# Posts by Elena

Total # Posts: 4,389

Physics
Motion of the particle without deflection => F(el) = F mag) qE =qvB B=E/v=2607/6.81•10³=0.38 T For the second particle F=qE-qv₁B v₁=(F+qE)/qB= =1.8•10⁻⁹•3.72•10⁻¹²•2607}/ 3.72•10⁻¹²&#...

Physics
mv²/R=qvB R=mv/qB ΔR=R₂-R₁= =m₂v/qB - m₁v/qB= = v(m₂ - m₁)/qB)= =3•10⁵•(238-235)•1.67•10⁻²⁷/92•1.6•10⁻¹⁹•0.6= =1.7•10⁻⁴ m

Physics
ΔV=VB-VA =kq/r(B)-kq/r(A)= =kq{r(a)-r(B)}/r(A)•r(B). q = r(A)•r(B)•ΔV/k{r(a)-r(B)}= =(3.4-4.5)•45/9•10⁹•3.4•4.5 = = - 9.5•10⁻¹⁰ C

Phyics
ΔQ =4KCal=16747.2 J ΔQ=ΔU+W ΔU=ΔQ-W =16747.2-1000=15747.2 J

physics
ΔPE=A= q •Δφ Δφ= A/q=6.7•10⁻³/1.85•10⁻⁴=36 V

Physics
h= vₒ²•sin²α/2•g = =L•g•sin²α/2•g•sin2α= = L•sinα/4•cos α= =L•tanα/4

Physics
The voltage induced around the Al ring is ΔV=NΔΦ/Δt=Δ(0.5•B•A) /Δt = =0.5•A •ΔB /Δt. B =μ₀nI, ΔB /Δt= Δ(μ₀•nvI)/ Δt, ΔB /Δt= μ₀•n•ΔI...

Physics
F=mg=100•9.8=980 N

Physics
C=q/U C=εε₀A/d q=CU= εε₀AU/d A) q1= εε₀A4V/2d= 2 εε₀AV/d B) q2 = εε₀2A2V/d= 4 εε₀AV/d max C) q3 = εε₀2A V/2d= εε₀AV/d D) q4 = εε₀A V/...

Physics
KE= m•v²/2 => v²= 2•(KE)/m a=v²/2•s =2(KE)/2•m•s = KE/m•s m•a=e•E E= m•a/e = m•KE/e•m•s = KE/e•s = =250•1.6•10⁻¹⁹/1.6•10⁻¹⁹•0.2 = 1250 V/m

Physics
A) 2ma=F-2T = > a=(F-2T)/2m = (12000-2•2100)/2•700 = 5.57 m/s² s=v²/2a = 40²/2•5.57 =143.6 m. B) T₁=ma +F(fr) =700•5.57 + 2100 =5999 N

physics
Angular magnification M= - F(objective)/F(eyepiece) F(objective)=1/1.2 = 0.833 F(eyepiece) = -1/220= - 0.0045 M= 0.833/0.0045 = 185

1) F=ma => a=F/m = 75/17 = 4.4 m/s² s=v²/2a=3.75²/2•4.4=1.6 m 2) p= mv=0.18•3.75 =... 3) Impulse =mv=0.18•3.75 =... 4) Q=mcΔT=2•2060•20 = ...

physics
Rate of cooling R= ΔT₁/t₁=(80-50)/5=6 degr/min t₂= ΔT₂/R=(60-30)/6=30/6=5 min

physics
F(hor) =Fcos45 F(vert) =Fsin45

physics
F=ILB I=F/LB

physics
h=gt²/2 t=2h/g=2•0.63/9.8 =0.13 s. L=vt=0.3•0.13 = 0.039 m

Physics
γ=(i+2)/i=5/3=1.67 W=νRT₁ /(γ-1)}•{1-(V₁/V₂)^(γ-1)} = ={p₁V₁/(γ-1)}•{1-(V₁/V₂)^(γ-1)} = =(330000•1.2/0.67) •(1-0.5^0.67)= =591044.8•0.37 =2.2•10⁵ J

1a) W1=W2 = mgh 2a) P1 > P2 , mgh/25 >mgh/35 2) W=Fs=80•10 =800 J

physics
I=(0.73/2)²(4+3+5+2) = 1.87 kg•m² M=Iε=1.87•0.63 =1.175 kg•m

physics

PHYSICS
mgh=mv²/2 v=sqrt(2gh)=...

