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3 years ago

How liquids are drown in to syringe tube when the plunger is pulled out?

Physics

1 answer:

jeka943 years ago

4 0

Answer:

Fluid, such as a drug or blood, is drawn up through a hollow needle into the main tube when the plunger handle is pulled back. As long as the needle tip remains in the fluid while the plunger handle is pulled, air will not enter.

I hope it's helpful!

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A large storage tank, open to the atmosphere at the top and filled with water, develops a small hole in its side at a point 15.6

VladimirAG [237]

To solve this problem it is necessary to take into account the kinematic equations of motion and the change that exists in the volume flow.

By definition the change in speed is given by

$v_f^2-v_i^2 = 2ax$

Where,

x= distance

$v_f =$ final velocity

$v_i =$ initial velocity

a = acceleration

On the other hand we know that the flow of a fluid is given by

$\dot{V} = Av$

Where,

A = Area

v = Velocity

PART A )

Applying this equation to the previously given values we have to

$v_f^2-v_i^2 = 2ax$

$v_f^2-0 = 2*(9.8)(15.6)$

$v_f^2=305.76$

$v_f = 17.48$

Therefore the velocity of the water leaving the hole is 17.48m/s

PART B )

In the case of the hole we take the area of a circle, therefore replacing in the flow equation we have to,

$\dot{V} = \pi r^2 v$

$r = \sqrt{\frac{\dot{V}}{\pi v}}$

$r = \sqrt{\frac{3*10^{-3}*\frac{1}{60}}{\pi (17.48)}$

$r = \sqrt{9.10*10^{-7}}$

$r = 0.54*10^{-4}$

The diameter is 2 times the radius, then is $1.91*10^{-3}$ m or 1.91mm

<em>Note: The rate flow was converted from minutes to seconds.</em>

8 0

3 years ago

A spring-loaded launcher has a mass of 0.60 kg and is placed on a platform 1.2m above the ground. The force of friction is negl

kobusy [5.1K]

Answer:

C The launcher will fall off the platform and land D/2 to the left of the platform because the launcher is twice the mass of the ball.

Explanation:

The figure is missing: you can find it in attachment.

We can apply the law of conservation of momentum to check that the launcher will leave the platform with a speed which is half the speed of the ball. In fact, the total initial momentum is zero:

$p_i = 0$

while the total final momentum is:

$p_f = m_l v_l + m_b v_b$

where

$m_l = 0.60 kg$ is the mass of the launcher

$m_b = 0.30 kg$ is the mass of the ball

$v_l$ is the velocity of the launcher

$v_b$ is the velocity of the ball

Since the total momentum must be conserved, $p_i=p_f$ , so

$0=m_l v_l + m_b v_b$

Therefore we find

$v_l = - \frac{m_b}{m_l}v_b = -\frac{0.30}{0.60}v_b = -\frac{v_b}{2}$

which means that the launcher leaves the platform with a velocity which is half that of the ball, and in the opposite direction (to the left).

Since the distance covered by both the ball and the launcher only depends on their horizontal velocity, this also means that the launcher will cover half the distance covered by the ball before reaching the ground: therefore, since the ball covers a distance of D, the launcher will cover a distance of D/2.