect to the angle:
[tex]E_i = mg{\frac{l}{2}}}sin(\theta)[/tex]

Its final energy:
[tex]E_f = \frac{1}{2}I\omega^2[/tex]

About the pivot point, a rod's moment of inertia (I) is 1/3ml².

[tex]E_f = \frac{1}{2}(\frac{1}{3}ml^2) = \frac{1}{6}ml^2\omega^2[/tex]

Now, solving for angular speed:

[tex]E_i = E_f\\\\mg\frac{l}{2}sin\theta = \frac{1}{6}ml^2 \omega^2\\\\3gsin\theta = l\omega^2 \\\\\omega = \sqrt{\frac{3gsin\theta}{l}}[/tex]

Plugging in the values and solving:
[tex]\omega = \sqrt{\frac{3(9.8)sin(70.4)}{2.1}} = \boxed{3.63 rad/s}[/tex]

2.

Let's determine where to place our pivot point. To minimize the number of torques to take into account, let's put the pivot point at the base of the rod.

Now, the only torque we must consider is that produced by the force of gravity.

Recall the equation for torque:
[tex]\tau = r \times F[/tex]

This is a cross-product, where 'r' is the lever arm of the force. This is the perpendicular distance between the line of action of the force and the pivot point.

In this instance, based on our reference angle, this would be the COSINE.

[tex]\tau = rFcos\theta[/tex]

r = distance from pivot point to center of the rod (1.05 m)

F = Force of gravity, mg (19.03 N)

θ = angle of rod from horizontal

Using the rotational equivalent of Newton's Second Law:
[tex]I\alpha = rFcos\theta[/tex]

'I' is the rod's moment of inertia about the pivot point. For a uniform rod, its moment of inertia at its end is 1/3ml².

Therefore:
[tex]I = \frac{1}{3}(1.94)(2.1^2) = 2.85 kgm^2[/tex]

Now, we can solve for the function for angular acceleration.

[tex]\alpha(\theta) = \frac{rF}{I}cos\theta\\\\\alpha(\theta) = \frac{(1.05)(19.03)}{2.85}cos\theta = 7cos\theta[/tex]

This is the angular acceleration of the rod as a function of theta.

Substituting in θ = 0°:
[tex]\alpha(0) = 7cos(0) = \boxed{7 rad/s^2}[/tex]

**Note, we couldn't just use rotational kinematics because the angular acceleration is NOT constant.

The horizontal component of the tension in the string is a centripetal force, so by Newton's second law we have

• net horizontal force

[tex]F_{\rm tension} \sin(\theta) = \dfrac{mv^2}R[/tex]

where [tex]m=4.10\,\rm kg[/tex], [tex]v=2.85\frac{\rm m}{\rm s}[/tex], and [tex]R[/tex] is the radius of the circular path.

As shown in the diagram, we can see that

[tex]\sin(\theta) = \dfrac Rr \implies R = r\sin(\theta)[/tex]

where [tex]r=1.69\,\rm m[/tex], so that

[tex]F_{\rm tension} \sin(\theta) = \dfrac{mv^2}R \\\\ \implies F_{\rm tension} = \dfrac{mv^2}{r\sin^2(\theta)}[/tex]

The vertical component of the tension counters the weight of the mass and keeps it in the same plane, so that by Newton's second law we have

• net vertical force

[tex]F_{\rm \tension} \cos(\theta) - mg = 0 \\\\ \implies F_{\rm tension} = \dfrac{mg}{\cos(\theta)}[/tex]

Solve for [tex]\theta[/tex] :

[tex]\dfrac{mv^2}{r\sin^2(\theta)} = \dfrac{mg}{\cos(\theta)} \\\\ \implies \dfrac{\sin^2(\theta)}{\cos(\theta)} = \dfrac{v^2}{rg} \\\\ \implies \dfrac{1-\cos^2(\theta)}{\cos(\theta)} = \dfrac{v^2}{rg} \\\\ \implies \cos^2(\theta) + \dfrac{v^2}{rg} \cos(\theta) - 1 = 0[/tex]

Complete the square:

