Electro statics ไฟฟ้าสถิต

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# h3phye1

ทรงกลมเล็ก ๆ แขวนแนวดิ่งไว้ด้วยเชือกเบาที่เป็นฉนวน จากนั้นค่อยๆ เพิ่มขนาดสนามไฟฟ้า สม่ำเสมอในแนวระดับทำให้ทรงกลมเล็กค่อย ๆ เคลื่อนที่ไปในทิศทางดังรูป ถ้าทรงกลมมีประจุ -2.5 ไมโครคูลอมบ์ และมีมวล 0.015 กรัม เชือกเบาสามารถทนแรงดึงได้สูงสุด 0.25 x 10-3 นิวตัน จงหาขนาดและทิศทางสนามไฟฟ้าพร้อมกับมุม θ ที่ทำให้เชือกเบาขาดพอดี (g = 10 m/s 2)

(1)	80 N/C ,ทิศจาก A ไป B และ θ=sin-1 0.6
(2)	80 N/C ,ทิศจาก B ไป A และ θ=cos-1 0.6
(3)	80 N/C ,ทิศจาก B ไป A และ θ=sin-1 0.6
(4)	80 N/C ,ทิศจาก A ไป B และ θ=cos-1 0.6

Strategy - the rule "Must have" of Mr.Zhang ® โจทย์ให้ แรงตึงเส้นเชือก 0.25 x 10-3 นิวตัน, มวล 0.015 g., ประจุ -2.5 ไมโครคูลอมบ์ โจทย์ถาม     ขนาดของสนามไฟฟ้า(E)และทิศทาง สิ่งที่ต้องรู้    (1)  F=qE  1.1  (q)ประจุลบวิ่งสวนสนามไฟฟ้า  1.2  (q)ประจุบวกวิ่งตามสนามไฟฟ้า  1.3  (F)แรงที่เกิดกับประจุลบมีทิศตรงข้ามกับ E  1.4  (F)แรงที่เกิดกับประจุบวกมีทิศตาม E (2)  การที่เชือกเบาขาดพอดีได้ แรงที่กระทำต้องไม่น้อยกว่าแรงตึงเชือกได้สูงสุด (3)  การแตกแรงตึงเส้นเชือกให้อยู่ ในแนวดิ่งและแนวราบ

Solution

หาแรงตึงเชือกจากสมดุลแนวดิ่ง ;

Tcosθ =mg

cosθ	=	mg T
cosθ	=	0.015 x 10-3 kg x 10  m/s-2 0.25 x 10-3N.
cosθ	=	3 5
∴   θ	=	cos-1 (0.6)

สมดุลแนวราบ ;
FE	=	Tsinθ
qE	=	Tsinθ

qE	=	T	√ (1-cos2θ)
E x 2.5 x 10-6C	=	0.25 x 10-3 N	√ (1-(0.6)2)
E	=	80 N/C
∴ ขนาดของสนามไฟฟ้า(E)	=	80 N/C
ประจุลบวิ่งสวนสนามไฟฟ้า แสดงว่าสนามไฟฟ้า(E) วิ่งจาก B ไป A
คำตอบ คือข้อ 2

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spring highschool

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# h1psp1

A 150 g. mass is suspended from a vertical spring and descends a distance of 4.6 cm, after which it hangs at rest. An additional 500 g. mass is then suspended from the first.   What is the total extension of the spring? (neglect the mass of the spring.)

Strategy - the rule "Must have" of Mr.Zhang ®
Find the value of k ;    Fs=kx=mg	→ (1)
And later find x(total stretch length) from ;   kx=(m1+m2)g	→ (2)

Solution

Find k(constant of spring) from equation(1)

k	=	m1g x1
	=	(0.15 kg)(9.8 m·s-2) 0.046 m
	=	32 N·m-1

Find x(total extension of the spring) from equation(2)

x	=	(m1+m2)g k
	=	(0.15 kg+0.50 kg)(9.8 m·s-2) 32 N·m-1
	=	0.20 m(or 20 cm.)