PHYSICS
v(fin)=v₀=20 m/s

PHYSICS
mgh₁-mgh₂=mv²/2 h₂=h₁-v²/2

PHYSICS
Simbol - Arial Unicode MS http://www.alanwood.net/unicode/fonts.html

PHYSICS
Q=ΔKE=mv²/2

PHYSICS
F=ma=mv²/2s v=sqrt(2Fs/m)

PHYSICS
W=kx²/2

PHYSICS
mgh=mv²/2 v=sqrt(2gh)=...

physics
Q=W=F(fr)s=μNs=μ•mg•s =…

physics
α=ΔL/L•ΔT α=3.6•10⁻⁵/3 ΔL = 3.6•10⁻⁵•17• (5-100)/3 = = -0.05814/3=0.01938 m

physics
α=ΔL/L•ΔT ΔL = α• L•ΔT=3.6•10⁻⁵•17• (5-100) = = -0.05814 m

physics
The time that the bullet spends in the building t=x/v(0x) =7/340 =0.02 s. The vertical displacement of the bullet in the building y= 0.56 m. The vertical component of the velocity of the bullet as it passes through the window is v(y) = (y- gt²/2)/t = y/t - gt/2= =0.5/0.02...

Physics
h= vₒ²•sin²α/2g, Solve for vₒ

Physics
T=F/cosφ =60/0.707 = =60/0.707 = 84.87 N

Physics
cos φ=15/40= 0.375 φ=68⁰ F=Tsin φ = 1000•sin68 =927 N

Physics
v₁= {+2m₂v₂₀ +(m₁-m₂)v₁₀}/(m₁+m₂) v₂={ 2m₁v₁₀ + (m₂-m₁)v₂₀}/(m₁+m₂)

Physics
L=v(x)•t => t=L/v(x) h=gt²/2 = gL²/2v(x)² v(x)= sqrt (gL²/2h)

physics
k=mg/x=2.5•9.8/0.0305 =803.3 N/m the natural frequency is ω =sqrt(k/m) = sqrt(803.3/2.5)=17.9 rad/s

physics

physics
(a) W(grav) = 0 since there is no change in height between the initial point and point A. (b) W(grav) – mgh –mg(h/2) = mgh/2 (c) W(grav) mgh -0 = mgh (d) mgh/2 (e) mgh

Physics
I=ℇ/(R+r) R= (ℇ-Ir)/I= (7.9-0.88•1.37)/0.88 = … P=I²r = …

Physics
x=Asinωt v=dx/dt =Aω cosωt= Aω•sqrt(1-sin²ωt) = =ω•sqrt(A² - A² sin²ωt)= ω•sqrt(A² -x²) A=0.076 m x= 0.045 m ω =17.8 rad/s v= ω•sqrt(A² -x²)= …

physics
ΔKE=-ΔPE KE2-KE1= -(PE2-PE1) ΔKE= KE2-KE1= m(v/2)²/2-mv²/2= = mv²/8-mv²/2= - 3mv²/8, ΔPE= PE2-PE1= k•q1q2/r-0=ke²/r - 3mv²/8 = - ke²/r, r=8ke²/3mv²= ... k =9•10⁹ N•m²/C² e =1.6...

Kepler’s Third Law. The square of the period is proportional to the cube of the distance from the star: T²∼R³ If the distance is increased by 25, then T² is increased by 25³, which means T is increased by the sqrt(25³)=125.

Physics
See the answer to a similar question below under related questions

Physics
Elongation of the spring is x=2.69 –(2.22+0.28) = 0.19 m mg= kx k=mg/x- 145/0.19=763.16 N/m

Physics
5T=68 => T=68/5=13.6 s λ=34.1 m v= λ/T f=1/T

Physics
v = sqrt(T/m₀) =sqrt(TL/m) λ =v/f => v= λf (λf)²=TL/m L=(λf)²m/T

Physics
I can't see your drawing but if there are no nods of standing wave between the ends λ =2L v = sqrt(T/m₀) λ =v/f => f=v/ λ

Physics
t = s/v =2.19/343 = 6.38•10⁻³s λ =v/f = 343/525 = 0.653 m N = s/ λ

Physics
f =f₀/(1-v/u) f₀=1390 Hz f=1430 Hz v/u=(f-f₀)/f= (1430-1390) /1430=2.8•10⁻² m/s v/u = 2.8%

physics
v1+v2=88+104 =192 km/h

Physics
Beat frequency f =| f1-f2| 242<f<288 f-242=26 288-f=20 f=268 Hz

physics
t=s/v=1/0.1•3•10⁸ =3.33•10⁻⁸ s. y=gt²/2=9.8•(3.33•10⁻⁸)²/2=5.44•10⁻¹⁵ m

Physics
f=80 Hz, V=150V, R=3200 Ω, C=3.5•10⁻⁶ F ω=2πf =2π•80=160π(rad/s) ω²=1/T² =1/(2π)²LC => L=1/(2π ω)²C= .... X(L) =ωL= ... X(C) =1/ωC= ... tan φ=(X(L) –X(C))/R = ...