[tex]\cos^2(\theta) + \dfrac{v^2}{rg} \cos(\theta) + \dfrac{v^4}{4r^2g^2} = 1 + \dfrac{v^4}{4r^2g^2} \\\\ \implies \left(\cos(\theta) + \dfrac{v^2}{2rg}\right)^2 = 1 + \dfrac{v^4}{4r^2g^2} \\\\ \implies \cos(\theta) + \dfrac{v^2}{2rg} = \pm \sqrt{1 + \dfrac{v^4}{4r^2g^2}} \\\\ \implies \cos(\theta) = -\dfrac{v^2}{2rg} \pm \sqrt{1 + \dfrac{v^4}{4r^2g^2}}[/tex]

Plugging in the known quantities, we end up with

[tex]\cos(\theta) \approx 0.784 \text{ or } \cos(\theta) \approx -1.27[/tex]

The second case has no real solution, since [tex]-1\le\cos(\theta)\le1[/tex] for all [tex]\theta[/tex]. This leaves us with

[tex]\cos(\theta) \approx 0.784 \implies \theta \approx \cos^{-1}(0.784) \approx \boxed{38.3^\circ}[/tex]

The specific gravity of the object’s material is 5.09.

Weight difference = 9 - 7.2 = 1.8 N = Buoyant force of water

Buoyant Force in water(Fb) = density of water x g x volume of the   body(Vb)

1.8 = 1000 x 9.81 x Vb

Vb = 1.8/9810 cubic meter

Now, in the air;

Weight of body = mg = 9 N

Mass of body,m = 9/9.81 Kg

So,

Density of body = m/ Vb

= 9/9.81 ÷ 1.8/9810

= 5094.44 kg per cubic meter

The specific gravity of body = density of body ÷ density of water

= 5094.44 ÷ 1000

= 5.09

Therefore, Specific gravity of body = 5.09

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Answer:

dependent variable

Explanation:

The amount that Mary will pay in social security taxes will be $1550.

It should be noted that the current tax for the social security is 6.2% for the employer and 6.2% for the employee.

In this case, let's assume that the income for Mary is $25000, the amount that Mary will pay will be:

= 6.2% × $25000

= 0.062 × $25000

= $1550

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The angle above the horizontal of the motion of mass A + B after the collision. as well as others are mathematically given as

According to the collision hypothesis of chemistry, chemical reactions happen when two molecules or atoms collide.

Generally, the equation for Collison equilibrium is  mathematically given as

For horizontal motion

mvAx+m2vAx = 2mvfx

For vertical motion

-mvAy+m2vAy = 2mvfy

vfy = 1/2 vAy

[tex]\tan \theta f = vfy/vfx \\\\\tan \theta f = 1/3 vAy/vAx \\\\\tan \theta f = vAy/vAx \\\\\theta f = tan-1 ( 1/3 tan (56) )\\\\[/tex]

[tex]\theta f = 26.3 \deg[/tex]

(b)

[tex]v fy = vf sin \theta f\\\\vAy = vA sin theta i\\\\vf/vA = vfy/vAy sin( \theta i)/sin(\theta f)\\\\vf/vA = 1/2 sin(56)/sin(26.3) \\\\[/tex]

vf/vA= 0.94

(c)

[tex]Ef/Ei = \frac{(1/2 * (2m) * (vf)^2) }{ (1/2 * m * (vA)^2) + 1/2*m*(2vA)^2) ]}[/tex]

2vf^2 / 5vA^2 = 0.35

In conclusion, The angle above the horizontal of the motion of mass A + B after the collision. as well as others

[tex]\theta f = 26.3 \deg[/tex]

vf/vA= 0.94

2vf^2 / 5vA^2 = 0.35

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Question :

Options given :

Solution :

As we know that,

When a circuit is connected in series resistance is calculated by :

Here,

As a circuit contains three resistors. So R1 , R2 and R3 would be :

Now,

>> Rs = 10 + 10 + 10

>> Rs = 30 Ω

Therefore,

The equivalent resistance in the circuit is 30.0 Ω. The correct option is D.

Ohm's law is a fundamental principle of electric circuits that describes the relationship between electric current, voltage, and resistance. It states that the electric current flowing through a conductor between two points is directly proportional to the voltage across the two points, and inversely proportional to the resistance between them.

