Wednesday

September 28, 2016
Total # Posts: 117

**physics**

How can there be acceleration if the body is moving at constant velocity?

*August 25, 2012*

**physics**

(PE+KE)initial = (PE+KE)final mgh + mv0^2/2 = 0 + mv^2/2 => h = (v^2-v0^2)/2g

*August 23, 2012*

**PHYSICS**

The current through each bulb is the voltage across it divided by its resistance. In this case the bulbs are in parallel and voltage across them is 2.4V.

*August 23, 2012*

**PHYSICS**

The series resistor R should be chosen so as to have a voltage drop of 7.6V(10-2.4)across it. The currents in the two bulbs: i1=2.4/3.0 = 0.8A i2=2.4/2.5 = 0.96A Total current in the circuit = i1+i2 = 1.76A R = 7.6V/1.76A = 4.3 Ohms

*August 23, 2012*

**physics**

First find K of the spring: Mg = Kx here, x = elongation of the spring = 13.40 - 12.00 = 1.40 cm = 0.014 m K = Mg/x = 3.15*9.8/0.014 = ?? Having got K, find X from: KX^2/2 = 10.0J where X is the elongation when stored PE is 10.0 Joules. The total length of spring then would be...

*August 22, 2012*

**physics**

use the formula: V^2 = U^2 + 2*a*s here, V=0(final speed); U= Initial speed (convert units mi/hr to m/s) a = acceleration s = 108 m Find a in m/s^2 - your answer will be a negative value indicating deceleration.

*August 22, 2012*

**Physics**

Initial momentum of boat+Sally system is=(125+65)*5= 190*5=950 Kg m/s Sally's speed is 2.0m/s relative to boat i.e. w.r.t.a stationary observer it is (2+V)m/s where V is the speed of the boat when Sally walks. Since there is no external force acting on our system, the ...

*August 6, 2012*

**physics**

F = k q^2/r^2 = 9x10^9*1x10^-12/16x10^-4 = 5.6 N 5.6=m*9.8 => m=5.6/9.8=5.7 Kg

*August 6, 2012*

**Physics**

Yes, you did it correctly. In order to understand phasor diagram, read your text book. In a purely inductive load, the current lags voltage by 90 deg. In this case (R+L load)the phase angle will be less than 90 - you can find it out once you read the text.

*August 5, 2012*

**physics**

The string closest to the axis of rotation would have the maximum tension. Draw the free body diagram and it will be clear that the tension in the closest string balances not only the centrifugal force on the attached ball but also the tension in the string next in the chain.

*August 5, 2012*

**Dynamics**

The kinetic energy of the car will be dissipated as heat due to the frictional forces. KE=(1/2)mV^2=1500*36/2=27000 J The work done by frictional forces=F*d Now, F*d=KE F=KE/d=27000/5=5400 N

*August 4, 2012*

**Physics**

Use PV=nRT to find the pressure of the gas in the container at 20 deg.C. All the variables except for P are given to you in the problem. The heating process is isobaric -so, the pressure would not change. But its volume would increase which you can find by using the equation: ...

*August 4, 2012*

**dyanamics**

(a) is correct. Force, momentum and velocity are vectors. Energy and work are scalar quantities.

*August 3, 2012*

**Physica**

i=q/t => q=i*t=5.5*120(sec) =660 Coul.

*August 3, 2012*

**science**

You have not mentioned the units in the problem - I trust the mass is 15.5 Kg and Force =38.0 N. Fx= 38*Cos42 .....in hor. direction Fy= 38*Sin42 .....in vertical downward direction For equilibrium in vertical dir.- Normal force N = mg+Fy N=15.5*9.8 + 38*Sin42 =151.9 + 25.4 =...

*August 3, 2012*

**Physica**

Pressure at the bottom of a water column of height h = d*g*h where, d=density of water(1000Kg/m^3) g=9.8 m/s^2 & h=30 m Calculate P yourself.

*August 3, 2012*

**Dynamics**

Wi= 1200 rev/min = 20 rev/sec = 20*2*pi rad/sec Wf = 0; alpha= -1.5 rad/s^2 Wf = Wi + alpha*t 0 = 40*pi - 1.5*t t = 40*pi/1.5 = 84 sec Now, theta = Wi*t + (1/2)*alpha*t^2 Plug in the values in this eqn. to get theta - the rotation of the wheel (in radians)before it comes to ...

*August 1, 2012*

**physics constant angular acceleration**

4 rev.= 4*2*pi radians use the formula for angular motion: theta = W0*t + (1/2)*alpha*t^2 .....(1) here theta = 4*2*pi W0 = 4.3 alpha = 3.75 Solve the quadratic eqn.(1)to get the value of t (time) The data on radius of the wheel has no use in solving this problem.

