Friday
April 25, 2014

Posts by Elena


Total # Posts: 4,312

Physics :)
Angular acceleration= torque/moment of inertia

physics
T + mg = mv²/R => T = m (v²/R -g)

physics
Tsin(α/2)=kq²/{2Lsin(α/2)}², Tcos(α/2)=mg. tan(α/2)= kq²/mg{2Lsin(α/2)}², m= kq²/g{2Lsin(α/2)}²•tan(α/2)=...

physics
m₁gh₁ =mv²/2, v=sqrt(2gh₁)=… m₁v=(m₁+m₂)u, u= m₁v/(m₁+m₂)=.. . (m₁+m₂)u²/2=(m₁+m₂)gh₂, h₂=(m₁+m₂)u²/2(m₁+m₂)=..

physics
h=gt²/2 t=sqrt(2h/g) =sqrt(2•200/9.8)=6.39 s

Physics
a) density b) acceleration c) - d) - e) refractive index f)capacity g)resistance h) pressure

physics
a=F/m=90/(40+3) = 2.1 m/s2

physics
ma=F(fr), mv²/R =μmg, v=sqrt(μRg) = sqrt(0.63•39.6•9.8)= …

physics
F(x) =-F(1x) +F(2x)+F(3x) – F(4x) = =(G•m₅•cos45°•2/s√2)•(-m₁+m₂+m₃-m₄)=... F(y) = F(1y) +F(2y)-F(3y) – F(4y) = =(G•m₅•sin45°•2/s√2)•( m₁+m₂-m₃-m₄)...

Physics!! HELP
λ=2•2.5=5 m s=14•5=70 m v=s/t=70/15=4.67 m/s

Physics
An object is placed 11 cm in front of a concave mirror whose focal length is 23 cm. The object is 3.6 cm tall. Determine (a) the location of the image, taking a real image as a positive value and a virtual image as a negative value. (b) Determine the height of the image, where...

Physics Hear
1) Q=c•m•ΔT=4183•10•(100-20)=3346400 J 2) Q=Lm = 2260000•0.01 = 22600 J 3) Q= Q₁+Q₂+Q₃= c₁•m•ΔT₁+rm+ c₂•m•ΔT₂= =2060•3•7+ 335000•3+4183•3•25 = =43260 ...

physics
http://en.wikipedia.org/wiki/Snell's_law

Physics
a=v/t=20/25 = ...

physics
Please see the Related Questions below

Physics
n=actual depth/apparent depth = =3/(3-1.1) = 3/1.9=1.58

physics
P=(m₁+m₂)g/A=(40+12) •9.8/2(6•4)•10⁻⁴=106167 Pa= 106.167 kPa

physics
v(ave) =total distance/ time taken = =(210+470+150)/(25+20+30) = 11.7 m/s

Physics
n₁ = 1.520 n₂= 1.538 n₁=sini₁/sinr₁ n₂=sini₂/sinr₂ sinr₁=sinr₂ sini₁/n₁=sini₂/n₂ sini₂=n₂•sini₁/n₁=1.538•sin 20.97/1.52 =0.3621 i₂=21.23°

physics
p=mv

physics
(a) v=at=0.3•15=0.45 m/s (b) v₁=30km/h=30000/3600 =833.3 m/s t₁=v₁/a=833.3/0.3 =2777.8 s. (c) s=v₁²/2a= 833.3²/2•0.3=1157314 m =1157 km

physics
F= F(fr) + mgsinα= =32.6 +63.8sin59=87.3 N

physics
1T=10000 Gauss B=0.02•10⁻⁴= 2•10⁻⁶ T v=8 km/s=8•10³ m/s U=BLv= 2•10⁻⁶•4828• 8•10³=77.25 V

