Ask question

Ask question

All categories

3 years ago

8

Air pressure is 1.0 · 105 N/m2, air density is 1.3 kg/m3, and the density of soft drinks is 1.0 · 103 kg/m3. If one blows carefu

lly across the top of a straw sticking up 0.100 m from the liquid in a soft drink can, it is possible to make the soft drink rise half way up the straw and stay there. How fast must the air be blown across the top of the straw?

1 answer:

natita [175]3 years ago

8 0

Answer:

v = 27.456 m/s

Explanation:

The support pressure needed of the water in the straw can be calculated by the formula

Given that,

P = r*g*h

= 1000*9.8*0.05 Pa.= 490 Pa

This pressure is compensated by 0.5*r*v^2 of the air,

Hence,

0.5*1.3*v^2 = 490

velocity of air blown into the straw =

v = 27.456 m/s

You might be interested in

Which of the following is a unit of acceleration?

olasank [31]

${\tt{\red{\underline{\underline{\huge{Answer:}}}}}}$

$\longrightarrow$ The rate of change of velocity per unit time is called acceleration.

$\longrightarrow$ Its SI unit is m/s².

$\huge\boxed{\fcolorbox{blue}{red}{Thank you}}$

7 0

2 years ago

What is the difference between biopsychology and neuropsychology?

devlian [24]

Neuropsychology was the study of brain-behavior relationships. <span>Biological psychology is concerned with both a molecular and global perspective</span>

7 0

3 years ago

Paki po plss po plssssss

slega [8]

Answer:A language is a structured system of communication. Language, in a broader sense, is the method of communication that involves the use of – particularly human – languages. The scientific study of language is called linguistics

Explanation:

7 0

2 years ago

An Atwood machine consists of two masses, mA = 6.8 kg and mB = 8.0 kg , connected by a cord that passes over a pulley free to ro

Lisa [10]

To solve this problem it is necessary to apply the concewptos related to Torque, kinetic movement and Newton's second Law.

By definition Newton's second law is described as

F= ma

Where,

m= mass

a = Acceleration

Part A) According to the information (and as can be seen in the attached graph) a sum of forces is carried out in mass B, it is obtained that,

$\sum F = m_b a$

$m_Bg-T_B = m_Ba$

$T_B = m_Bg-m_Ba$

In the case of mass A,

$\sum F = m_A a$

$T_A = m_Ag-m_Aa$

Making summation of Torques in the Pulley we have to

$\sum\tau = I\alpha$

$T_BR_0-T_AR_0=I\alpha$

$T_B-T_A=I\frac{a}{R^2_0}$

Replacing the values previously found,

$(m_Bg-m_Ba )-(m_Ag-m_Aa )=I\frac{a}{R^2_0}$

$(m_B-m_A)g-(m_B+m_A)a=I\frac{a}{R^2_0}$

$a = \frac{(m_B-m_A)g}{\frac{I}{R_0^2}+(m_B+m_A)}$

$a = \frac{(m_B-m_A)g}{\frac{MR^2_0^2/2}{R_0^2}+(m_B+m_A)}$

$a =\frac{(m_B-m_A)g}{\frac{M}{2}+(m_B+m_A)}$

Replacing with our values

$a =\frac{(8-6.8)(9.8)}{\frac{0.8}{2}+(8+6.8)}$

$a=0.7736m/s^2$

PART B) Ignoring the moment of inertia the acceleration would be given by

$a' =\frac{(m_B-m_A)g}{(m_B+m_A)}$

$a' =\frac{(8-6.8)(9.8)}{(8+6.8)}$

$a' = 0.7945$

Therefore the error would be,

$\%error = \frac{a'-a}{a}*100$

$\%error = \frac{0.7945-0.7736}{0.7736}*100$

$\%error = 2.7%$

8 0

2 years ago

During a lab investigation, students added four 50 g masses to two boxes and arranged the boxes so that they were motionless on

IceJOKER [234]

When the resultant force is not equal to zero termed an unbalanced force. By procedures 4 and 5 students observe an unbalanced upward force on Box 1. Hence option 1 is right for the problem.

<h3>What is an unbalanced force?</h3>

The forces operating on a body are known as unbalanced forces when the resulting force exerted on it is not equal to zero.

Unbalanced forces acting on the body, causing it to modify its state of motion. To further grasp the nature of imbalanced forces.

<h3 />

The following reasons by which we can understand the unbalanced force caused by the box.

Due to these two reasons, books will move up.

By adding another mass to box 2. The box becomes lighter. As the box becomes lighter the gravity force acting on the box will be less due to which the box easily can move up.

By removing the two masses from box 1. Due to which other become heavier other becomes heavier pulling it down causing box 1 one to go up.

Hence by procedures 4 and 5 students observe an unbalanced upward force on Box 1. Hence option 1 is right for the problem.

To learn more about the unbalanced force refer to the link;

brainly.com/question/227461

3 0

2 years ago