the acceleration of a car is 10m/s².if the starts from rest , what will be its speed after 10sec.​

Answer:

Convex lens and convex mirrors are similar that

1. They have the same image characteristics at various object positions

2. They possess a positive focal length

3. Both their ray lines converge to a particular focal point

Convex lens and concave mirror have certain similarities that are explained below.

The shape of concave mirror is spherical.

Convex lens is the combination of two concave mirrors.

When the ray lines are drawn at the convex lens, then the coverage of the rays leads parallel to its principal axis.

When the ray lines are drawn at the concave mirror, then the coverage of the rays leads parallel to its focal point.

The focal length of concave mirror and convex lens is positive.

Concave mirror and convex lens both project the real and inverted image of the object.

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Answer:

124 just subtract only one

Explanation:

Answer:

The wavelength of the waves created in the swimming pool is 0.4 m

Explanation:

Given;

frequency of the wave, f = 2 Hz

velocity of the wave, v = 0.8 m/s

The wavelength of the wave is given by;

λ = v / f

where;

λ is the wavelength

f is the frequency

v is the wavelength

λ = 0.8 / 2

λ = 0.4 m

Therefore, the wavelength of the waves created in the swimming pool is 0.4 m

Answer:

The maximum speed the ball can have is 15.37 m/s

Explanation:

Given;

mass of the ball, m = 0.4 kg

radius of the chord, r = 1.5 m

maximum tension on the chord, T = 63 N

The maximum tension on the chord is given by;

[tex]T_{max} = \frac{mv_{max}^2}{r} \\\\v_{max}^2 = \frac{T_{max} *r}{m}\\\\ v_{max} = \sqrt{\frac{T_{max} *r}{m}} \\\\ v_{max} =\sqrt{\frac{63 *1.5}{0.4}}\\\\v_{max} = 15.37 \ m/s[/tex]

Therefore, the maximum speed the ball can have is 15.37 m/s

Answer:

   v = 127.66 m / s      θ’= 337.59

Explanation:

For this exercise we must use the speed composition of the drone.

The first speed is 63 m / s in direction 29, three minutes later the speed reaches 22 m / s and direction 233, finally it returns to the launch point with 98 m / s in direction 321, in the attachment you can see a diagram of these speeds.

To find the resulting average velocity, the easiest thing is to decompose each velocity into the x and y coordinate system, then add each velocity

let's break down the speeds

             cos 29 = v₁ₓ / v₁

              sin 29 = [tex]v_{1y}[/tex] / v₁

             v₁ₓ = v₁ cos 29

             v_{1y} = v₁ sin 29

             v₁ₓ = 63 cos 29 = 55.10 m / s

             v_{1y} = 63 sin 29 = 30.54 m / s

speed 2

              cos 22 = v₂ₓ / v₂

              sin 22 = v_{2y} / v₂

               v₂ₓ = v₂ cos 233

               v_{2y} = v₂ sin 233

               v₂ₓ = 22 cos 233 = -13.24 m / s

                v_{2y} = 22 sin 233 = -17.57 m / s

speed 3

              cos 321 = v₃ₓ / v₃

              sin 321 = v_{3y} / v₃

               v₃ₓ = v₃ cos 321

               v_{1y} = vₐ sin 321

               v₃ₓ = 98 cos 321 = 76.16 m / s

               v_{3y} = 98 sin 321 = -61.67 m / s

We already have all the component of the speeds, the resulting speed is

               vₓ = v₁ₓ + v₂ₓ + v₃ₓ

               vₓ = 55.10 -13.24 +76.16

               vₓ = 118.02 m / s

               v_{y} = v_{1y} + v_{2y} + v_{3y}

                v_{y} = 30.54 -17.54 - 61.67

                v_{y} = -48.67 m / s

there are two ways to give the result

               v = (118.02 i -48.67 j) m / s

or in the form of magnitud and angle.