*July 31, 2012*

**Physics**

Assume a thermal equilibrium temp. T and find the heat absorbed by ice (say Q1)to raise its temp.from -35 deg. to T deg. Similarly, find heat released by water (say Q2) when its temp. lowers from 45 deg. to T deg. Putting Q1 = Q2 will give you T in deg. C. Note: Do not forget ...

*July 31, 2012*

**Physics**

Follow the following steps: 1. Calculate the ice block's mass from its volume and density. 2. Compute heat required to raise its temp. from -19.3 degC to 0 deg - Q1 3. Consider latent heat required to change the state from ice at 0 deg. to water at 0 deg. - Q2 3. Compute ...

*July 31, 2012*

**Physics**

Q=m*s*T I hope you can compute Q (heat required)

*July 31, 2012*

**physics**

use the formula: V^2=U^2-2*a*s where V is final velocity =0 U is initial velocity a is accn. = 12 m/s^2 s is height attained Compute s from the formula.

*July 29, 2012*

**PHY: To AJAYB, follow up on previous response**

I just answered your questions in the previous post. You are welome to ask again in case of any doubt.

*July 29, 2012*

**PHysics**

w^2 = k/m => k = m*w^2 =15*32^2 =15,360 n/m E =(1/2)k*A^2 =(1/2)*15360*(.20)^2 =307 J

*July 28, 2012*

**Physics**

You've to consider the motion in vertical and horizontal directions seperately. A)Vertical (y) dir.=> Vy^2 = Uy^2 + 2*(-g)*s = (10Sin45)^2 + 2*(-9.8)(-10) = 50 + 196 = 246 So, Vy= 15.68 m/s (Vy is the vertical component of the projectile when it hits the ground) B) ...

*July 28, 2012*

**Physics**

Remember Newton's third law of motion. The reactionary force on the car would be of the same magnitude i.e. 1580N.

*July 28, 2012*

**Physics- Repost (Answer did not make sense)**

My mistake - I wrongly considered the battery voltage to be 6V instead of 12V. B) The capacitor voltage will reach 6V (half of its final voltage) in 0.693*TC time. You can also use the equation: V = V0{1-e^(-t/RC)} where V=6 and V0=12. You'd get t= 8.316 ms C)Find q0 by ...

*July 28, 2012*

**Physics- Repost (Answer did not make sense)**

A) Your answer is correct - the time constant of the circuit is indeed 12ms. B) The time required to reach the final voltage by a capacitor is roughly 4*TC. In this case,the capacitors would charge up to full battery voltage(6V)in 48 ms. C) t=10 nano-seconds is such a short ...

*July 28, 2012*

**physics**

T = 2*pi*sqrt(m/k) so k = 4*pi^2*m/T^2 = 444.5 N/m E = (1/2)*k*A^2 = (1/2)*444.5*0.155^2 = 5.34 J

*July 27, 2012*

**Physics**

You might have forgotten to include the latent heat - heat required to convert the state from ice to water w/o change in temp. Consider this factor and you could get the correct answer.

*July 20, 2012*

**physics**

You have been given the linear expansion coefficient - find increase in radius R and then find the disk's MI.

*July 20, 2012*

**physics**

1.You apply a force F on the book in horizontal direction towards the wall. 2. The normal reaction of the wall on the book will be (say) N. For equilibrium in horizontal direction => F = N 3. The weight of the book M*g will act on the book in downward direction. 4. The ...

*July 19, 2012*

**Physica**

Work done by the man = F*d = 10*5 = 50 J work done by frictional force = mu*mg*d = - 0.2*5*9.8*5 = - 49 J Net work done on the box = 50-49= 1 J

*July 17, 2012*

**Physics**

You can consider the surface area of the sphere (4*pi*r^2) as the surface area of the steel plate. Multiply it with the thickness i.e. 0.5in to get volume of the steel material.

*July 9, 2012*

**Physics**

In the case of completely elastic collision, not only momentum is conserved but KE is also conserved. Considering these two principles, you get two equations from which you can find V1 and V2 in terms m1,m2 and U1.

