Bailey wants to find out which frozen solid melts the fastest: soda, ice, or orange juice. She pours each of the three liquids into the empty cubes of an ice tray, and then places the ice tray in the freezer overnight. The next day, she pulls the ice tray out and sets each cube on its own plate. She then waits and watches for them to melt. When the last part of the frozen liquid melts, she records the time.

Answer:

The distance covered by the cheetah during the motion is 77.75 m

Explanation:

Given;

initial velocity of cheetah, u = 0

final velocity of cheetah, v = 25 m/s

time of acceleration, t = 6.22 s

Apply kinematic equation;

[tex]s = (\frac{v+u}{2} )t[/tex]

where;

s is the distance covered by the cheetah during the motion

[tex]s = (\frac{v+u}{2} )t\\\\s = (\frac{25+0}{2} )6.22\\\\s = 77.75 \ m[/tex]

Therefore, the distance covered by the cheetah during the motion is 77.75 m

Answer:

  Q₁ = 2.4 10⁻⁴ C

Explanation:

We have a circuit with several capacitors, let's find the equivalent capacitor of the parallel

    [tex]C_{eq1}[/tex] = C₂ + C₃

    C_{eq1} = (10 +30) 10⁻⁶

    C_{eq1} = 40 10⁻⁶ F

There remains a series system between C₁ and C_{eq1}, let's find the equivalent capacitor

     1/C_{eq2} = 1 / C₁ + 1 / C_{eq1}

     1 /C_{eq2} = 1 / 20 10⁻⁶ + 1/40 10⁻⁶

     1 / C_{eq2} = 0.075 10⁶

     C_{eq2} = 13.33 10⁻⁶ F

let's use the relationship

        V = Q / C_{eq2}

        Q = V C_{eq2}

        Q = 18  13.33 10⁻⁶

        Q = 2.4 10⁻⁴ C

In a combination of capacitors in series the charge is constant, so the charge on C₁ is the same

        Q₁ = 2.4 10⁻⁴ C

Answer:

Answer:

43.4J

Explanation:

We know that

Work done = total heat energy

But work done is force x distance

=> F = 19 x8.6 = 163.4 J

So the total heat. Will be Heat of cube + heat of floor = 163.4J

So that heat of floor will now be

floor = 163.4 J - 120 J = 43.4 Joules

Explanation:

Answer:

Mass of flake = 67.716 kg

Explanation:

Given:

Length of lawn = 21 ft

Width of lawn = 20 ft

Note:

Each snow flake mass = 1.90 mg

Find:

Weight of total flake in a minute

Computation:

Area of lawn = 21 × 22

Area of lawn = 440 ft²

Amount of flake per minute = 440 × 1350

Amount of flake per minute = 594000 flake/ minutes

Mass of flake = 594000 × 1.90 mg × 60 minutes

Mass of flake = 67,716,000 mg

Mass of flake = 67.716 kg

Answer: 1798

Explanation:

Given that there is no dielectric material between the plates,

the permittivity of free space = 9 × 10^-12 f/m

The frequency F = 10 MHz

The ratio of conduction current JC to the displacement current Jd is also known as loss tangent.

Please find the attached file for the solution

Answer:

Explanation:

d )

power in an electrical loop = volt x current

= 5.067sin(140πt)  x 2.03sin(140πt)

= 10.286 sin²(140πt) .

10.286 ( 1 - cos 280πt ) / 2

= 5.143 ( 1 - cos 280πt )

e )

Torque on a current carrying loop in a magnetic field

= M B sinθ

M is magnetic moment of coil , B is magnetic field , θ is angle between area vector of coil and direction of  B .

Here magnetic moment of coil

= A i where A is area of coil and i is current .

= .12² x 2.03sin(140πt)

= .029 sin(140πt)

B = .8 T

MB = .8 x .029 sin(140πt)

= .0232  sin(140πt) x sin(140πt)

=.0232 sin²(140πt).