We use the Pythagorean theorem for the module

             v = √ (vₓ² + v_{y}²)

             v = RA (118.02² + 48.67²)

             v = 127.66 m / s

let's use trigonometry for the angle

             tan θ = v_{y} / vₓ

             θ = tan⁻¹ (v_{1} / vₓ)

             θ = tan⁻¹ (-48.67 / 118.02)

             θ = -22.41

if we want to measure the angles with respect to the positive side of the x axis

               θ’= 360 - 22.41

               θ’= 337.59

Answer:

velocity^2=force*length/mass

Explanation:

velocity^2=force*length/mass  is dimensionally consistent

as we know that

unit of force=Newton

unit of length=meter

unit of mass=kilogram so

velocity^2=newton*meter/kilogram

newton=kg*m/sec^2 therefore

v^2=kg*m*m/sec^2*kilogram

v^2=kg*m^2/sce^2*kg

v^2=m^2/sec^2

taking square root on both side

√ v²=√m²/sec²

v=m/s  now the dimensions of V is

v=(L/T)

so option A is dimensionally consistent

It should be related like  velocity^2=force*length/mass

Since we know that

unit of force=Newton

unit of length=meter

And,

unit of mass=kilogram

Therefore,

velocity^2=newton*meter/kilogram

Now

newton=kg*m/sec^2

So,

v^2=kg*m*m/sec^2*kilogram

v^2=kg*m^2/sce^2*kg

v^2=m^2/sec^2

Now we take square root on both side

So,

√ v²=√m²/sec²

v=m/s  

v=(L/T)

Therefore, option A is correct

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Answer:

b. henry must have stopped for one hour during his trip

This question is incomplete, the complete question is;

A student dropped a textbook from the top floor of his dorm and it fell according to the formula s(t) = -16t² + 8√t, where t is the time in seconds and s(t) is the distance in feet from the top of the building.

(a) Write a formula for the average velocity of the ball for t near 4.

(b) Find the average velocity for the time interval beginning when t = 4 with duration 1 seconds, 0.5 seconds, and 0.05 seconds

(c) What is your estimate for the instantaneous velocity of the ball at t = 4

Answer:

a)

Average velocity, (Vavg)  of the ball for t near 4.

Vavg = [s(4) - s(0)] / (4 - 0)

Where s(4) = -16 × 4² + 8 × √4= - 240 m

s(0) = -16 × 0 + 8 * 0 = 0

b)

duration = 1 sec

Vavg = [s(5) - s(4)] / (5 - 4)

s(5) = -16 × 52 + 8 × √5 = - 382 m

s(4) = -16 × 42 + 8  √4 = - 240 m

Vavg = (-382 - (-240)) / (5 - 4)

Vavg = - 142.1 m/s

duration = 0.5 sec

Vavg = [s(4.5) - s(4)] / (4.5 - 4)

s(4.5) = -16 × 4.52 + 8 × √4.5 = - 307 m

s(4) = -16 × 42 + 8 × √4 = - 240 m

Vavg = (-307 - (-240)) / (4.5 - 4)

Vavg= - 134.1 m/s

duration = 0.05 sec

Vavg = [s(4.05) - s(4)] / (4.05 - 4)

s(4.05) = -16 × 4.052 + 8 × √4.05 = - 246 m

s(4) = -16 × 42 + 8 × √4 = - 240 m

Vavg = (-246 - (-240)) / (4.05 - 4)

Vavg= - 126.8 m/s

c)

Instantaneous velocity, v = ds/dt

= - 16 × 2 × t + 8 ×× (0.5 / √t )

= - 32 × t + 4/√t

ds/dt at t = 4 is,

v = - 32 × 4 + 4 / √4

= - 126 m/s

Answer:

Weight

Explanation:

Weight is the downward pull on an object due to gravity.

For example, the moon has less gravity than Earth so we would weigh less on the moon. Our Mass and volume always stay the same but our weight could change.

Answer:

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Explanation:

Answer:

projectiles

electromagnetic

Answer:

Explanation:

física cuántica y  Quantum Moves

Answer:

Explanation:

The question lacks the appropriate diagram. Find the diagram attached below:

Let the mass of his friend be M.

Using the principle of moment which states that the sum of clockwise moment is equal to the sum of its anticlockwise moment to solve the problem,.

Since Moment = Force * perpendicular distance.

The smaller boy will move in the clockwise direction and his friend will move in the anti clockwise direction.

Clockwise moment = 40 * 4 = 160kgm

anticlockwise moment = M * 2 = 2Mkgm

Equating both moments to get the mass M of his friend

160 = 2M

Divide both sides by 2

2M/2 = 160/2

M = 80kg

Hence the mass of his friend that will keep the seesaw balanced horizontally is 80kg.