Ans.    The total extension of spring is 20 cm

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Projectile motion

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Physics High School
# h2phyt1 - Projectile motion

A stone is thrown from the edge of a vertical cliff with a velocity of 50 m•s -1

at an angle tan-1

7 24

above the horizontal.    The stone strikes the sea at

a point 240 m from the foot of the cliff.    Find the time for which the stone

is in the air and the height of the cliff.

Strategy - the rule "Must have" of Mr.Zhang ®

1. Convert tan-1

7 24

to be sinθ and cosθ

2. Formulae : Sy = vyt +

1 2

gt2

(vertical) → (1)

S x= vxt (horizontal) → (2)

Horizontal motion (time) = Vertical motion (time)

Solution

Compute the time to find the height

∵   tan-1	7 24
Therefore,tanθ=	7 24	,  sinθ=	7 25	,  cosθ=	24 25

Horizontal motion from equation (2);

S x	=	vxt
S x	=	vxcosθt
240 m	=	50 m•s-1	24 25	t
t	=	5 s

We bring the horizontal time to calculate the vertical
motion to get the height value.

Sy = vyt +

1 2

gt2

→ (1)

Sy = 50 m•s-1sinθ(5) +

1 2

(-9.8 m•s-2)(52)

Gravitational against downward direction of the earth, therefore, we take the minus sign.

-h = 50 m•s-1

7 25

(5) +

1 2

(-9.8 m•s-2)(52)

Displacement direction required is lower the level of the cliff. Hence we take the minus sign.

Ans. 1. The stone is in the air for 5 second.

2. The height of the cliff 52.5 m.

Fuel Injector

Pressure Volume Temperature

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Temperature vs Pressure vs Volume Problems
# h2phyt1 - Pressure and Volume

Photo credited by en/wiki

The chamber in a bicycle pump contains air at standard atmospheric
pressure. The valve is closed so no air escapes, and a downward push
of the handle decreases the volume of the air by 30 percent.  What is the
new pressure of air in the chamber (assume no temperature change)

Strategy - the rule "Must have" of Mr.Zhang ®

Boyle's law;
V	∝	1 P
V	=	C x	1 P
PV	=	C
P1V1	=	P2V2	→ (1)

***Both sides must be equal to the same constant.

P1 is standard atmospheric pressure	=	1.013x105 pa
	=	1 atm
	=	1.013x105 N.m-2

                                                                                 = 14.7  lb.in.^-2
                                                                                 = 760  mmHg
                                                                                 = 76  cmHg
                                                                                 = 29.9  in.Hg
                                                                                 = 760  torr
                                                                                 = 1013.25  mb

Solution

Compute the new pressure from equation(1)

V2

=

(100-30)%V1

=

70%V1

=

0.70V1

P1V1

=

P2V2

(1.013x105 pa)V1

= P2(0.70V1)

(1.013x105 pa) 0.70

=

P2

P2

=

0.144x106 pa

Ans. New pressure of air in the chamber is

1.4x105 pa

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Physics - Fluid - Surface Tension Force

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Fluid Problems
# h2phy1 - Pressure converted inside Aorta
# h2phy2 - Pressure inside vein
# h2phy3 - Pressure on the floor
# h2phy4 - Pressure under the sea	# h2phym1- Hydraulic Lift - 1
# h2phy5 - Pressure between oil and water	# h2phym2- Hydraulic Lift - 2
# h2phy6 - Buoyant Force	# h2phym3- Hydraulic Lift - 3
# h2phy7 - Flow rate & Velocity
# h2phy8 - Surface Tension Force(1)
# h2phy9 - Volume & Mass Flow rate
# h2phy10 - Surface Tension Force(2)	↰

(a) To what height will water (at 20 ^oC) rise in a glass tube
       with a bore radius of 0.1 mm?
       The contact angle for water in a glass tube is 0^o
       so the cos θ factor in figure 7, is equal to unity.  Then,
(b) To what depth will mercury ( at 15^oC) be
       depressed in the same tube?  In this case the contact
       angle is 135^o ; therefore,
(density of Mercury=13.6x10³ kg.m^-3)

Strategy - the rule "Must have" of Mr.Zhang ®

Coefficient of surface tension(γ) = Tension Force(FT) Circumference(L)	→	SI unit N.m-1
Ftension → (Forced required to stop movable side from sliding)
L(circumference ) =2πr ; because of the round shape of the surface.