Answer:

Given that

V2/V1= 0.25

And we know that in adiabatic process

TV^န-1= constant

So

T1/T2=( V1 /V2)^ န-1

So = ( 1/0.25)^ 0.66= 2.5

Also PV^န= constant

So P1/P2= (V2/V1)^န

= (1/0.25)^1.66 = 9.98

A. RMS speed is

Vrms= √ 3RT/M

But this is also

Vrms 2/Vrms1= (√T2/T1)

Vrms2=√2.5= 1.6vrms1

B.

Lambda=V/4π√2πr²N

So

Lambda 2/lambda 1= V2/V1 = 0.25

So the mean free path can be inferred to be 0.25 times the first mean free path

C. Using

Eth= 3/2KT

So Eth2/Eth1= T2/T1

So

Eth2= 2.5Eth1

D.

Using CV= 3/2R

Cvf= Cvi

So molar specific heat constant does not change

The rms speed of the system is 1.6 while Mean free path is 0.25 times the first mean free path.

Given that  

[tex]\bold{\dfrac {V_2}{V_1}= 0.25}[/tex]

In adiabatic process  

[tex]\bold {TV^\gamma^-^1}[/tex]= constant  

So  

[tex]\bold {\dfrac {T1}{T2}=(\dfrac {V1 }{V2})^ \gamma^-^1}}\\[/tex]  

So,

[tex]\bold {\dfrac {T_1} {T_2} = (\dfrac { 1}{0.25})^ 0^.^6^6= 2.5}[/tex]  

for pressure,

[tex]\bold {PV^\gamma = \ Constant }[/tex]  

So

[tex]\bold {\dfrac {P1}{P2}=(\dfrac {V1 }{V2})^ \gamma}}\\\\\bold {\dfrac {P1}{P2}=(\dfrac {1}{0.25})^ 0^.^6^6} = 9.98 } }[/tex]  

A. rms speed can be calculated as

[tex]\bold {\dfrac {Vrms 2}{Vrms1}= \sqrt {T2T1})}\\\\\bold {Vrms2 =\sqrt {2.5} = 1.6\ Vrms1 }[/tex]

B.  The mean free path can be calculated as

[tex]\bold {\dfrac {\lambda_1 }{\lambda_2} = \dfrac {V_1}{V_2} = 0.25 \\ }[/tex]

Mean free path is 0.25 times the first mean free path.

C.

[tex]\bold {Eth= \dfrac {3}{2}kT}}\\\\\bold {\dfrac {E_t2}{E_t1} = \dfrac {T_2}{T_1}}\\\\\bold {E_t2= 2.5\ E_t1}[/tex]  

D.  the molar specific at a constant volume can be calculated a using,

[tex]\bold {CV= \dfrac 3{2}R }[/tex]  

 [tex]\bold {Cv_f= Cv_i }[/tex]

So, molar specific heat constant will not change.

Therefore, the rms speed of the system is 1.6 while Mean free path is 0.25 times the first mean free path.

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Answer: P = 4.86 × 10⁻²

Therefore, the particle's quantum number is 4.86 × 10⁻²

Explanation:

The expression of wave function for a particle in one dimensional box is given as;

φ(x) = ( √2/L ) sin ( nπx/L )

now we input our given figures, the limit of the particle to find it within the center of the box is

xₓ = L/2 + 20% of L/2

xₓ = L/2 + (0.2)L/2

xₓ = 3L/5

And the lower limit is,

x₁ = L/2 - 20% of L/2

x₁ = L/2 - (0.2) L/2

x₁ = 2L / 5

The expression for the probability of finding the particle within the center of the box is

P = ∫ˣˣₓ₁ ║φ(x)║² dx

P = ∫ ³L/⁵ ₂L/₅║(√2/L) sin ( nπx/L)║²dx

= 2/L ( ∫ ³L/⁵ ₂L/₅║sin ( nπx/L)║²dx

= 2/L ( ∫ ³L/⁵ ₂L/₅ (( 1 - cos ( 2πnx/L)/2) dx)