Answer:

Shakina can sayThe car moves initially to the right with decreasing speed. Eventually, the car stopsand moves to the left with increasing speed.

Explanation:

Because initially the acceleration is negative during which the car is moving to the right and then it slows down and stops, and once it begins moving to the left, it begins to accelerate faster

Answer:

[tex]\frac{kq^{2} }{2a^2}[/tex]

Explanation:

The magnitude on the charge at the bottom-left corner due to the charge on the top vertex of the triangle will act along the +ve x-axis and the +ve y-axis.

From Coulomb's law the magnitude of the forces on the charge at the bottom-left corner, due to the charge on the top vertex of the triangle are [tex]\frac{kq^{2} }{a^2}[/tex]cos 60, and

The magnitude on the charge due to the charge at the bottom-right corner will only act in the -ve x-axis, since they repel each other (like charges repel). The magnitude is  [tex]\frac{kq^{2} }{a^2}[/tex]

The angle made by the upper charge to the charge we're considering is 60° with the horizontal.

The total force on the charge along the x-axis is

[tex]F_{x}[/tex] = [tex]\frac{kq^{2} }{a^2}[/tex]cos 60 -

[tex]F_{x}[/tex]  = [tex]\frac{kq^{2} }{2a^2}[/tex] - [tex]\frac{kq^{2} }{a^2}[/tex]

==> -[tex]\frac{kq^{2} }{2a^2}[/tex]

For the y-axis, we have

[tex]F_{y}[/tex] = [tex]\frac{kq^{2} }{a}[/tex]sin 60

[tex]F_{y}[/tex] = [tex]\frac{\sqrt{3}* kq^{2} }{2a^2}[/tex]

The resultant force is

[tex]|F| = \sqrt{F_{x}^{2}+ F_{y}^2 }[/tex]

The common factors between the two x-axis force, and the y-axis force is

[tex]\frac{kq^{2} }{2a^2}[/tex], we put this outside the square root (squaring this and square rooting will give us the initial value)

[tex]|F|[/tex] = [tex]\frac{kq^{2} }{2a^2}[/tex][tex]\sqrt{(\frac{1}{2})^2 + (\frac{\sqrt{3} }{2})^2 }[/tex]

[tex]|F|[/tex] =  [tex]\frac{kq^{2} }{2a^2}[/tex][tex]\sqrt{\frac{1}{4} +\frac{3}{4} }[/tex]

==>  [tex]\frac{kq^{2} }{2a^2}[/tex][tex]\sqrt{1}[/tex]

the magnitude of the electric force on the charge at the bottom left-hand vertex of the triangle due to the other two charges is

[tex]|F|[/tex]  =  [tex]\frac{kq^{2} }{2a^2}[/tex]

Answer:

0.004394Joules

Explanation:

The question is incomplete. Find the complete question in the attachment below;

Energy stored in a capacitor = 1/2CV² where;

C is the capacitance of the capacitor

V is the potential difference across the plates

Given parameters

C = 13.0μF = 13 * 10⁻⁶F

V = 26.0V

Required

Energy stored in the capacitor

Substituting the values into the equation given;

E = 1/2 * 13 * 10⁻⁶ * 26²

E = 1/2 * 13 * 10⁻⁶ * 676

E = 1/2* 8788 * 10⁻⁶

E = 4394  * 10⁻⁶

Hence the Energy stored in the capacitor in Joules is 0.004394Joules

Answer:

the instantaneous velocity of P is greater that Q because P obviously passes Q at that instant in time t

2. P has greater velocity at that instant (to) because P will need to be at a greater velocity to pass Q

Answer:

The work must be negative or positive depending on direction of motion of the object

Explanation:

This is because

W= force x distance

And = F x S cosစ

And this is positive when

Theta is less than π/2 and negative when theta is less than or equal to π/2

If a force acts in a specific direction on a thing, the work it puts on it can be positive or negative, regardless of the direction the object moves.

A force is indeed an influence which can alter an angular velocity. An object can alter its velocity, or speed, as a result of a force. Naturally, force can be characterized as a push or just a pull.

If a force acts in a specific direction on a thing, the work it puts on it can be positive or negative, regardless of the direction the object moves.

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Answer:

Uses of concave mirror:

Shaving mirrors.