Physical Science
1B 2C 3A 4C

Physics
T=F=mv²/R=mω²R ω=sqrt(T/mR) f=ω/2π

PHYSICS
T=F=mv²/R=mω²R ω=sqrt(T/mR) ω = εt t=ω / ε 2πN= εt²/2 N= εt²/4π

Physics
W=KE2-KE1=KE2 - mv²/2= =7.14 - 0.67•2.2²/2 = 5.52 J

physics
for springs connected in tandem 1/k=1/k1 +1/k2 => k=1200 N/m the stretch is x=mg/k=5.9•9.81/1200 =

Physics
The total charge Q=2q=2•3•10⁻⁶ C=6•10⁻⁶ C. The total area of two spheres A=4π(r₁²+r₂²) = 4π•149•10⁻⁴ m² q₁=QA₁/A= 6•10⁻⁶•4πr₁²/A...

Physics
(a) C=εε₀A/d Energy=Q²/2C (b) C₁=εε₀A/2d=C/2 Energy=Q²/2(C/2)= Q²/C

Physics
r=0.15 m, E< E₀ The dielectric strength of air E₀ =3 •10⁻³ V/m E=q/4πε₀r² => q=4πε₀r²E = 4πε₀r²E₀= = 4π•8.85•10⁻¹²•0.15²•3 •...

Physics
E= - gradφ= = - {(δφ/δx)i̅+(δφ/δy)j̅ + (δφ/δz)k̅}= = 2yz• i̅ –(4z-2xz)• j̅–(4y-2xy)•k̅.

Physics
E= - gradφ= = - {(δφ/δx)i̅+(δφ/δx)j̅ + (δφ/δx)k̅}= = 2yz• i̅ –(4z-2xz)• j̅–(4y-2xy)•k̅.

physics
v=const => a=0 N =F(x)•cosα F(y)•sinα=mg+F(fr) F(y)=(mg+μN)/sinα F=sqrt(F(x)²+F(y)²)

Physics
L=vₒ²•sin2α/g, B. 45°

physics
http://en.wikipedia.org/wiki/Electric_power

Physics
ε₀=8.85 •10⁻¹² F/m ε=5.4 e =1.6•10⁻¹⁹ C A=4.50•10⁻⁹ m² d=1.08•10 ⁻⁸ m U=54.6•10⁻³V C=ε•ε₀•A/d C=q/U ε•ε₀•A/d =q...

PLEASE COULD SOMEONE HELP ME OUT Physics
Separation between two wires is b=2Lsinα=2•0.5•sin3⁰=5.23•10⁻² m The mass of 1meter of the wire is m₀=ρπd²/4=2700•3.14•(0.3•10⁻³)²/4=1.9•10⁻⁴ kg where ρ =2700 kg/m&#...

Physics ~really need help!~
2. v₁= {-2m₂v₂₀ +(m₁-m₂)v₁₀}/(m₁+m₂)= =(m₁-m₂)v₁₀/(m₁+m₂)= v₂={ 2m₁v₁₀ - (m₂-m₁)v₂₀}/(m₁+m₂)= = 2m₁v₁₀...

PLEASE COULD SOMEONE HELP ME OUT
Separation between two wires is b=2Lsinα=2•0.5•sin3⁰=5.23•10⁻² m The mass of 1 meter of the wire is m₀=ρπd²/4, where ρ is the density of Al The force between two wires (per 1 meter) is F=μ₀I²/2π...