Physics
ℰ=-dΦ/dt=-d(NBAcosωt)/dt= =NBAωsinωt = =500•0.05•0.05•20•sin30° =12.5 V

Physics (Algebra-based)
PE → KE1 KE1 → Work(fr) +KE2 => PE → Work(fr) +KE2 If the initial height is ‘h₁’, the height of the horizontal section is ‘h₂’, the length of the rough section is ‘s’, then mg (h₁-h₂)=μmgs +mv²...

physics
vt=at²/2 t=2v/a=2•19.4/3.2=121.25 s=2.02 min

physics
ℰ=-dΦ/dt=-d( BAcosα )/dt cosα=1 ℰ =BA/t= Bab/t=Bb(a/t)=Bbv= =0.125•0.045•4.3=0.024 V

physics
ma=F => a=F/m=2/2.5=0.8 m/s² v=v₀-at v=0 t=v₀/a=0.6/0.8=0.75 s.

physics
During acceleration (or deceleration) of the bus, she continues to move in previous derection due to inertia, and, therefore, she may fall down.

physics
Energy =Power x time =680 W/m2 x 1 second =680 Joule/m2 Energy of 1 photone = hf =hc/λ Number of photones =Energy/energy of 1 photon = N=680• λ/hc=680•478•10⁻⁹/6.63•10⁻³⁴•3•10⁸=…

Physics
(a) Torque is M=p⒨•B•sinα p⒨=NIA If M(max) =sin α=1 M(max)=NIAB, I= M(max)/NAB = =40/250•0.12•0.04•1=33.3 A (b) ℰ=-dΦ=-d(NBAsinωt)/dt= =-NBAωcosωt ℰ(max)=NBAω=2πNBAf= =2π•250...

physics
(a) I =cε₀E²/2 =3•10⁸•8.85•10⁻¹²•(7.9•10⁻⁶)²/2=8.2•10⁻¹⁴W/m² (b) P=I•A=I•πD²/4=8.2•10⁻¹⁴•π•2.1²/4=2.84•...

physics
PE=mgh=10•2=20 J. PE->KE

physics
Question????

Physics
The speed at the 5-m level is of the same magnitude on the way up and on the way down=> s=(v²-v₀²)/2g Solve for v₀

Physics 2
tanα=x/L=3.2•10⁻³/9=3.56•10⁻⁴ tanα≈sinα bsinα=kλ b= kλ/sinα=1•530•10⁻⁹/3.56•10⁻⁴=1.5•10⁻³ m

Physics
M=Iε ε=M/I = 22.9/0.168 = 136.3 rad/s² 2πN=2πn₀t + εt²/2 2πn=2πn₀ + εt n₀=0 => 2πN= εt²/2 2πn= εt => t=2πn/ε N= πn²/ε n=sqrt {εN/π}= =sqrt{136....

physics
Δp(rocket)=Δp(gas)= mv=26•182=4732 kg•m/s

Physics
2s=ct s=ct/2=3•10⁸•2.56/2=3.84•10⁸ m

Physics
2s=vt s=vt/2=345•4.4/2=759 m

Physics
t=2s/v=2•150/345=0.87 s

Physics
Speed of sound at 20⁰ v=343.1 m/s λ=v/f=343.1/512 =0.67 m

Physics
M=M(fr) M=p⒨Bsinα sinα=sin90=1, M=NIAB M(fr)=F(fr)R=μF⒩R NIAB =μF⒩R F⒩= NIAB/μR= =425•0.27•3•10⁻³•0.21/0.81•0.011= =8.11 N

Physics
λ=v/f Speed of sound at 20⁰ v=343.1 m/s f=v/ λ=343.1/0.667=517.4 Hz

machanics
A vertical distance covered for t₁=10 s of accelerated motion is h₁=at₁²/2. The speed at this height is v₁=at₁. The distance covered at decelerated motion during t₂=15- t₁=15-10=5 s is h₂=v₁t₂-gt₂²/2. ...

physics
h=at²/2 a=2h/t²=2•0.76/5.2²=0.056 m/s² Two equations for the loads -m₁a=m₁g-T₁ => T₁=m₁(g+a) ….(1) m₂a=m₂g-T₂,=> T₂=m₂ (g+a) ….(2) and the 3rd equation (on the base of Newton&#...