Head mirrors.

Ophthalmoscope.

Astronomical telescopes.

Headlights.

Solar furnaces.

Uses of convex mirror:

Convex mirrors always form images that are upright, virtual, and smaller than the actual object. They are commonly used as rear and side view mirrors in cars and as security mirrors in public buildings because they allow you to see a wider view than flat or concave mirrors.

please give me full points.

Answer:

(a) [tex]\vec F_{\parallel} = -\frac{102}{13}\,i-\frac{103}{13}\,j[/tex] , (b) [tex]\vec F_{\perp} = \frac{102}{13}\,i -\frac{68}{13}\,j[/tex], (c) [tex]W = -51[/tex]

Explanation:

The statement is incomplete:

The force on an object is [tex]\vec F = -17\,j[/tex]. For the vector [tex]\vec v = 2\,i +3\,j[/tex]. Find: (a) The component of [tex]\vec F[/tex] parallel to [tex]\vec v[/tex], (b) The component of [tex]\vec F[/tex] perpendicular to [tex]\vec v[/tex], and (c) The work [tex]W[/tex], done by force [tex]\vec F[/tex] through displacement [tex]\vec v[/tex].

(a) The component of [tex]\vec F[/tex] parallel to [tex]\vec v[/tex] is determined by the following expression:

[tex]\vec F_{\parallel} = (\vec F \bullet \hat {v} )\cdot \hat{v}[/tex]

Where [tex]\hat{v}[/tex] is the unit vector of [tex]\vec v[/tex], which is determined by the following expression:

[tex]\hat{v} = \frac{\vec v}{\|\vec v \|}[/tex]

Where [tex]\|\vec v\|[/tex] is the norm of [tex]\vec v[/tex], whose value can be found by Pythagorean Theorem.

Then, if [tex]\vec F = -17\,j[/tex] and [tex]\vec v = 2\,i +3\,j[/tex], then:

[tex]\|\vec v\| =\sqrt{2^{2}+3^{3}}[/tex]

[tex]\|\vec v\|=\sqrt{13}[/tex]

[tex]\hat{v} = \frac{1}{\sqrt{13}} \cdot(2\,i + 3\,j)[/tex]

[tex]\hat{v} = \frac{2}{\sqrt{13}}\,i+ \frac{3}{\sqrt{13}}\,j[/tex]

[tex]\vec F \bullet \hat{v} = (0)\cdot \left(\frac{2}{\sqrt{13}} \right)+(-17)\cdot \left(\frac{3}{\sqrt{13}} \right)[/tex]

[tex]\vec F \bullet \hat{v} = -\frac{51}{\sqrt{13}}[/tex]

[tex]\vec F_{\parallel} = \left(-\frac{51}{\sqrt{13}} \right)\cdot \left(\frac{2}{\sqrt{13}}\,i+\frac{3}{\sqrt{13}}\,j \right)[/tex]

[tex]\vec F_{\parallel} = -\frac{102}{13}\,i-\frac{153}{13}\,j[/tex]

(b) Parallel and perpendicular components are orthogonal to each other and the perpendicular component can be found by using the following vectorial subtraction:

[tex]\vec F_{\perp} = \vec F - \vec F_{\parallel}[/tex]

Given that [tex]\vec F = -17\,j[/tex] and [tex]\vec F_{\parallel} = -\frac{102}{13}\,i-\frac{153}{13}\,j[/tex], the component of [tex]\vec F[/tex] perpendicular to [tex]\vec v[/tex] is:

[tex]\vec F_{\perp} = -17\,j -\left(-\frac{102}{13}\,i-\frac{153}{13}\,j \right)[/tex]

[tex]\vec F_{\perp} = \frac{102}{13}\,i + \left(\frac{153}{13}-17 \right)\,j[/tex]

[tex]\vec F_{\perp} = \frac{102}{13}\,i -\frac{68}{13}\,j[/tex]

(c) The work done by  [tex]\vec F[/tex] through displacement [tex]\vec v[/tex] is:

[tex]W = \vec F \bullet \vec v[/tex]

[tex]W = (0)\cdot (2)+(-17)\cdot (3)[/tex]

[tex]W = -51[/tex]

Answer:

for every decrease in potential energy, a kinetic energy proportional to 2gΔh is gained.