γ 2πr	=	FT
We break down FT to FT (cosθ);
γ 2πr(cos θ)	=	FT (cosθ)	→ (1)	Upward force  ↑
mg	=	ρvg

                                                                 =(πr²h)(ρg)→ (2)Downward force  ↓

Table 8 Values of the Surface Tension for various liquids in contact with air
Liquid	Temperature(oC)	-γ-(J/m2 or N/m)
Acetone	20	0.0237
Alcohol,methyl	20	0.0226
Benzene	20	0.0288
Water	0	0.0756
Water	20	0.0728
Water	30	0.0712
Water	100	0.0589
Mercury	15	0.487

Solution

(a) Compute to find the height by bringing equation (1)=(2) according to the equlibrium

γ 2πr(cos θ)

=

(πr2h)(ρg)

γ 2πr(cos θ)

=

(πr2h)(ρg)

h

=

2γcos θ ρgr

h	=	2(0.0728 N.m-1) (103kg/m3)(9.8m/s2)(10-4m)
	=	0.15 m
Ans. Water (at 20 oC) rise 15 cm height in a glass tube.

(b) To what depth will mercury ( at 15oC) be        depressed in the same tube?  In this case          the contact angle is 135o ; therefore,        (density of Mercury=13.6x103 kg.m-3)

h	=	2γcos135o ρgr
h	=	2(0.487 N.m-1)(-0.707) (13.6x103 kg.m-3)(9.8 m.s-2(10-4 m)
h	=	-5.16 cm.
Ans. Mercury (at 15 oC) will be depressed 5.16 cm.

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Physics volume & mass flow rate

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Fluid Problems
# h2phy1 - Pressure converted inside Aorta
# h2phy2 - Pressure inside vein
# h2phy3 - Pressure on the floor
# h2phy4 - Pressure under the sea	# h2phym1- Hydraulic Lift - 1
# h2phy5 - Pressure between oil and water	# h2phym2- Hydraulic Lift - 2
# h2phy6 - Buoyant Force	# h2phym3- Hydraulic Lift - 3
# h2phy7 - Flow rate & Velocity
# h2phy8 - Surface Tension Force(1)
# h2phy9 - Volume & Mass Flow rate ↰
# h2phy10 - Surface Tension Force(2)

(a) A water line necks down from a pipe with a 12.5 mm.
       radius to a pipe with a 9 mm radius. If the speed of the
       water in the 12.5 mm pipe is 1.8 m.s^-1, what is the speed
       in the smaller pipe?
(b) What is the volume flow rate?
(c) What is the mass flow rate?

Strategy - the rule "Must have" of Mr.Zhang ®

ΔM	=	ρΔV	=	ρAV&#916t
ΔM &#916t	=	ρAV	→	(1) The rate of flow of mass
ρV1A1 = ρV2A2			→	The equation of continuty.
V1A1 = V2A2			→	(2) Density is constant.

Solution

(a) Compute the speed in the smaller pipe by solving the equation (2)

V2

=

V1A1 A2

V2

=

V1 π(r1)2 π(r2)2

V2	=	1.8 m.s-1(12.5 x 10-3)2 (9 x 10-3)2
Ans. The speed in the smaller pipe is 3.47 m.s-1,therefore, it make sense.
How doest it make sense?

(b) Compute the volume flow rate is;
V1A1	=	V2A2
	=	π(12.5x10-3m)2(1.8 m.s-1)
	=	8.8 x 10-4 m3.s-1

Ans. The volume flow rate is 8.8 x 10^-4m³s^-1.

