PHYSICS COLLEGE

The pressure in a section of horizontal pipe with a diameter of 2.5 cm is 139 kPa. Water ï¬ows through the pipe at 2.9 L/s. If the pressure at a certain point is to be reduced to 101 kPa by constricting a section of the pipe, what should be the diameter of the constricted section? The acceleration of gravity is 9.81 m/s2 . Assume laminar nonviscous ï¬ow.

Answers

Answer 1

Answer:

Answer:

d = 2*0.87 = 1.75 cm

Explanation:

by using flow rate equation to determine the speed in larger pipe

$\phi =\pi r^2 v$

$v = (\phi)/(\pi r^2)$

$= (2900 cm^3/s)/(3.14(1.25cm)^2)$

= 591.10 cm/s

= 5.91 m/s

by Bernoulli's EQUATION

$p1 +(1)/(2) \rho v1^2 = p2 +(1)/(2) \rho v2^2$

$139000+ (1)/(2)*1000*5.91^2 = 101000 +(1)/(2)*1000* v2^2$

solving for v2

v2 = 10.53 m/s

diameter can be determine by using flow rate equation

$q = v \pi r^2$

$r^2 = (q)/(\pi v)$

$= (2900)/(3.14*1053)$

r = 0.87 cm

d = 2*0.87 = 1.75 cm

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A battery lighting a bulb is an example of _ energy converting to _ energy.

Answers

the power source is electrical ( whether the light is plugged in or has a battery)

the light bulb converts the electricity to light and heat.

in a fluorescent bulb, it is different, but the electricity is again converted to light, very little heat, though.

Electric energy converting into light and heat energy.

So, for the first blank electric, and the second blank the better answer is light.

A disgruntled autoworker pushes a small foreign import offacliff with a height of h. the vehicle lands a distance away
fromthe cliff. Determine how fast the vehicle was pushed off
thecliff.

Answers

Answer:

v = a/√(2h/g) m/s

Explanation:

Lets say the distance away from the cliff is a.

then, a = v t

where v is velocity with which it was thrown and t is time taken to fall.

Using equations of motion, we can also say that

h=1/2gt^2

where h is the height of the cliff

Thus, t^2 = 2h/g and t = √(2h/g)

Thus, v = a/√(2h/g).

the vehicle was pushed off the cliff with the velocity , v = a/√(2h/g). m/s

You are standing on the edge of a ravine that is 17.5 m wide. You notice a cave on the opposite wall whose ceiling is 7.3 m below your feet. The cave is 4.2 m deep, and has a vertical back wall. You decide to kick a rock across the ravine into the cave. What initial horizontal velocity must you give the rock so that the rock barely misses the overhang? The acceleration of gravity is 9.8 m/s 2 .

Answers

Answer:

v₀ₓ = 14.34 m / s

Explanation:

We can solve this problem using the projectile launch equations.

Let's look for the time it takes to descend to the height of the cave

y = $v_(oy)$ t - ½ g t²

As it rises horizontally the initial vertical speed is zero

y = 0 - ½ gt²

t = √2 y / g

t = √2 7.3 / 9-8

t = 1.22 s

This is the same time to cross the ravine

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v₀ₓ = x / t

v₀ₓ = 17.5 / 1.22

v₀ₓ = 14.34 m / s

This is the minimum speed.

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My opinion is Hinata...just saying

Answers

Answer:

hinata for sure

Explanation:

seems reasonable

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