5.0 L/s water flows through a horizontal pipe that narrows smoothly from 10.0 cm diameter to 5.0 cm diameter. A pressure gauge in the narrow section reads 50 kPa. What is the reading of the pressure gauge in the wide section

Answer:

The variables that can be measured in joules are

B. Energy

D. Work

Hope it will help :)

Answer:

0.546 ohm / μm

Explanation:

Given that :

N = 1.015 * 10^17

Electron mobility, u = 3900

Hole mobility, h = 1900

Ng = 4.42 x10^22

q = 1.6*10^-19

Resistivity = 1/qNu

Resistivsity (R) = 1/(1.6*10^-19 * 1.015 * 10^17 * 3900)

= 0.01578880889 ohm /cm

Resistivity of germanium :

R = 1 / 2q * sqrt(Ng) * sqrt(u*h)

R = 1 / 2 * 1.6*10^-19 * sqrt(4.42 x10^22) * sqrt(3900*1900)

R = 1 /0.0001831

R = 5461.4964 ohm /cm

5461.4964 / 10000

0.546 ohm / μm

Answer:

There is net loss of gravitational energy .

Explanation:

When Xanaxa is on the ground , her potential energy is assumed to be zero . When she leaps to a height of 153 m , she gains gravitational energy . When she dives and reaches the surface , she loses potential energy and on reaching the ground her potential energy becomes zero . When she further goes down inside ground to a depth of 17.5 m , she loses potential energy further . Her potential energy becomes less than zero or negative .

Ultimately her potential energy changes from zero to negative in the whole process . So there is net loss of potential energy .

Answer:150M

Explanation:

Answer:

the work done by the field is positive and the potential energy of the electron field system decreases

Explanation:

This exercise asks to find the work and the potential energy of an electron in an electric field.

Work is defined by

         W = F .d = F d cos θ

the electric force is

          F_e = q E

         W = q E d cos θ

since the charge of the electron is negative the force is in the opposite direction to the electric field

          W = - e E d

we select the direction to the right is positive, point i is to the left of point f,

therefore the work moving from point i to point F has two possibilities

* The electric field lines go from i to f point , so that point i is on the side of the positive charges, so the electron approaches them, This movement is opposite to that indicated

* the field line reaches point i, this implies that the charges are negative, so the electrioc field is then negativeand the electron charge is negative too.  The electron moves away from this point, this is in accordance with the indicated movement

In the latter case the electric field lines go from f to i point, therefore the Work is positive

Now let's examine the potential energy

            ΔU = - q E .d

so we see that this definition is related to work,

            ΔU = -W

Therefore, as the work is positive, the power energy must decrease

When reviewing the different answers, the correct ones are:

the work done by the field is positive and the potential energy of the electron field system decreases

The work done by the electron while moving from point [tex]i[/tex] to point [tex]f[/tex] in the direction of uniform electric field is negative and the potential energy of the electron increases.

An electron moves from point i to point f, in the direction of a uniform electric field, then  the potential energy of the electron can be calculated s given below.

[tex]\Delta V=-qEd[/tex]

Where [tex]\Delta V[/tex] is the potential energy, [tex]E[/tex] is the electric field, [tex]q[/tex] is the charge and [tex]d[/tex] is the displacement of the electron.

The work done by the electron in the uniform electric field can be calculated as,

[tex]W = F\times d \times cos\theta[/tex]

Where [tex]W[/tex]is the work done by electron, [tex]F[/tex] is the electric force, [tex]d[/tex] is the displacement of the electron and  for uniform electric field, the value of [tex]\theta[/tex] is zero.

Hence  [tex]W=F\times d\times 1\\W=F \times d[/tex]

Electric force  [tex]F = q E[/tex]

By substituting the value of electric force on the above formula,

[tex]W = qEd[/tex]

Hence, the relation between the work done the electron in an uniform electric field and potential energy of the electron can be given below.

[tex]W = -\Delta V[/tex]

The work done by the electron is negative and the potential energy of the electron increases.

For more information, follow the link given below.

https://brainly.com/question/8666051

Time taken : 4.77 s

Given

a = 13 m/s²

vf = final velocity = 62 m/s

Required

time taken

Solution

An equation of uniformly accelerated motion  

[tex]\large {\boxed {\bold {x=xo+vo.t+\frac {1} {2} at ^ 2}}}[/tex]

vf = vo + at  

vf² = vo² + 2a (x-xo)  

x = distance on t  

vo / vi = initial speed  

vt / vf = speed on t / final speed  

a = acceleration  

Input the value :

vo =0 ⇒from rest

vf=at

62 m/s = 13 m/s².t

t = 62 : 13

t = 4.77 s

Answer - B. Betelguese.

I really hope this helps!!

Answer:

WE NEED TO ADD ALL 40+2.38 +75+5

Explanation:

PLSE GIVE SOME POINTS DUDE

Answer:

6 kg

Explanation:

F=ma

F is Force(newtons)

m is mass(kg)

a is acceleration(m/s^2)

Plug in the numbers

3000 = m(500)

divide both sides by 500 to cancel out the 500.

