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3 years ago

How fast does water flow from a hole at the bottom of a very wide, 3.2 m deep storage tank filled with water

Physics

1 answer:

Nezavi [6.7K]3 years ago

5 0

Answer:

the velocity of the water flow is 7.92 m/s

Explanation:

The computation of the velocity of the water flow is as follows

Here we use the Bernouli equation

As we know that

$V_1 = \sqrt{2g(h_2 - h_1)}\\\\=\sqrt{2g(\Delta h)} \\\\= \sqrt{2\times9.8\times3.2} \\\\= \sqrt{62.72}$

= 7.92 m/s

Hence, the velocity of the water flow is 7.92 m/s

We simply applied the above formula so that the correct value could come

And, the same is to be considered

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A flywheel of diameter 1.2 m has a constant angular acceleration of 5.0 rad/s2. the tangential acceleration of a point on its ri

vodka [1.7K]

We know that tangential acceleration is related with radius and angular acceleration according the following equation:
at = r * aa
where at is tangential acceleration (in m/s2), r is radius (in m) aa is angular acceleration (in rad/s2)
So the radius is r = d/2 = 1.2/2 = 0.6 m
Then at = 0.6 * 5 = 3 m/s2
Tangential acceleration of a point on the flywheel rim is 3 m/s2

5 0

3 years ago

Practice Exercises Name: : Billy-Joe stands on the Talahatchee Bridge kicking stones into the water below a) If Billy-Joe kicks

babymother [125]

Answer:

a) The answer is 11,7m

b) The time it takes to fall will be shorter

Explanation:

We will use the next semi-parabolic movement equations

$H=Hi+Viy*t+1/2*g*t$

$X=Xi+Vx*t$

Where g(gravity acceleration)=9,81m/s^2

Also Xi, Hi and Viy are zero, as the stones Billy-Jones is kicking stay still before he moves them, so we take that point as the reference point

The first we must do is to find how much time the stones take to fall, this way:

$t=(5.40m)/(3.50m/s)$

Then t=1,54s

After that we need to replace t to find H, this way

$H=(1/2)*(9,81m/s^2)*(1,54s)^2$

Then H=11,7m

b) The stones will fall faster as the stones will be kicked harder, it will cause the stones move faster, it means, more horizontal velocity. In order to see it better we could assume the actual velocity is two times more than it is, so it will give us half of the time, this way:

$t=(5,40m)/(2*3,50m/s)$

Then, t=0,77

4 0

3 years ago

How and why does air pressure change with altitude in the atmosphere?

wlad13 [49]

Pressure with Height: pressure decreases with incrementing altitude. The pressure at any caliber in the atmosphere may be interpreted as the total weight of the air above a unit area at any elevation. At higher elevations, there are fewer air molecules above a given surface than a homogeneous surface at lower calibers.

5 0

3 years ago

Having difficulty finding the PE and KE for these values no mass is given. Does anyone know to go solve these?

Alexandra [31]

11) $1.04\cdot 10^7 J$

12) $1.04\cdot 10^7 J$

13) 50.0 m/s

14) 41.6 m/s

Explanation:

11)

The potential energy of an object is the energy possessed by the object due to its position relative to the ground. It is given by

$PE=mgh$

where

m is the mass of the object

g is the acceleration due to gravity

h is the height relative to the ground

Here in this problem, when the train is at the top, we have:

m = 8325 kg (mass of the train + riders)

$g=9.8 m/s^2$ (acceleration due to gravity)

h = 127 m (height of the train at the top)

Substituting,

$PE=(8325)(9.8)(127)=1.04\cdot 10^7 J$

12)

According to the law of conservation of energy, the total mechanical energy of the train must be conserved (in absence of friction). So we can write:

$KE_t + PE_t = KE_b + PE_b$

where

$KE_t$ is the kinetic energy at the top

$PE_t$ is the potential energy at the top

$KE_b$ is the kinetic energy at the bottom

$PE_b$ is the potential energy at the bottom

The kinetic energy is the energy due to motion; since the train is at rest at the top, we have

$KE_t=0$

Also, at the bottom the height is zero, so the potential energy is zero

$PE_b=0$

Therefore, we find:

$KE_b=PE_t=1.04\cdot 10^7 J$

13)

The kinetic energy of an object is the energy of the object due to its motion. Mathematically, it is given by

$KE=\frac{1}{2}mv^2$

where

m is the mass of the object

v is the speed of the object

From question 12), we know that the kinetic energy of the train at the bottom is

$KE=1.04\cdot 10^7 J$

We also know that the mass is

m = 8325 kg

Therefore, we can calculate the speed of the train at the bottom:

$v=\sqrt{\frac{2KE}{m}}=\sqrt{\frac{2(1.04\cdot 10^7)}{8325}}=50.0 m/s$

14)

At the top of the second hill, the total mechanical energy of the train is still conserved.

Therefore, we can write again:

$KE_1 + PE_1 = KE_2 + PE_2$

where

$KE_1$ is the kinetic energy at the top of the 1st hill

$PE_1$ is the potential energy at the top of the 1st hill

$KE_2$ is the kinetic energy at the top of the 2nd hill

$PE_2$ is the potential energy at the top of the 2nd hill

From the previous questions, we know that

$KE_1=0$

and

$PE_1=1.04\cdot 10^7 J$

The height of the second hill is

h = 39 m

So we can also find the potential energy at the second hill:

$PE_2=mgh=(8325)(9.8)(39)=3.2\cdot 10^6 J$

So, the kinetic energy at the second hill is

$KE_2=PE_1-PE_2=1.04\cdot 10^7 - 3.2\cdot 10^6 =7.2\cdot 10^6 J$

And so, the speed is

$v=\sqrt{\frac{2KE_2}{m}}=\sqrt{\frac{2(7.2\cdot 10^6)}{8325}}=41.6 m/s$

4 0

3 years ago

Science and math people where you at

PilotLPTM [1.2K]

For the second question you’re solving for resistance. resistance= voltage/ current. 120/0.5= 240. the answer is 240 ohms
for the third question you would do 2*4 since it’s asking for voltage, the answer is 8 volts :)

8 0

4 years ago