Explanation:

Let us consider a ball falling from its maximum height

For a body falling from its maximum height to a point p

change in height = Δh

The potential energy decrease is then proportional to

ΔPE = mgΔh

where

ΔPE is the decrease in kinetic energy

m is the mass of the ball

g is acceleration due to gravity

Δh is the change in height

For a body falling from its maximum height, the increase change in velocity

Δv = u + 2gΔh    (at maximum height u = 0)

where

u is the initial kinetic energy of the ball

Δv = 0 + 2gΔh

Δv = 2gΔh

The kinetic energy increases by

ΔKE = [tex]\frac{1}{2}[/tex]m(Δv)^2

but Δv = 2gΔh

therefore

ΔKE = [tex]\frac{1}{2}[/tex]m(2gΔh)^2 = 2m(gΔh)^2

comparing the increase in kinetic energy to the decrease in potential energy, we have

(2m(gΔh)^2)/(mgΔh) = 2gΔh

This means that for every decrease in potential energy, a kinetic energy proportional to 2gΔh is gained.

Answer:

Explanation: Thermals are created by the irregular heating of Earth's surface from solar radiation, and are an example of convection, specifically atmospheric convection.

At this point, air is converging on low pressure and rising air forms clouds.

Answer:

8.25 V

Explanation:

We can ignore the 22Ω and 122Ω resistors at the bottom.  Since there's a short across those bottom nodes, any current will go through the short, and none through those two resistors.

The 2Ω resistor and the 44Ω resistor are in parallel.  The equivalent resistance is:

1 / (1 / (2Ω) + 1 / (44Ω)) = 1.913Ω

This resistance is in series with the 12Ω resistor.  The equivalent resistance is:

1.913Ω + 12Ω = 13.913Ω

This resistance is in parallel with the 24Ω resistor.  The equivalent resistance is:

1 / (1 / (13.913Ω) + 1 / (24Ω)) = 8.807Ω

Finally, this resistance is in series with the 4Ω resistor.  The equivalent resistance of the circuit is:

8.807Ω + 4Ω = 12.807Ω

The current through the battery is:

12 V / 12.807Ω = 0.937 A

The voltage drop across the 4Ω resistor is:

(0.937 A) (4Ω) = 3.75 V

So the voltage between the bottom nodes and the top nodes is:

12 V − 3.75 V = 8.25 V

Answer:

A. x = -㏑(1/2)/μ B. x = -㏑(I/I₀)/μ

Explanation:

A. Since the intensity I = I₀exp(-μx) where I₀ = intensity at x = 0.

When I = I₀/2,  

I = I₀exp(-μx)

I₀/2 = I₀exp(-μx)

dividing through by I₀, we have

1/2 = exp(-μx)

taking natural logarithm of both sides, we have

㏑(1/2) = ㏑[exp(-μx)]

㏑(1/2) = -μx

dividing both sides by -μ

x = -㏑(1/2)/μ

where x is the thickness of the material that would absorb half of the gamma rays

B. Since the intensity I = I₀exp(-μx) at thickness x, where I₀ = intensity at x = 0

I = I₀exp(-μx)

dividing through by I₀

I/I₀ = exp(-μx)

taking natural logarithm of both sides, we have

㏑(I/I₀) = ㏑exp(-μx)

㏑(I/I₀) = -μx

dividing both sides by -μ

x = -㏑(I/I₀)/μ

where x is the thickness of the material that would reduce the radiation intensity to a fraction of the initial intensity.

Answer:

The acceleration will be a

Explanation:

Using F = ma

But we are given that F2= 2F1 and

m2= 2m1

So

a2= F2/m2= F1/m1

and

F1/m1=a

To contrast inner and outer planets we will start with the climate of the planets and then move on to there lighting. To start the planets closet to the sun, mercury, venus, earth and mars, are all hot compared to the further one, jupiter, saturn, uranus, neptune. This distance also makes the farthe away planets darker than the ones closer. Now to compare all the planets vary from either gass or solid, rocky or icy. All of them spin around the sun and all have objects spinning around them, moons.

Answer: table

Explanation:

Answer:Table

Explanation:

A table is a way to represent data , but does not identify trends in the data

Answer:

Total speed of the insect as seen by a person standing on the side of the road is 22 m/s

Explanation:

There i a composition of velocities, since the air inside the bus is moving at the same speed as the bus, and the insect flying relative to it.