3000/500=6

6 = m

6kg is the mass

Answer:The Moon has a smaller mass than the Earth. If you were to travel to the moon your weight would..

Explanation: It would decrease.

The Moon has a smaller mass than the Earth. If you were to travel to the moon your weight would decrease because the acceleration due by gravity on the moon is less than the acceleration due to gravity on the earth, therefore the correct answer is option C.

It can be defined as the force by which a body attracts another body towards its center as the result of the gravitational pull of one body and another.

In comparison to the Earth, the Moon is less massive. Your weight would drop if you traveled to the moon because the acceleration caused by gravity there is lower than that caused by gravity here on Earth.

As a result of the less gravity on the moon, the weight would decrease.

Thus, the correct answer is option C.

To learn more about gravity from here, refer to the link;

brainly.com/question/4014727

#SPJ2

Answer:

heyy

Explanation:how r uuu

Answer:

a) v₀ₓ = 62.76 m / s, b)   θ₁ = 17.6º,   θ₂ = 67.0º

Explanation:

We can solve this exercise using the projectile launch ratios

a) Let's find the time it takes for the bullet to reach the water level

       y = y₀ + v_{oy} t - ½ g t²

when it reaches the water its height is zero y = 0, as the bullet is fired horizontally its initial vertical velocity is zero

       0 = y₀ + 0 - ½ g t²

       t =[tex]\sqrt{2y_o/g}[/tex]

       t = [tex]\sqrt{2 \ 7 /9.8}[/tex]          

       t = 1,195 s

now we can calculate the speed with the horizontal movement

        x = v₀ₓ t

        v₀ₓ = x / t

        v₀ₓ = 75.0 / 1.195

        v₀ₓ = 62.76 m / s

b) if the speed of the bullets is half of that found

         v₀ = 62.76 / 2 = 31.38 m / s

let's write the expressions for the distance

          x = v₀ cos θ t

          y = y₀ + v_{oy} sin θ t - ½ g t²

          t = [tex]\frac{x}{v_o \ cos \theta}[/tex]

we substitute

          [tex]0 = y_o + v_o sin \theta \ \frac{x}{v_o \cos \thetay} - 1/2 g \ (\frac{x}{v_o \ cos \theta})^2[/tex]

          [tex]0 = y_o + x tan \theta - \frac{1}{2} g \ \frac{x^2}{ v_o^2 \ cos^2 \theta}[/tex]    

let's use the identified trigonometry

          sec² θ = 1 + tan² θ

         sec θ = 1 / cos θ

we substitute

          [tex]0 = y_o + x tan \theta - \frac{g x^2}{2 v_o^2} ( 1 + tan^2 \theta)[/tex]

          [tex]\frac{g x^2}{2v_o^2} tan^2 \theta - x tan \theta + \frac{gx^2}{2v_o^2} - y_o = 0[/tex]

we change variable

         tan θ = H

         [tex]\frac{gx^2}{2 v_o^2 } H^2 - x H + \frac{gx^2}{2v_o^2}-y_o =0[/tex]

we subtitle the values

         [tex]\frac{9.8 \ 75^2}{2 \ 31.38^2} H^2 - 75 H + \frac{9.8 \ 75^2}{2 \ 31.38^2}-7 =0[/tex]

         27.99 H² - 75 H + 20.99 = 0

         H² - 2.679 H + 0.75 = 0

we solve the quadratic equation

         H = [2.679 ± [tex]\sqrt{2.679^2 - 4 0.75}[/tex]] / 2

         H = [2,679 ± 2,044] / 2

         H₁ = 0.3175

         H₂ = 2.3615

now we can find the angles

          H₁ = tan θ₁

          θ₁ = tan⁻¹ H₁

          θ₁ = tan⁻¹ 0.3175

          θ₁ = 17.6º

          θ₂ = 67.0º

for these two angles the bullet hits the boat

Answer:

(A)

A cup placed on a table won't float away

Answer:

m = 4.11 grams

Explanation:

Given that,

The specific heat of Iron, c = 0.45 J/g°C

Heat removed, [tex]Q=1.16\times 10^2\ J[/tex]

Initial temperature, [tex]T_i=89^{\circ} C[/tex]

Final temperature, [tex]T_f=26.41^{\circ} C[/tex]

We need to find the mass of the iron. We know that the heat removed in terms of specific heat is given by :

[tex]Q=mc\Delta T\\\\m=\dfrac{Q}{c\Delta T}\\\\m=\dfrac{-1.16\times 10^2}{0.45\times (26.41-89)}\\\\m=4.11\ g[/tex]

So, the mass of the iron is 4.11 grams.

Answer:

A concave lens is exactly the opposite with the outer surfaces curving inward, so it makes parallel light rays curve outward or diverge. That's why concave lenses are sometimes called diverging lenses.

Answer: A concave lens

A concave lens is exactly the opposite with the outer surfaces curving inward, so it makes parallel light rays curve outward or diverge. That's why concave lenses are sometimes called diverging lenses.