Mathematically,

V = IR

Where V = is the voltage across the conductor in volts (V),

I =  is the electric current flowing through the conductor in amperes (A),

R =is the resistance of the conductor in ohms (Ω).

This relationship can also be expressed in terms of power, using the formula:

P = VI

Where P = is the power in watts (W),

V = is the voltage in volts (V),

I = is the current in amperes (A).

Ohm's law is named after the German physicist Georg Simon Ohm, who first published it in 1827 as a result of his experiments with electric circuits. Ohm's law is widely used in electrical engineering and is essential for understanding and designing electronic devices and circuits.

Here in this question,

The equivalent resistance of resistors in series is the sum of their individual resistances.

Now, the equivalent resistance of three 10.0 Ω resistors connected in series is:

Req = R1 + R2 + R3

Req= 10.0 Ω + 10.0 Ω + 10.0 Ω

Req= 30.0 Ω

Using Ohm's law, we can find the current flowing through the circuit:

I = V/R

= 6.0 V/30.0 Ω

I = 0.2 A

Therefore, the equivalent resistance of the circuit is 30.0 Ω (Option D).

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Answer:

patron

Explanation:

 The patron is something or someone who defends some cause or point of view. In the art field, for example, the patron may be considered a sponsor, that is, someone who is known for defending a particular group of people or specific situation.

  The patron is one who advocates, advises and directs. In the military, patrons are heroic figures who are chosen to defend a military unit, for example.

Explanation:

patron

Explanation:

 The patron is something or someone who defends some cause or point of view. In the art field, for example, the patron may be considered a sponsor, that is, someone who is known for defending a particular group of people or specific situation.

  The patron is one who advocates, advises and directs. In the military, patrons are heroic figures who are chosen to defend a military unit, for example.

Boiling points are raised by hydrogen bonds because they make different molecules desire to "attach" to one another, which requires more energy to do so. In water, for instance, the hydrogen proton is in a state that resembles ionization because the connections between oxygen and hydrogen, while covalent, are strongly polar. The oxygen also receives a partial negative charge. Therefore, hydrogen bonds are formed when the electro-positive H in one molecule is strongly electrostatically attracted to the electro-negative O in nearby molecules. Despite being weak links, they are powerful enough to significantly alter the liquid's characteristics.  

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This question is health related and seeks to compute the details surrounding a traumatic head injury. See the explanation below.

Traumatic brain injury (TBI), a form of acquired brain injury, occurs when a sudden trauma causes damage to the brain.

TBI can result when the head suddenly and violently hits an object, or when an object pierces the skull and enters brain tissue

Final velocity before hitting the ground

V² = 2g*0.56

V² = 2 * 9.8 * 0.56

V = √10.976

V = 3.312 m/s

*** Note that g which is the acceleration of gravity is about 9.8 meters.

1a) Where there is a hardwood floor

-u^2 = -2as

a = 3.312^2/2*0.0019

a = 1,042.08 m/s^2

1b) Duration

t = u/a

= 3.312/1,042.08

t = 3.18 ms

2a) Where there is a carpet on the floor

a = 3.312^2/2*0.013

a = 713.01 m/s^2

t = u/a

=  3.312/713.01

t = 4.65 ms

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The  energy was lost to air friction is determined as 42.56 J.

The energy lost due to friction is calculated as follows;

ΔE = ΔK.E + ΔP.E

where;

ΔE = ¹/₂(0.45)(19.89² - 15.1²) + (0.45)(9.8)(12 - 30.2)

ΔE = 37.7 - 80.26

ΔE = -42.56 J

Thus, the  energy was lost to air friction is determined as 42.56 J.

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The statement that water boils at a temperature below 100 degrees in a pressure cooker is correct

The term freezing refers to the temperature at which a liquid is turned into a solid.

The following is the sorting of the statements into those that are correct and incorrect.

Water boils at a temperature below 100 degrees at high altitudes - correct

Water expands on freezing - correct

Water contracts on freezing - incorrect

Water boils at a temperature below 100 degrees in a pressure cooker - correct

Water boils at 100 degrees under normal pressure - correct

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