The particle is in its first excited state, then

n =2

Then calculate  the particle's quantum number as follows;

= 2/L ( ∫ ³L/⁵ ₂L/₅ (( 1 - cos ( 2π(2)x/L)/(2)) dx)

= 1/L ( ∫ ³L/⁵ ₂L/₅ (( 1 - cos ( 4πx/L)/2) dx)

= 1/L ( x - (L/4π)sin (4πx/L)) ³L/⁵ ₂L/₅

= 1/L ((3L/5) - (L/4π) sin (( 4π(3L/5)/L)) - (( 2L/5) - (L/4π)sin  ( 4π(2L/5)/L)))

= 1/L ( L/5 + L/4π (sin(8π/5) - sin ( 12π/5)))

Use the trigonometric formula to solve the above equation

sinA - sinB = 2sin ( A-B/2) cos (A+B/2)

Calculate the particle's quantum number as follows

P = 1/L ( L/5 + L/4π (sin(8π/5) - sin ( 12π/5)))

= 1/5 + 1/4π ( 2sin( (8π/5 -12π/5 ) / 2 ) cos ( (8π/5 + 12π/5) / 2 ))

= 1/5 + 1/2π ( -sin(2π/5) cos2π

= 1/5 - 1/2π ( sin (2π/5)(1))

= 0.0486 (10⁻²)(10²)

= 4.86 × 10⁻²

Therefore, the particle's quantum number is 4.86 × 10⁻²

Answer:

D. zero

Explanation:

For momentum of an isolated or closed system to be conserved (initial momentum must equal final momentum), the net external force acting on the system must be zero.

There is always external forces acting on a system, for this system’s momentum to remain constant, all the external forces acting on the system must cancel out, so that the net external force on the system is zero.

[tex]F_{ext} = 0[/tex]

Therefore, the correct option is "D"

D. zero

Answer:

18808.7 m/s^2

Explanation:

Given

Length of the pendulum L = 1.44 m

Number of complete cycles of oscillation n = 1.10 x 10^2

total time of oscillation t = 2.00 x 10^2 s

The period of the T = n/t

T = (1.10 x 10^2)/(2.00 x 10^2) = 0.55 ^-s

The period of a pendulum is gotten as

T = [tex]2\pi \sqrt{\frac{L}{g} }[/tex]

where g is the acceleration due to gravity

substituting values, we have

0.55 = [tex]2\pi \sqrt{\frac{1.44}{g} }[/tex]

0.0875 = [tex]\sqrt{\frac{1.44}{g} }[/tex]

squaring both sides of the equation, we have

7.656 x 10^-3 = 144/g

g = 144/(7.656 x 10^-3) = 18808.7 m/s^2

Answer:

The greater of the two currents is 0.692 A

Explanation:

Given;

distance between the two parallel wires; r = 6 mm = 6 x 10⁻³ m

let the current in the first wire = I₁

then, the current in the second wire = 2I₁

length of the wires, L = 3.0 m

magnitude of force on the wires, F = 8 μN = 8 x 10⁻⁶ N

The magnitude of force on the two parallel wires is given by;

[tex]F = \frac{\mu_o I_1(2I_1)}{2\pi r}\\\\F = \frac{\mu_o 2I_1^2}{2\pi r}\\\\I_1^2 = \frac{F*2\pi r}{2\mu_o} \\\\I_1^2 = \frac{8*10^{-6}*2\pi (6*10^{-3})}{2(4\pi*10^{-7})}\\\\I_1^2 = 0.12\\\\I_1 = \sqrt{0.12}\\\\ I_1 =0.346 \ A[/tex]

the current in the second wire = 2I₁ = 2 x 0.346 A = 0.692 A

Therefore, the greater of the two currents is 0.692 A

Answer:

The  current is [tex]I = 0.5425 \ A[/tex]

Explanation:

From the question we are told that

   The  magnetic field is  [tex]B = 1.2 \ T[/tex]