Physics
B=sqrt{B(x)²+B(y)²} = … F=qvBsinα sinα=sin90=1 => F=qvB = …

Physics
Problem #60 http://drjj.uitm.edu.my/DRJJ/Lecture/PHY407/Sample%20problems%20Chap%2021%20Cutnell.pdf

Physics
The field of the wire has a magnitude B(wire) = µ₀•I/(2πd) = … B(wire)/B(hor) = tan11⁰ ==> B(hor) = B(wire)/ tan11⁰ =

physics
P =I²R R =ρL/A= ρL/πr² P =I² ρL/πr² r=sqrt(I² ρL/πP)=... L=2 m, I=23.8 A, P=2.6 W, ρ= 1.72 x 10 ⁻⁸ Ω

physics
R=R₁+R₂+R₃=1.17+2.84+3.65 =7.66 Ω I=U/R = 20.3/7.66=2.65 A P₁ =I²R₁=2.65²•1.17= P₂ =I²R₂=2.65²•2.84= P₃=I²R₃=2.65²•3.65=

Physics
I=317 A B=7.69•10⁻³ T B(wire)=B B(wire)=μ₀I/2πa =B a= μ₀I/2πB= =4π•10⁻⁷•317/2π•7.69•10⁻³=8.24•10⁻³ m

Physics
A₁=(L/4)² M₁=p₁•B=I•A₁•B=I•L²•B/16=… A₂=(L/16)² M₂=2p₂•B=2I•A₂•B=2I•L²•B/256=…

physics
n=4 rev/s angular velocity ω =2πn =8π rad/s linear velocity v= ωR

PHYSICS
I've corrected my mistake in the 1st equation V(x) = - v•cos α +u•cosβ = - 190cos25 + 45cos15=-172.2+43.5= -128.7 km/h, V(y)= v•sinα+u•sinβ= 190sin25 + 45sin15= 80.3 + 11.6 = 91.9 km/h, V=sqrt{V(x)²+V(y)²}= 158.1 km/h I ...

Physics
V(x) = - v•cos α +u•cosβ = - 190cos25 + 45cos15=172.2+43.5= -128.7 km/h V(y)= v•sinα+u•sinβ= 190sin25 + 45sin15= 80.3 + 11.6 = 91.9 km/h V=sqrt{V(x)²+V(y)²}= 158.1 km/h

Physics
v=190 km/h, α= 25°, u=45 km/h, β =15°. V(x) =v•cos α - u•sin β =… V(y)= v•sinα+u•sinβ=… V=sqrt{V(x)²+V(y)²}=... Sine Law: V/sin(α+ β) =u/sinγ Solve for γ. V directed (α+γ...

Physics
PE=CU²/2 U=sqrt{2•PE/C}

physics
A= 4πr² => r=sqrt(A/4π )= sqrt(1.2/4•3.14)= 0.31 m φ=q/4πε₀r => q= 4πε₀rφ= =4π•8.85•10⁻¹²•0.31•590=2.03•10⁻⁸ C

Physics
k =9•10⁹ N•m²/C², q=4.8•10⁻⁶ C m=5.5•10⁻³ kg r=0.88 m PE₁-PE₂=2KE -kq²/r +3kq²/r =2mv²/2 2kq²/r =2mv²/2 kq²/r =mv²/2 v=sqrt( kq²/rm) = =sqrt{9•10⁹•(4....

Physics
E= 8.18•10⁴ V/m d=2.2 mm q=0.775 μ C ε (paper)= 2.3 (a) C=q/U E=U/d C=q/Ed = ... (b) ε₀=8.85•10⁻¹² F/m C= ε₀εA/d => A=Cd/ε₀ε=...

Physics
Problem #16 http://drjj.uitm.edu.my/DRJJ/Lecture/PHY407/Sample%20problems%20Chap%2019%20Cutnell.pdf

physics science
Since v=const, F(motor) =F(fr) P=F•v=680•27 =18360 W

Physics !
P=Fv=Fs/t=267•493/9•60 = … Watt

Physics
s=at^2/2

Physics
p =F/A =(m+M)g/4πr² = (6.3+67.8)9.8/4π(0.024)²= …

physics!
m=10000 kg KE=mv²/2= ...

Physics help!!
The total range of the arrow is L=vₒ²•sin2α/g =45²•sin100°/9.8 =203.5 m R=150 m > L/2=101.75 m => the arrow is at its descending path The time of the motion to the point of 'arrow-apple' meeting is t₁=R/v(x)=R/v₀(x...

PHY
(a) qB =–2.0 µC and qD = +12.0µ C. q=(- 2 + 12)/2 = 5 µC (b) qA = –8.0 µC, qC = +5.0µ C, and qD = +12.0µ C. q=(- 8 + 5+12)/3 = 9/3=3 µC

Physics
KE=mv²/2 v=sqrt(2KE/m)

Physics
A=0.75 m, m=8 kg, k= 400 N/m ω=sqrt(k/m) = … x=Acosωt v= - Aωsinωt v(max) = Aω KE=mv²/2 =0.5m A²ω²sinωt KE(max) = 0.5m A²ω²