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3 years ago

can you suggest improvement that can be made towards the design of siphon so that the transfer of liquid is much higher.

Physics

1 answer:

ahrayia [7]3 years ago

3 0

Answer:

A siphon is a tube that makes use of the potential energy of fluid at an elevated level to transfer the fluid to a lower level, due to pressure differences between the inlet and the outlet points of the tube, such that the pressure at the outlet is higher than the pressure at the inlet

The pressure energy is converted into velocity (kinetic) energy, and therefore, in other to increase the flow rate through the tube of a siphon, with constant diameter, the level of the fluid in the container at the inlet (supply) of the siphon is raised higher than the level at the outlet receiving) container or the outlet point of the siphon tube

The larger the difference between the inlet and outlet levels, the faster the transfer of fluid by the siphon

Explanation:

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Consider a father pushing a child on a playground merry-go-round. The system has a moment of inertia of 84.4 kg.m^2. The father

Sophie [7]

Answer:

Explanation:

Given that:

the initial angular velocity $\omega_o = 0$

angular acceleration $\alpha$ = 4.44 rad/s²

Using the formula:

$\omega = \omega_o+ \alpha t$

Making t the subject of the formula:

$t= \dfrac{\omega- \omega_o}{ \alpha }$

where;

$\omega = 1.53 \ rad/s^2$

∴

$t= \dfrac{1.53-0}{4.44 }$

t = 0.345 s

Using the formula:

$\omega ^2 = \omega _o^2 + 2 \alpha \theta$

here;

$\theta$ = angular displacement

∴

$\theta = \dfrac{\omega^2 - \omega_o^2}{2 \alpha }$

$\theta = \dfrac{(1.53)^2 -0^2}{2 (4.44) }$

$\theta =0.264 \ rad$

Recall that:

2π rad = 1 revolution

Then;

0.264 rad = (x) revolution

$x = \dfrac{0.264 \times 1}{2 \pi}$

x = 0.042 revolutions

Here; force = 270 N

radius = 1.20 m

The torque = F * r

$\tau = 270 \times 1.20 \\ \\ \tau = 324 \ Nm$

However;

From the moment of inertia;

$Torque( \tau) = I \alpha \\ \\ Since( I \alpha) = 324 \ Nm. \\ \\ Then; \\ \\ \alpha= \dfrac{324}{I}$

given that;

I = 84.4 kg.m²

$\alpha= \dfrac{324}{84.4} \\ \\ \alpha=3.84 \ rad/s^2$

For re-tardation; $\alpha=-3.84 \ rad/s^2$

Using the equation

$t= \dfrac{\omega- \omega_o}{ \alpha }$

$t= \dfrac{0-1.53}{ -3.84 }$

$t= \dfrac{1.53}{ 3.84 }$

t = 0.398s

The required time it takes= 0.398s

5 0

2 years ago

When talking about the speeds of atoms in a gas you must refer to the average speed because

cluponka [151]

Answer:

An Atom's individual speed will change as it collides with other atoms, so we have to use an average.

Explanation:

In a gas a single atoms does an assortment of things during its time in the gas—sometimes it collides with an other atom gaining a lot of speed, sometimes losing a lot of speed in the collision, and sometimes just moving freely. Therefore: the motion of one individual atom is unpredictable, and it cannot be representative of all the the atoms in a gas, which is why we must average over all speeds of all atoms to find an average speed that allows us to calculate other quantities like temperature and pressure of the gas.

Hence, the second option <em>"an Atom's individual speed will change as it collides with other atoms, so we have to use an average" </em>stands correct.

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3 years ago

What are things thats causing climate change

ollegr [7]

Greenhouse gases, eg. carbon dioxide and methane.

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3 years ago

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What is the resultant displacement of the two vectors below?

andriy [413]

Answer:

C 3.6 cm, 56 degrees North of the East axis

Explanation:

The two vectors are perpendicular to each other, so we can find the magnitude of their resultant simply by using the Pythagorean theorem:

$R=\sqrt{A^2+B^2}$

where

A = 2.0 cm is the magnitude of the first vector

B = 3.0 cm is the magnitude of the second vector

Substituting,

$R=\sqrt{2^2+3^2}=3.6 cm$

Now we have to find the angle. If we measure the angle as North of East, the tangent of the angle is equal to the ratio between the component along North and the component along East. Therefore, in this case:

$tan \theta = \frac{B}{A}=\frac{3}{2}=1.5\\\theta = tan^{-1}(1.5)=56.3 \sim 56^{\circ}$