   The first length is  [tex]a = 8 \ cm = 0.08 \ m[/tex]

    The  second length is  [tex]a^* = 16 \ cm = 0.16 \ m[/tex]

    The first width is  [tex]b = 17 \ cm = 0.17 \ m[/tex]

     The second  width is  [tex]b^* = 22 \ cm = 0.22 \ m[/tex]

    The time interval  is  [tex]dt = 0.04 \ s[/tex]

     The resistance is  [tex]R = 1.2 \ \Omega[/tex]

Generally the first area is

     [tex]A = a * b[/tex]

=>    [tex]A = 0.08 * 0.17[/tex]

=>     [tex]A = 0.0136 \ m^2[/tex]

The second area is  

      [tex]A^* = a^* * b^*[/tex]

=>   [tex]A^* = 0.16 * 0.22[/tex]

=>     [tex]A^* = 0.0352 \ m^2[/tex]

Generally the induced emf is mathematically represented as

       [tex]\epsilon = - \frac{ B * [A^* - A]}{dt}[/tex]

This negative show that it is moving in the opposite direction of the motion producing it

=>   [tex]|\epsilon | = \frac{ 1.2 * [ 0.0352-0.0135]}{0.04}[/tex]

=>    [tex]|\epsilon | = 0.651 \ V[/tex]

The induced current is

     [tex]I = \frac{|\epsilon|}{R}[/tex]

=>   [tex]I = \frac{ 0.651}{1.2}[/tex]

=>   [tex]I = 0.5425 \ A[/tex]

Answer:

Explanation:

force constant of spring  k = force / extension

= 35.6 / 0.5

k = 71.2 N / m

angular frequency ω of oscillation by spring mass system

[tex]\omega = \sqrt{\frac{k}{m} }[/tex]

where m is mass of the body attached with spring

Putting the values

[tex]\omega = \sqrt{\frac{71.2}{5} }[/tex]

ω = 3.77 radian / s

The oscillation of the mass will be like SHM having amplitude of 0.5 m and angular frequency of 3.77 radian /s . Initial phase will be π / 2

so the equation for displacement from equilibrium position that is middle point can be given as follows

x = .5 sin ( ω t + π / 2 )

= 0.5 cos ω t

= 0.5 cos 3.77 t .

x = 0.5 cos 3.77 t .

Answer:

The speed must a ball be thrown vertically from ground level to rise to a maximum height is 28.35 m/s.

Explanation:

Given;

maximum vertical height of the throw, H = 41 m

Apply the following kinematic equation;

V² = U² + 2gH

where;

V is the final speed with which the ball will rise to a maximum height

U is the initial speed of the ball = 0

g is acceleration due to gravity = 0

V² = U² + 2gH

V² = 0² + 2gH

V² =  2gH

V = √2gH

V = √(2 x 9.8 x 41)

V = 28.35 m/s

Therefore, the speed must a ball be thrown vertically from ground level to rise to a maximum height is 28.35 m/s.

Answer:

Explanation:

Without wearing glasses , her near point is 20 cm .

for correction of eye

u = infinity ,

v = - 150 cm

f = ?

[tex]\frac{1}{v} -\frac{1}{u} = \frac{1}{f }[/tex]

[tex]\frac{1}{-150} -0 = \frac{1}{f }\\[/tex]

f = - 150 cm

He must be wearing glass of focal length of 150 cm .

If near point be x after wearing glass ,

u = x

v = - 20 cm

f = - 150 cm

[tex]\frac{1}{v} -\frac{1}{u} = \frac{1}{f }[/tex]

[tex]\frac{1}{-20} -\frac{1}{x} = \frac{1}{-150 }[/tex]

[tex]\frac{1}{-20} + \frac{1}{150 }= \frac{1}{x}[/tex]

x = 23 cm .

While wearing the glasses, Yang's near point will be 23.08 cm.

Given information:

Yang can focus on objects 150 cm away with a relaxed eye.

With full accommodation, she can focus on objects 20 cm away.