(c) The mass flow rate is;
Compute it from eqaution(2);
ρV1A1	=	ρV2A2
	=	(1.0x103 kg.m-3)(8.8x10-4 m3.s-1)

Ans. The mass flow rate is 0.88 kg.s^-1.

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Physics;Surface Tension High School

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Problem# h2phy1 - Pressure converted inside the Aorta

Problem# h2phy2 - Pressure inside vein

Problem# h2phy3 - Pressure on the floor

Problem# h2phy4 - Pressure under the sea

Problem# h2phy5 - Pressure between oil and water

Problem# h2phy6 - Buoyant Force

Problem# h2phy7 - Buoyant Force

Problem# h2phy8 - Surface Tension

A thin,circular wire ring (radius = 0.04 m) and total mass 0.0007 kg
is gradually drawn apart from the water surface in a vertical direction by means of a sensity spring with spring constant k=0.75 N.m^-1.  When
the spring is stretched 0.034 m from its equilibrium extension in air,
the ring is on the verge of being pulled free from the water surface.
Find the coefficient of surface tension of water.
Neglect the mass of the lifted water.

Strategy - the rule "Must have" of Mr.Zhang ®

Coefficient of surface tension(γ) = Tension Force(Ft) Circumference(L)	→	SI unit N.m-1
Ftension → (Forced required to stop movable side from sliding)
L(circumference ) =2(2πr) ; because of the ring shape.

Equilibrium ;
↑  Fspring	=	↓ mg  +  γl ↓	→(1)

Solution

Compute the coefficient of surface tension (γ) of water by solving the equation (1)

kx	=	mg  +  γl
kx	=	mg  +  γ2(2πr)
γ	=	kx - mg 4πr
γ	=	(0.75 N.m-1)(3.4x10-2m)  -  (0.7x10-3kg)(9.81 m.s-2) 4π(2x10-2m)
γ	=	0.026 N - 0.0069 N 0.25 m
γ	=	0.076 N.m-1
Ans.   The coefficient of surface tension of water is 0.076 N.m-1

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Physics;Flow rate, high school

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Problem# h2phy1 - Pressure converted inside the Aorta

Problem# h2phy2 - Pressure inside vein

Problem# h2phy3 - Pressure on the floor

Problem# h2phy4 - Pressure under the sea

Problem# h2phy5 - Pressure between oil and water

Problem# h2phy6 - Buoyant Force

Problem# h2phy7 - Flow rate; area vs velocity

The stream of water emerging from a faucet
"necks down" as it falls. The cross sectional area
A_o  is 1.2 cm², and A is 0.35 cm².  The two levels
are seperated by a vertical distance h=45 mm.
At what rate does water flow from the tap?

Strategy - the rule "Must have" of Mr.Zhang ®

Flow Rate Equations;
Flow Rate (Q)	=	Area x Velocity     → (1)
	=	AV
V2	=	U2+2gs     → (2)
	=	Vo2+2gs

Solution
Compute the initial velocity ;
From equation (1) ;
AoVo	=	AV

AoVo A

=V     → (3)

We bring equation(3) substitue in equation(2) ;

The Volume flow rate is [R]	=	AoVo
R	=	1.2 cm2x28.6 cm/s
Ans.  Water flow from the tap	=	34 cm3/s

Water falls from a tap,whether its speed increases or not?

If "yes", why?

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physics fluid buoyant force

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Problem# h2phy1 - Pressure converted inside the Aorta

Problem# h2phy2 - Pressure inside vein

Problem# h2phy3 - Pressure on the floor

Problem# h2phy4 - Pressure under the sea

Problem# h2phy5 - Pressure between oil and water

Problem# h2phy6 - Buoyant Force

The density of fluid "A" is 1.2 times of density fluid "B".
When we bring an object lower into the liquid "B", and
find that volume of the object sunk down is 0.6 times of total volume.
If we take this object lower into the liquid "A", and sunk down, what is
the ratio of lower portion into liquid "A" comparing with the total volume?