physics
See the Related Questions below.

physics...
v=sqrt(v₁²+v₂²) = sqrt(4+1) =2.236 m/s s=vt=2.236•5=11.18 m

PHYSIcs!!!
h=gt²/2=> t=sqrt(2h/g). s=vt=> v=s/t=s/ sqrt(2h/g)= =20/sqrt(2•2/9.8)=31.3 m/s

simple PHYISCS
h=gt²/2 t=sqrt(2•5/g)=1.01 s

physics
N= mgcosα F(fr)=μN (a) T₁= mgsinα -F(fr) = = mgsinα - μ(s)mgcosα (b) T₂=F(fr) +mgsinα= =μ(s)mgcosα +mgsinα (c) T₃=F(fr) +mgsinα= =μ(k)mgcosα +mgsinα

Physics
R₂= 116 Ω, ℰ=120 V, P₁=20.7 W, The resistance of light buble R₁=? R=R₁+R₂, I=V/R, R₁=P₁/I²=P₁R²/V²=P₁(R₁+R₂)²/V². R₁²+(2R₂ - V²/P₁)R₁+R₂...

Physics
80•110•10⁶•5.9 =5.192•10¹⁰ W=5.192•10⁷ kWh, 5.192•10⁷•0.11=5.71•10⁶ dollars

Physics
ℰ=IR=IρL/A => I₁ρL₁A = I₂ρL₂A L₂=I₁L₁/I₂=2.3•52/2.9=41.2 m

Physics
R=4.9•10⁹ Ω V=68•10⁻³V I=V/R=68•10⁻³/4.9•10⁹=1.39•10⁻¹¹ A I=q/t =Ne/t N=It/e=1.39•10⁻¹¹•0.45/1.6•10⁻¹⁹=3.9•10⁷

Physics
http://basketball.lifetips.com/cat/59049/vertical-jump/index.html Ans. b

Physics
M=F(fr)•(d/2) =μ•F•(d/2)= 0.15•0.62•1.2•10⁻²/2 =0.000558 N•m

Physics
(a)KE=mv²/2 (b) PE=mgh (c) KE1=KE+PE (d) mv₁²/2= mv²/2+mgh v₁= sqrt(v² +2gh)

CHM 1010
p₁V₁=p₂V₂ p₂=p₁V₁/V₂

Physics-friction problem
ma=F(fr) =μN=μmg a= μg. s=(v₀²-v²)/2a= =(v₀²-v²)/2μg => v=sqrt{v₀²-2μgs}= =sqrt{21.2²-2•0.005•9.8•58.5} = =21.06 m/s. For the 2nd case: s=(v₀²-v²)/2μ₁g= =(21.2...

physics
h=v₀t+gt²/2, gt²+2v₀t-2h=0 Solve for t. v= v₀+gt

Physics - Electric Potential
ΔPE= -Work=-qΔφ= -{(-0.15)•(13-60)} = =-7.05 J

Physics
F=ILBsinα B=F/ILsinα=0.15/67•45•sin51.5⁰=6.35•10⁻⁵ T

PHYSICS
Work =500mgh=500•12•9.8•2=117600 J. You need 117600•100/20 =588000 J 1 kg gives 30•10⁶J, mass of burning fat is 588000/30•10⁶ = 0.0196 kg

Physics
h=at²/2=20•16/2=160 m

Physics
(a) (m₁+m₂)v=m₁v₁+m₂v₂ v₁(a)= {(m₁+m₂)v-m₂v₂}/m₁ (b) (m₁+m₂)v=m₁v₁-m₂v₂ v₁(b)= {(m₁+m₂)v+m₂v₂}/m₁ (c) (m₁+m₂)v=m₁v&...

physics
L=2mvx=2•m•v•(b/2)=2•3•5.41•3.15/2= ...

physics
m₁=m= 70 kg m₃=0.175 kg, v₃=10 m/s Let v₃ be positive and directed to the right 0= - m₁v₁+m₃v₃ v₁=m₃v₃/m₁=0.175•10/70=0.025 m/s (directed to the left) m₃v₃ =(m₂+m₃)v v= m₃...

physics
Q1=Q2+cm•Δt c=(Q1-Q2)/ m•Δt= =(12000-1300)/0.185•(73-22)= =1134 J/kg•℃

Physics
F=qvB sin12⁰ 1.8F=qvBsinφ sinφ/sin12⁰ =1.8 sinφ =sin12⁰•1.8=0.208•1.8=0.374 φ=sin⁻¹0.374=22⁰

PHYSICS help please!!!!!!!!!!
Q(gained) =Q(lost) Q(lost) =Q₁+Q₂+Q₃ Q₁ = c₁m(Δt₁)=2020•1•(120-100) = 40400 J Q₂ = Lm=2256000•1= 2256000 J Q₃=c₂m(Δt₂)=4186•1•(100-85) = 62790 J Q(lost) =40400 +2256000+62790 =235919...

PHYSICS help please!!!!!!!!!!
Q(gained) =Q(lost) Q(gained) =Q₁+Q₂+Q₃ Q₁ = c₁m(Δt₁)=2020•1•(120-100) = 40400 J Q₂ = Lm=2256000•1= 2256000 J Q₃=c₂m(Δt₂)=4186•1•(100-85) = 62790 J Q(gained) =40400 +2256000+62790 =23...

physics
After the 1st rebounce the speed is v/2, after the 2nd rebounce – v/4, after the third rebounce –v/8, => m(v/8)²/2=mgh₁ v=sqrt(128gh₁) =141.7 m/s mv²/2=mgh h= v²/2g=141.7²/2•9.8=1024 m

math
An elementary school collected a total of 240 cans during a food drive. Grade 3 students collected 1/3 of all the cans grade 4 students collected 52 cans, and the rest of the cans were collected by grade 5 students. How many cans did grade 5 collect? A.28 B.80 C.108 D.188

physics
Assume g=10 m/s². The ball upward motion 0=v₀-gt, t=v₀/g=10/10=1 s. The train motion for time 1 s 0= v-at, a=v/t=5/1 = 5 m/s², the distance s= v t -at²/2 =5•1 -5•1²/2 = 2.5 m

Physics. LAST PROBLEM!!
A small metal ball with a mass of m = 82.5 g is attached to a string of length l = 1.72 m. It is held at an angle of θ = 58.5° with respect to the vertical. The ball is then released. When the rope is vertical, the ball collides head-on and perfectly elastically with ...

physics
y=gt²/2 t=sqrt(2y/g) =sqrt(2•920/9.8)=13.7 s

Physics
mv²/2=qU v=sqrt(2qU/m) R=mv/qB= (m/qB)• sqrt(2qU/m)= R3/R4 = sqrt{ 2Um q₄B²/q₃B²2Um)= =sqrt(q₄/q₃) sqrt(4e/3e)=1.15

Physics
T=t/N=11.8/10=1.18 s ω=2π/T=2π/1.18=5.32 rad/s T=mv²/R=mω²R=0.013•5.32 ²•0.93=0.34 N

physics
s=2.08 m m=0.592 kg R=0.0786 m a is the acceleration of the system (m1+m2) m₁a= T₁ m₂a= m₂g-T₂ Iε=M => mR²•a/2R = (T₂-T₁)R, T₂-T₁= ma/2, a= m₂g/( m₁+m₂+m/2)=... a=v²/2s v=sqrt(2as)=....

physics
s=2.08 m m=0.592 kg R=0.0786 m a is the acceleration of the system (m1+m2)

physics
m₁a= T₁ m₂a= m₂g-T₂ Iε=M => mR²•a/2R = (T₂-T₁)R, T₂-T₁= ma/2, a= m₂g/( m₁+m₂+m/2)=... a=v²/2s v=sqrt(2as)=... ω=v/R=...

Physics
Magnetic field of the current-carring wire at distance b from it is B=1/2πb If B=B(Earth)/3= 5•10⁻⁵/3 T, I=2πbB= 2π•0.3•5•10⁻⁵/3 = …

Physics
x=0.45 m, D=0.12 m, I= 2 A, B=2.1 kG=2100G= 0.21 T μ₀=4π•10⁻⁷ H/m B=μ₀nI = μ₀NI/x N=Bx/ μ₀I The length of the wire L=N•πD= Bx πD / μ₀I= =0.21•0.45•π•0.12/4π...

Physics
r1=r2=3 m

Physics
The speed is v=s/t=186/3=62 km/h <65 km/h He can arrive on time.

Physics
1. g =GM/R² g₁=GM/(R+h)² g₁/g=30/100= GM R²/(R+h)²GM => 0.3= R²/(R+h)² Solve for “h” (Earth’s radius is R = 6.378•10⁶ m) 2. mv²/(R+h)= GmM/(R+h)² v=sqrt { GM/(R+h)}

physics
Tsinα =ma Tcosα =mg tanα=a/g=3.2/9.8 = 0.33 α = 18°

Physics
E =kq/r² (a) E(x) = E₁-E₃=kq₁/r² - kq₃/r² = =9•10⁹{(1•10⁻⁶/0.23²)- (7•10⁻⁶/0.23²)}= -1.02•10⁶ V/m (b) An altitude of a triangle is h=0.46cos30=0.46•0.866=0.398 m E(y) ...

physics
1. Flowrate=A•v=21•0.89 = 18.69 m³/s 2. A₁v₁=A₂v₂ v₂=A₁v₁/A₂=21•0.89/17=1.1 m/s

Math/Physics
Q=55034175 J c(ice)=2060 J/kg•℃ λ=330000 J/kg c(water)=4183 J/kg•℃ Q=Q₁+Q₂+Q₃ Q=c(ice)m(12.15-0)+λm+c(water)m(t-0), Q/m=12.15•c(ice) +λ+c(water)•t. Solve for “t”

physics
R=ρL/A R =4ρL/πd² R₁ =4ρL/ (π4d²- πd²) =4ρL/ 3πd² R₁=R/3

Physics
sinα=R/L=0.3/1=0.3 α=17.45° mv²/R=Tsinα …..(1) mg=Tcosα …..(2) Divide (1) by (2) v²/Rg=tanα v=sqrt(R•g•tanα)= =sqrt(0.3•9.8•0.31)=0.27 m/s ω=v/R=0.27/0.3=0.9 rad/s f= ω/2π=0.9/2•3.14=...

physics impulse
1) p=mv=2•4= 8 kg•m/s 2) KE=mv²/2 3) At elastic collision p= - mv 4) Δp= p1-(-p2) =mv – (-mv)= 2mv 5) p(wall)= Δp=2mv 6) F•Δt=mv F= mv/ Δt

physics
We’ll use Kirchhoff’s Current Law: At any node (junction) in an electrical circuit, the sum of currents flowing into that node is equal to the sum of currents flowing out of that node. Node: I₁, I₂ and I₁₂ currents: I₁+I₂-I₁&...

Physics
sini/sinr=n2/n1 n2=n1(sini/sinr)=1.45(sin64.8/sin68.0)=...

Ecology
The effect of humans on the environment is more drastic than that of other species because people have a greater ability to 1. adapt to seasonal changes. 2. learn new skills. 3. alter the environment. 4. eat different types of food.

Biology
Eutrophication refers to 1. the concentration of a toxic substance in the tissues of organisms. 2. depletion of mineral resources as a result of leaching. 3. the warming of the earth’s surface. 4. the damage to the ozone layer that protects the earth from UV light. 5. nut...

Physics
http://en.wikipedia.org/wiki/Torque

Physics
F=Rmv=140•0.0026•510/60=3.1 N

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