For correction, we have to use a concave lens such that it can make the image of a distant object at 150 cm.

So, the object distance will be infinity, and the image distance will be [tex]v=-150[/tex] cm.

So, the focal length of the lens can be calculated by lens formula as,

[tex]\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}\\\dfrac{1}{-150}-\dfrac{1}{\infty}=\dfrac{1}{f}\\f=-150\rm\;cm[/tex]

Now, after using the lens, the image distance will be [tex]v=-20[/tex] cm. Let u be the near point.

The near point, after correction, can be calculated as,

[tex]\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}\\\dfrac{1}{-20}-\dfrac{1}{u}=\dfrac{1}{-150}\\\dfrac{1}{u}=\dfrac{1}{150}-\dfrac{1}{20}\\u=23.08\rm\; cm[/tex]

Therefore, while wearing the glasses, Yang's near point will be 23.08 cm.

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Answer:

126 g of water, H2O.

Explanation:

First, we'll begin by calculating the number of atoms in 7 mole oxygen gas.

This is illustrated below:

From Avogadro's hypothesis, we understood that 1 mole of any substance contains 6.02×10²³ atoms. This means that 1 mole of O2 also contains 6.02×10²³ atoms.

Now, if 1 mole of O2 contains 6.02×10²³ atoms, then 7 moles of O2 will contain = 7 × 6.02×10²³ = 4.214×10²⁴ atoms.

Finally, we shall determine the mass of H2O that will contain 4.214×10²⁴ atoms.

This is illustrated below:

6.02×10²³ atoms is present in 1 mole of any substance contains

1 mole of H2O = (2x1) + 16 = 18 g

6.02×10²³ atoms is present in 18 g of H2O.

Therefore, 4.214×10²⁴ atoms will be present in = (4.214×10²⁴ × 18)/6.02×10²³

= 126 g of water, H2O.

Therefore, 126 g of water, H2O will contain the same number of oxygen atoms as 7.0 mol of oxygen gas.

7.0 mol of oxygen gas contain the same number of oxygen atoms as 250 g of water.

We want to know the mass of water (H₂O) that has the same number of oxygen atoms as 7.0 mol of oxygen gas (O₂).

First, we will calculate the number of oxygen atoms in 7.0 mol of O₂ considering the following relations.

[tex]7.0 mol O_2 \times \frac{6.02 \times 10^{23}molecule O_2 }{1molO_2} \times \frac{2atomO}{1molecule O_2} = 8.4 \times 10^{24} atomO[/tex]

Now, we want to calculate the mass of H₂O that contains 8.4 × 10²⁴ atoms of O. We will consider the following relations.

[tex]8.4 \times 10^{24} atomO \times \frac{1moleculeH_2O}{1atomO} \times \frac{1molH_2O}{6.02 \times 10^{23}moleculeH_2O } \times \frac{18.02gH_2O}{1molH_2O} = 250 gH_2O[/tex]

7.0 mol of oxygen gas contain the same number of oxygen atoms as 250 g of water.

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Answer:

50 m/s

Explanation:

[tex]Distance = 100m\\Time = 2 secs\\\\Velocity = \frac{Distance}{Time} \\\\V = \frac{100m}{2s} \\\\= 50m/s[/tex]

Answer:

The maximum angle does not depend on the mass

Explanation:

This is because In as much as the force acting on the normal which is the the maximum force of static friction increases as the mass increases, the component of the force of gravity parallel to the ramp increases at the same rate. Thus The maximum angle is independent of the mass.

Answer:

70 N

Explanation:

Draw a free body diagram of the boy.  There are four forces:

Weight force mg pulling down,

A 300 N normal force pushing up,

A 75 N applied force pulling right,

and a 5 N friction force pushing left.

The boy's acceleration in the y direction is 0, so the net force in the y direction is 0.

The net force in the x direction is 75 N − 5 N = 70 N.

Answer:

The velocity of any point on a rigid body is ___Always perpendicular______ to the relative position vector extending from the IC to the point

Explanation:

This is because The instantaneous center (IC) of zero velocity for the rigid body is at the point in contact with ground. The velocity direction at any point on the body is always perpendicular to the line connecting the point to the IC.

The velocity of any point on a rigid body is : Perpendicular   to the relative position vector extending from the IC to the point.

The IC ( instantaneous center ) is the point of the rigid body in contact with the ground when the rigid body is at zero velocity.

The velocity direction from any point of the rigid body will always be perpendicular to the ( IC ) since the IC is the reference point of the rigid body when the body is at rest ( zero velocity ).

Hence we can conclude that the velocity of any point on a rigid body will be  perpendicular.

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Although the options related to your question is missing an accurate answer is provided within the scope of your question

Answer:

[tex]\mathtt{Q_{sh} = 600.75 \ vars}[/tex]

Explanation:

Given that:

A circuit with a lagging 0.7 pf delivers 1500 watts and 2100VA

Here:

the initial power factor  i.e cos θ₁ = 0.7 lag

θ₁ = cos⁻¹ (0.7)

θ₁ = 45.573°

Active power P = 1500 watts

Apparent power S = 2100 VA

What amount of vars must be added to bring the pf to 0.85

i.e the required power factor here is cos θ₂ = 0.85 lag

θ₂ =  cos⁻¹   (0.85)

θ₂ = 31.788°

However; the initial reactive power [tex]Q_1[/tex] = P×tanθ₁

the initial reactive power [tex]Q_1[/tex] = 1500 × tan(45.573)

the initial reactive power [tex]Q_1[/tex] = 1500 × 1.0202

the initial reactive power [tex]Q_1[/tex] =  1530.3 vars

The amount of vars that must therefore be added to bring the pf to 0.85

can be calculated as:

[tex]Q_{sh} = P( tan \theta_1 - tan \theta_2)[/tex]

[tex]Q_{sh} = 1500( tan \ 45.573 - tan \ 31.788)[/tex]

[tex]Q_{sh} = 1500( 1.0202 - 0.6197)[/tex]

[tex]Q_{sh} = 1500( 0.4005)[/tex]

[tex]\mathtt{Q_{sh} = 600.75 \ vars}[/tex]

Answer:

B. 7.4 x 10⁻⁶ C, 0.6 x 10⁻⁶ C

Explanation:

From Coulomb's Law the electrostatic repulsive force is given by the following formula:

F = kq₁q₂/r²

where,

F = Repulsive Force = 0.15 N

k = Coulomb's Constant = 9 x 10⁹ N.m²/C²

q₁ = Magnitude of 1st Charge = ?

q₂ = Magnitude of 2nd Charge = ?

r = Distance between Charges = 0.5 m

Therefore,

0.15 N = (9 x 10⁹ N.m²/C²)q₁q₂/(0.5 m)²

q₁q₂ = (0.15 N)(0.5 m)²/(9 x 10⁹ N.m²/C²)

q₁q₂ = 4.17 x 10⁻¹²

q₁ = (4.17 x 10⁻¹²)/q₂   -------------------- equation (1)

The sum of charges is given as:

q₁ + q₂ = 8 μC

q₁ + q₂ = 8 x 10⁻⁶

using equation (1):

(4.17 x 10⁻¹²)/q₂ + q₂ = 8 x 10⁻⁶

(4.17 x 10⁻¹²) + q₂² = 8 x 10⁻⁶ q₂

q₂² - (8 x 10⁻⁶) q₂ + (4.17 x 10⁻¹²) = 0

Solving this quadratic equation:

q₂ = 7.4 x 10⁻⁶ C   (OR)   q₂ = 0.56 x 10⁻⁶ C

q₂ = 7.4 μC   (OR)   q₂ = 0.6 μC

Therefore,

q₁ = (4.17 x 10⁻¹² C)/(7.4 x 10⁻⁶ C)

q₁ = 0.6μC

Now, if we solve with q₂ = 0.6 μC, we will get q₁ = 7.4 μC.

Therefore, the correct option will be:

B. 7.4 x 10⁻⁶ C, 0.6 x 10⁻⁶ C

Answer:

11 mm

Explanation:

Original length L' = 26 m

Increase in temperature dT = 36.0°C

The gap l =?

The coefficient of linear expansion for steel & = 12 × 10^−6 per °C

The gap will be gotten from the equation of linear expansion.

l = L' x & x dT

substituting, we have

L = 26 x 12 × 10^−6 x 36

L = 0.011 m = 11 mm

Answer:

Posterior of the body

Explanation:

Gluteal region is located at the proximal end of the femur and posterior to the pelvic girdle.  The gluteal muscles help to move the lower limb at the hip joint. The gluteal region is divided into two groups: Deep lateral rotators and superficial abductors and extenders.  

The lumbar is the lower region of the spine commonly known as lower back, it has five vertebrates.  The lumbar contain tissue and nerves that control communication between legs and brain. In anatomical terms they are located inferior to the rib cage, at the bottom section of the vertebral column and superior to sacrum and pelvis.

Answer:

4km

Explanation:

d = st

d=0.25*16

d=4km

Answer:

1)  speed of a cyclist = 100 m/min.

2) speed of a car moving = 3 km/min.

Explanation:

1)  speed of a cyclist = distance over time

                                  = 2000 m / 20 min

                                  = 100 m/min.

2) speed of car moving = distance over time

                                       = 60 km / 20 min

                                        = 3 km/min.

Answer:

r = 4.747 cm and h = 9.4925 cm

Explanation:

We know that volume of a cylinder is given as:

V = πr²h

Also, surface area is given as;

S = 2πr² + 2πrh

Where r is radius and h is height

Now, we are told that the volume is 672 cm³

Thus, πr²h = 672

Making h the subject gives;

h = 672/πr²

Putting 672/πr² for h in the surface area equation gives;

S = 2πr² + 2πr(672/πr²)

Factorizing gives;

S = 2π[r² + 672/πr]

Differentiating to get first derivative gives;

S' = 2π[2r - (672/πr²)]

Equating to zero gives;

2π[2r - (672/πr²)] = 0

4πr - 1344/r² = 0

4πr = 1344/r²

r³ = 1344/4π

r³ = 106.95212175775

r = ∛106.95212175775

r = 4.747 cm

So, since h = 672/πr²

Then, h = 672/π(4.747)²

h = 9.4925 cm

Answer:

Hey mate ,

Area of rectangle = l×b

1.82×1.5

2.73cm2

Answer:

P = 198.118 kPa

Explanation:

Given:

Atmospheric pressure = P[tex]_{atm\\}[/tex] = 1.01×10⁵

depth = h = 9.91 m

To find:

Absolute pressure P[tex]_{abs}[/tex]

Solution:

Density of water = ρ = 1.000x10 ³kg/m ³

acceleration due to gravity = ρ = 9.8 m/s²

P[tex]_{abs}[/tex] = P[tex]_{atm\\}[/tex] + ρgh

      = 1.01×10⁵ + 1.000x10 ³x 9.8 x 9.91

      = 101000 + 1000(9.8)(9.91)

      = 101000 + 97118

      = 198118 Pa

      = 198.118 kPa

P[tex]_{abs}[/tex] = 198.118 kPa

The absolute pressure at a depth below the surface of this deep lake is 198.118 kPa.

Given the following data:

Scientific data:

To calculate the absolute pressure at a depth below the surface of a deep lake:

Mathematically, absolute pressure is given by this formula:

[tex]P_{abs} = P + \rho gh[/tex]

Substituting the given parameters into the formula, we have;

[tex]P_{abs} = 1.01 \times 10^5 + 1000 \times 9.8 \times 9.91 \\\\ P_{abs} = 101000+ 97118 \\\\[/tex]

Absolute pressure = 198118 Pa

Note: 1 kPa = 1000 Pa

Absolute pressure = 198.118 kPa

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