Strategy - the rule "Must have" of Mr.Zhang ®

Fb= ρvg

Solution

Consider, force acts against object when 1st. lowering into the “B” liquid ;

ΣFup=ΣFdown

Fb=mg

ρbg(0.6V)=mg

ρb(0.6V)=m     → (1)

Consider, force acts against object when 2nd. lowering into the “A” liquid ;

ΣFup=ΣFdown

Fb=mg

We represents Vx  with the portion which sunk down into liquid "A"

ρag(vx)=mg     → (2)

From the problem gives ρa=1.2ρb,and we also take the equation(1) substitue in equation(2) , therefore,

1.2ρbg(vx)=0.6ρbvg

1.2ρb(vx)=0.6ρbv

1.2(vx)=0.6v

vx= 0.6v 1.2

vx=0.5v

Ans.  The ratio of lower portion into liquid "A" comparing with the total volume is 0.5v

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Pressure between oil and water

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Problem# h2phy1 - Pressure converted inside the Aorta

Problem# h2phy2 - Pressure inside vein

Problem# h2phy3 - Pressure on the floor

Problem# h2phy4 - Pressure under the sea

Problem# h2phy5 - Pressure between oil and water

The U-tube contains water of density&nbspρ_w in the right arm
and oil of unknown density ρ_x in the left.   Measurement gives
L = 135 mm. and d = 12.3 mm. What is the density of the oil ?

Strategy - the rule "Must have" of Mr.Zhang ®

Interface area is joint of the two substances. P = (Po) + (Pg) Absolute pressure = Atmospheric pressure + Gauge pressure       Material or Object         Density (kg/m3)     Water: 20 οC and 1 atm 0.998x103   1 atm = 760 mmHg = 760 torr = 29.9 in.Hg =101.325 kPa = 14.7 lb/in.2

Solution Pint = Po +&nbspρwgl          (right side)  →(1) Pint = Po +&nbspρoilg(d+l)    (left side)  →(2) (1)=(2);   Po +&nbspρwgl = Po +&nbspρoilg(d+l)     ρwgl = ρoilg(d+l)     ρwl = ρoil(d+l)     (ρw)/(d+l) = ρoil ∴The density of the oil is 916.5 kg/m3

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Pressure under the water

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Problem# h2phy1 - Pressure converted inside the Aorta

Problem# h2phy2 - Pressure inside vein

Problem# h2phy3 - Pressure on the floor

Problem# h2phy4 - Pressure under the sea

An enterprising diver reasons that if a typical 20 cm.
long snorkel tube works, a 6.0 m. long tube should
also. If he foolishly uses such a tube, what is the pressure
different Δp between the external pressure
on him and the air pressure in his lungs? Why is he in danger?

Strategy - the rule "Must have" of Mr.Zhang ®

Absolute pressure (P) The difference between absolute pressure and atmospheric pressure is called gauge pressure. P =  (Pg) + (Po) Absolute pressure - Atmospheric pressure = Gauge pressure       Material or Object         Density (kg/m3)     Water: 20 οC and 1 atm 0.998x103 Sea water: 20 οC and 1 atm 1.024x103   1 atm = 760 mmHg = 760 torr = 29.9 in.Hg =101.325 kPa = 14.7 lb/in.2

Solution The diver at depth L = 6.0 m without the snorkel tube. The external pressure on him is given by ; P =  Po + ρgh ΔP=P-Po=ρgL ; because L=h     =(1000 kg/m3)(9.8 m/s2)(6.0m)     =5.9x104 Pa ∴The pressure difference is about 0.6 atm, is sufficient to collapse the lungs and force the still pressurized blood into them,a process known lung squeeze. "thoracic squeeze, also called Lung Squeeze, compression of the lungs and thoracic (chest) cavity that occurs during a breath-holding dive under water. During the descent, an increase in pressure causes air spaces and gas pockets within the body to compress." credited : Britannica

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