Hopes this helps :)

Answer:

I answered Number 4 (Solids and Elasticity)

Explanation:

solids and elasticity

Answer:

PRECAUTIONS

-The refracting faces of the glass prism should be smooth, transparent and without any air bubble or broken edge. ...

-Use a sharp pencil to draw boundary of the prism and rays of light.

-The alpins should have sharp tip and should be fixed exactly vertical to the plane of the paper.

Explanation:

Please give thanks to all my answers and please mark as brilliant and please follow me

Answer:

  λ = 440 nm

Explanation:

The phenomenon of constructive interference is described by the expressions

          d sin θ= m λ                    (1)

where d is the separation of the slits d = 0.0250 mm = 2.50 10⁻⁵ m, lam is the wavelength of the incident radiation and m is an integer indicating the order of interference

let's use trigonometry to find the angle

          tan θ = y / L

where L is the distance to the screen L = 1.25 m

in general interference experiments angles are very small

         tan θ = [tex]\frac{sin \ \theta }{cos \ \theta}[/tex]

         ten θ = sin θ

substituting

         sin θ = y / L

we substitute 1

          d y / L = m λ

          λ = [tex]\frac{ d \ y }{m \ L}[/tex]

in the exercise indicate

        m = 3

         y = 6.6 cm = 6.6 10-2 m

we calculate

          λ = 2.50 10⁻⁵ 6.6 10⁻² /( 3   1.25)

          λ = 4.4 10⁻⁷ m

let's reduce to nm

          λ = 4.4 10⁻⁷ m (10⁹ nm / 1 m)

          λ = 440 nm

Answer:

a = 30.832 ft/s²

Explanation:

To solve this problem let's start by finding the braking acceleration using kinematics, where the distance is x = 408 ft, the initial velocity vo = 100 mi / h and the final velocity is zero v = 0

           v² = v₀² - 2 a x

           0 = v₀² - 2ax

           a = [tex]\frac{v_o^2}{2x}[/tex]

Let's start by reducing the magnitudes to ft / s              

            v₀ = 100 mi / h (5280 foot / 1 mile) (1h / 3600 s) = 146.666 ft / s

let's calculate

         a = [tex]\frac{146.66^2}{2 \ 408}[/tex]

         a = 26.36 ft / s²

Let's call this acceleration a_effective, this acceleration is in the opposite direction to the speed of the vehicle.  

Let's use a rule of three (direct proportions) to find the acceleration applied by the brake system (a1) which has an efficiency of 95%. or 0.95

                 a₁ = [tex]\frac{a_e}{0.95}[/tex]

Let's use another direct proportion rule If the acceleration of the brake system (a₁) for an applied acceleration (a) with an efficiency of 0.90

           a = [tex]\frac{a_1}{0.90}[/tex]

we substitute

           a = [tex]\frac{a_e}{0.95 \ 0.90}[/tex]

let's calculate

           a = [tex]\frac{26.36}{ 0.95 \ 0.90}[/tex]

           a = 30.832 ft/s²

This is the maximum relationship that the vehicle can have for when it brakes to stop at the given distance

Answer:

first is Atoms

4) is True

Answer:

Only the car transforms electrical energy into more than one form of energy.

Explanation:

The motion of the car is mechanical energy but it can also transform into another energy witch is electrical energy

Answer:

4363.3231 feets²

Explanation:

Given that :

Distance, r = 50 ft

θ = 200°

The arc length of area covered :

Arc length = θ/360° * πr²

Arc length = (200/360) * 50 ft ^2 * π

Arc length = 0.5555555 * 2500 * π

Arc length = 4363.3231 feets²

Answer:

Almost all machines require energy to offset the effects of gravity, friction, and air/wind resistance. Thus, no machine can continually operate at 100 percent efficiency.

Answer:

38.09 m

Explanation:

We'll begin by calculating the distance travelled by the car during the reaction time. This can be obtained as follow:

Reaction time (tᵣ) = 0.404 s

Initial velocity (u) = 22.8 m/s,

Distance travelled during the reaction time (sᵣ) =?

sᵣ = utᵣ

sᵣ = 22.8 × 0.404

sᵣ = 9.21 m

Next, we shall determine the distance travelled by the car when the brake was applied. This can be obtained as follow:

Initial velocity (u) = 22.8 m/s

Acceleration (a) = –9 m/s² (since the car is decelerating)

Final velocity (v) = 0 m/s

Distance travelled when the brake was applied (s₆) =?

v² = u² + 2as₆

0² = 22.8² + (2 × –9 × s₆)

0 = 519.84 – 18s₆

Collect like terms

0 – 519.84 = –18s₆

–519.84 = –18s₆

Divide both side by –18

s₆ = –519.84 / –18

s₆ = 28.88 m

Finally, we shall determine the stopping distance of the car, as measured from the point where the driver first notices the red light. This can be obtained as follow:

Distance travelled during the reaction time (sᵣ) = 9.21 m

Distance travelled when the brake was applied (s₆) = 28.88 m

Stopping distance =?

Stopping distance = sᵣ + s₆

Stopping distance = 9.21 + 28.88

Stopping distance = 38.09 m

Answer:

Its nymber 2

Explanation: