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What is the velocity at discharge if the nozzle of a hose measures 68 psi? 100.25 ft./sec 10.25 ft./sec 125.2 ft./sec 11.93 ft./sec

Answers

Answer 1

Answer:

Answer:

The velocity at discharge is 100.46 ft/s

Explanation:

Given that,

Pressure = 68 psi

We need to calculate the pressure in pascal

$P=68*6894.74\ Pa$

$P=468842.32\ Pa$

We need to calculate the velocity

Let the velocity is v.

Using Bernoulli equation

$P=(1)/(2)\rho v^2$

$468842.32=0.5*1000* v^2$

$v=\sqrt{(468842.32)/(0.5*1000)}$

$v=30.62\ m/s$

Now, We will convert m/s to ft/s

$v =30.62*3.281$

$v=100.46\ ft/s$

Hence, The velocity at discharge is 100.46 ft/s

Answer 2

Answer:

Final answer:

The speed of water discharged from a hose depends on the nozzle pressure and the constriction of the flow, but the specific speed cannot be determined from pressure alone without additional parameters.

Explanation:

The question is asking about the velocity or speed achieved by water when it is forced out of a hose with a nozzle pressure of 68 psi. To understand this, we need to know that the pressure within the hose is directly correlated with the speed of the water's exit. This is due to the constriction of the water flow by the nozzle, causing speed to increase.

However, the specific velocity at discharge can't be straightforwardly calculated from pressure alone without knowing more details, such as the dimensions of the hose and nozzle, and the properties of the fluid. Therefore, based on the provided information, a specific answer in ft/sec can't be given.

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wo parallel plates of area 100cm2are given charges of equal magnitudes 8.9 ×10−7C but opposite signs. The electric field within the dielectric material filling the space between the plates is 1.4 ×106V/m. (a) Calculate the dielectric constant of the material. (b) Determine the magnitude of the charge induced on each dielectric surface.
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A Porsche sports car can accelerate at 8.8 m/s^2. Determine its acceleration in km/h^2.

Answers

Answer:

The acceleration expressed in the new units is $114.048 Km/h^(2)$

Explanation:

To convert from $m/s^(2)$ to $Km/h^(2)$ it is necessary to remember that there are 1000 meters in 1 kilometer and 3600 seconds in 1 hour:

Then by means of a rule of three it is get:

$8.8(m)/(s^(2)).((1Km)/(1000m)).((3600s)/(1h))^(2)$

$8.8(m)/(s^(2)).((1Km)/(1000m)).((12960000s^(2))/(1h^(2)))$

Hence, the units of meters and seconds will cancel. Notice the importance of square the ratio 3600s/1h, so that way they can match with the other units:

$114.048 Km/h^2$

So the acceleration expressed in the new units is $114.048 Km/h^2$ .

A charge of uniform volume density (40 nC/m3) fills a cube with 8.0-cm edges. What is the total electric flux through the surface of this cube?

Answers

Answer:

The flux through the surface of the cube is $2.314\ Nm^(2)/C$

Solution:

As per the question:

Edge of the cube, a = 8.0 cm = $8.0* 10^(- 2)\ m$

Volume Charge density, $\rho_(v) = 40 nC/m^(3) = 40* {- 9}\ C/m^(3)$

Now,

To calculate the electric flux:

$\phi = (q)/(\epsilon_(o))$ (1)

where

$\phi$ = electric flux

$\epsilon_(o) = 8.85* 10^(- 12)\ F/m$ = permittivity of free space

Volume Charge density for the given case is given by the formula:

$\rho_(v) = (Total\ charge, q)/(Volume of cube, V)$ (2)

Volume of cube, $V = a^(3)$

Thus

$V = (8.0* 10^(- 2))^(3) = 5.12* 10^(- 4)\ m^(3)$

Thus from eqn (2), the total charge is given by:

$q = \rho_(v)V = 40* {- 9}* 5.12* 10^(- 4)$

$q = 2.048* 10^(-11)\ F = 20.48\ pF$

Now, substitute the value of 'q' in eqn (1):

$\phi = (2.048* 10^(-11))/(8.85* 10^(- 12)) = 2.314\ Nm^(2)/C$

What is the most important safety rule to remember during lab activities

Answers

To follow instructions

A twirlers baton is 0.76 m long and spins around its center. The end of the baton has a centripetal acceleration of 47.8 m/s2?A 0.31 s
B 0.56 s
C 4.3s
D 70s

Answers

B 0.56 s is the time period of a twirlers baton.

What is Centripetal Acceleration?

Centripetal acceleration is defined as the property of the motion of an object which traversing a circular path.

Any object that is moving in a circle and has an acceleration vector pointed towards the center of that circle is known as Centripetal acceleration.

The centripetal acceleration is given by:

a = 4π²R/T²

Given values are:

a = 47.8 m/s²

D = 0.76 m so , R = 0.76/2 = 0.38m

Using this formula,

47.8*T² = 4π² x0.38

T² = $(4*3.14^2*0.38)/(47.8)$

T = 0.56 s

Therefore,

A twirlers baton is 0.76 m long and spins around its center. The end of the baton has a centripetal acceleration of 47.8 m/s2 which have time period of 0.56 s.

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Answer:

C. 4.3 seconds

Explanation:

Dave rows a boat across a river at 4.0 m/s. the river flows at 6.0 m/s and is 360 m across.a. in what direction, relative to the shore, does dave’s boat go?
b. how long does it take dave to cross the river?
c. how far downstream is dave’s landing point?
d. how long would it take dave to cross the river if there were no current?

Answers

a) Let's call x the direction parallel to the river and y the direction perpendicular to the river.

Dave's velocity of 4.0 m/s corresponds to the velocity along y (across the river), while 6.0 m/s corresponds to the velocity of the boat along x. Therefore, the drection of Dave's boat is given by:

$\theta= arctan((v_y)/(v_x))=arctan((4.0 m/s)/(6.0 m/s))=arctan(0.67)=33.7^(\circ)$

relative to the direction of the river.

b) The distance Dave has to travel it S=360 m, along the y direction. Since the velocity along y is constant (4.0 m/s), this is a uniform motion, so the time taken to cross the river is given by

$t=(S_y)/(v_y)=(360 m)/(4.0 m/s)=90 s$

c) The boat takes 90 s in total to cross the river. The displacement along the y-direction, during this time, is 360 m. The displacement along the x-direction is

$S_x = v_x t =(6.0 m/s)(90 s)=540 m$

so, Dave's landing point is 540 m downstream.

d) If there were no current, Dave would still take 90 seconds to cross the river, because its velocity on the y-axis (4.0 m/s) does not change, so the problem would be solved exactly as done at point b).

You are given a parallel plate capacitor with plates having a rectangular area of 16.4 cm2 and a separation of 2.2 mm. The space between the plates is filled with a material having a dielectric constant κ = 2.0.Find the capacitance of this system

Answers

Answer: C = 1.319×10^-11 F

Explanation: The formulae that relates the capacitance of a capacitor to the area of the plates, distance between the plates and dielectric constant is given as

C = kε0A/d

Where C = capacitance of plates =?

k = dielectric constant = 2.0

Area of plates = 16.4cm² = 0.00164 m²

d = distance between plates = 2.2 mm = 0.0022m

By substituting the parameters, we have that

C = 2 × 8.85×10 ^-12 ×0.00164/ 0.0022

C = 0.029028 × 10^-12/ 0.0022

C = 13.19× 10^-12

C = 1.319×10^-11 F

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