So for an observer on the side of the road, the insect is flying at a speed of :

20 m/s + 2 m/s = 22 m/s

Answer:

Explanation:

1) Acceleration is the change in velocity of a body with respect to time.

Acceleration a = vf - vi/t

vf is the final velocity

vi is the initial velocity.

t is the time taken

Since the body accelerates from rest, vi = 0m/s

a = 27-0/3

a = 27/3

a = 9m/s²

2) Given

u = 0m/s (accelerates from rest)

a = 190m/s²

t = 2.4seconds

v = ?

Using v = u+at

v = 0+190(2.4)

v = 190×2.4

v = 456m/s

The final velocity of the car is 456m/s

3) Given

u = 15m/s

a = 3.5m/s

t = 5seconds

Using the relationship

S = ut+1/2at²

S is the distance covered by the car.

S = 15(5)+1/2(3.5)×5²

S = 75+25×3.5/2

S = 75+43.75

S = 118.75m

4) Given

initial velocity u = 23m/s

Deceleration a = -0.25m/s²(negative acceleration)

Final velocity v = 0m/s

Using the relationship

V² = u²+2as

0² = 23²+2(-0.25)s

-23² = -0.5S

23² = 0.5S

S = 529/0.5

S = 1058m

The distance required for the train to stop is 1058m.

5) Given

initial velocity u = 12m/s

Final velocity v = 26m/s

time = 14sec

Acceleration a = v-u/t

a = 26-12/14

a= 14/14

a = 1m/s²

For Distance covered

v² = u²+2as

26² = 12²+2(1)S

676 = 144 +2S

2S = 676-144

2S = 532

S = 532/2

S = 266m

Distance that the car will cover is 266m

6) Given

Initial velocity u = 0m/s (person starts from rest)

acceleration a = 3.2m/s²

time t = 3.0s

To get the distance;

S = ut + 1/2at²

S = 0(3)+1/2(3.2)×3²

S = 0+1.6×9

S= 9×1.6

S = 14.4m

The distance that the person covered in 3.0s is 14.4m

7) initial velocity of train u = 12m/s

Distance covered S = 541m

Final velocity = 0m/s (on stopping)

acceleration a= ?

Acceleration will be negative since the train is coming to a stop (decelerating)

Using the formula v² = u²+2as

0² = 12² - 2a(541)

-12² = -1082a

144 = 1082a

a = 1082/144

a = 7.51m/s²

Hence the acceleration of a train in order for it to stop from 12 m/s in a distance of 541 m is 7.51m/s²

Answer:

option B is correct

Explanation:

according to law of conservation of momentum the total momentum in a closed system remains constant before and after collision

If the skaters are in a closed system the total momentum remains the same.

In a closed system the total momentum is always conserved.

By applying the principle of conservation of linear momentum, the total momentum before the push of skater A, will be equal to the total momentum after the push.

The equation is given as;

Initial momentum = final momentum

[tex]m_au_a + m_bu_b = m_av_a + m_bv_b[/tex]

where;

Thus, we can conclude that if the skaters are in a closed system the total momentum remains the same.

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Answer:

1200 lb-ft

Explanation:

Weight of diver W = 150 lb

length of board L = 8 ft  

This board is pivoted at one end therefore, the perpendicular distance from the pivot = 8 ft

Torque = weight x perpendicular distance from pivot

T = WL

Torque = 150 x 8 = 1200 lb-ft

Answer:

b. False

Explanation:

The Bernoulli's equation is given by;

[tex]P + \rho gh + \frac{1}{2} \rho v^2 = constat[/tex]

Divide through by the density of the fluid and acceleration due to gravity (ρg), to obtain energy per unit weight of the fluid known as head;

[tex]\frac{P}{\rho g} + h + \frac{v^2}{2g} = constant[/tex]

where;

[tex]\frac{P}{\rho g}[/tex] is pressure head

h is datum head (mechanical energy due to height of the fluid)

[tex]\frac{v^2}{2g}[/tex] is velocity head (mechanical energy due to velocity of the fluid)

Thus, the mechanical energy of a fluid per unit weight of the fluid, measured relative to the bottom of the channel is datum head.

The correct option is "b. False"