*July 9, 2012*

**Physics**

E=V/d = 60/.03 = 2000 v/m W = qV = 5*12.45 = 62.25 J

*July 5, 2012*

**Physics**

during the slide down the hill: use a=gSin9.8-mu*gCos9.8 to find vel. at the base of the incline. For the horizontal part accleration is -mu*g - to find the distance covered. Figure it out based on the above hint

*July 5, 2012*

**physics**

That is correct. This is a case of inelastic collision where linear momentum is conserved because there is no external force. So, Pi = Pf

*July 4, 2012*

**physics**

1. Max. compression is 20cm - so the amplitude A is 0.2m. 2.(1/2)mV^2 = (1/2)kA^2 V^2 = kA^2/m = 5.2*0.04/0.7 V=0.54m/s 3.Speed at X = sqrt(k/m)sqrt(A^2-X^2) =2.72sqrt(.04-0.0144) = 2.72*0.16= 0.43m/s Acc. at X= (k/m)x= (5.2/0.7)*0.12 =0.89 m/s^2

*July 4, 2012*

**Physics**

h+g*t^2/2 t=sqrt(2h/g) =sqrt(2*51/9.8) = 3.2 sec

*July 3, 2012*

**physics**

F=Pressure*area = 2*10^5(N/m^2)*pi*1.5*10^-6(m^2) = 3*pi*10^-1 N

*July 3, 2012*

**Physics**

Work reqd.(W)= gain in potential energy = m*g*h Solve it to get W in joules

*July 3, 2012*

**Physics**

F1*Cos32= F2*Cos45 F2=145*Cos32/Cos45 = 174N Any value of F2 less than 174N would result in the net force in forward direction.

*July 3, 2012*

**Physics (work and springs)**

A) The elastic potential energy stored in the spring would convert into KE of the box. This KE would be dissipated while working against the frictional force. So work done = (1/2)Kx^2 = 800*0.5^2/2 = 100 Joules B) Frictional force F= mu*mg F = 0.4*4*10 = 16 N Work done = F*d d...

*July 2, 2012*

**Physics**

Assume: P0-pressure at 4500m above sea level. P1-pressure at sea-level P2- pressure at depth of 20.6m P2=P1+dgh = P1 + 1024*9.8*20.6 = P1+2.06x10^5 Pa .......(1) P0=P1-dgh =P1 - 1.3*9.8*4500 =P1 - 0.57x10^5 Pa .......(2) [It is assumed density of air near earth's surface ...

*July 2, 2012*

**Engineering Physics**

PEi = - GMem/4Re = KEf + PEf = KEf + [- GMem/2Re] So KEf = GMem/2Re - GMem/4Re or (1/2)m v^2 = GMem/2Re v^2 = G*Me/Re

*July 2, 2012*

**College Physics**

U^2 = 2*a*s s = U^2/2*a = (16.66)^2/(2*30*9.8) = 0.47m or 47 cm

*July 2, 2012*

**Physics**

Kg is unit for mass and Newton is the unit for weight.( weight is a force W=mg) Max. wt.that can be lifted = Pressure * area = 1.01*10^5*16.7*3.14*0.123^2 N =80126 N

*July 2, 2012*

**Physics**

KE gained = PE lost (1/2)mv^2 = mgh v = sqrt(2gh)= sqrt(2*9.8*0.54)m/s

*July 2, 2012*

**Physics**

(1/2)mV^2 = (1/2)kx^2 V = x*sqrt(k/m) = 0.6*sqrt(650/0.1)m/s

*July 2, 2012*

**Physics**

Work required = Gain in PE = m*g*h = 1460*9.8*14Sin12.5 J (h-vertical height gained)

*July 2, 2012*

**physics**

Loss in KE = Gain in elastic PE of the spring. (1/2)mV^2 = (1/2)kx^2 find K from this equation. Remember to convert the units of V to m/s

*July 2, 2012*

**Physics**

consider a sytem comprising of Sunbather + raft. No external force on this system.Therefore thr Center of mass of the system will remain stationary. If she walks to the west, the raft will move to the east by a distance that will keep the CM at the original location. No ...

*July 1, 2012*

**Physics Angular displacement**

Car's displacement= S = ut + at^2/2 = 25*6 + 2*6^2/2 = 150 + 36 = 186m One rev. of wheel covers = 2*pi*r No. of revolutions = 186/(2*pi*0.25) = 118.4 rev.

*July 1, 2012*

**science**

Do vector addition of the two velocities to obtain boat's velocity V w.r.t.river bank: V = sqrt(12^2+5^2) = 13 Km/hr and tan theta = 5/12; (theta - angle between V and north)

*July 1, 2012*

**science**

1) Speed of man w.r.t.escalator = L/t1 2) Speed of escalator w.r.t.ground = L/t2 3)Speed of man w.r.t.ground = (L/t1)+(L/t2) Time taken by man to cover distance(L) = L/(L/t1)+(L/t2) = t1*t2/(t1+t2)

*July 1, 2012*

**science**

1) change in speed: 12-5 = 7m/s 2) V1=5i & V2=12j Change in vel.= V2-V1 = 12j-5i its magnitude = sqrt(12^2+5^2) = 13m/s^2 3) x-dir.: 0=5+Ax*5 => Ax= -1m/s^2 y-dir.: 12=0+Ay*5 => Ay= +12/5=2.4m/s^2 So A = -1i+2.4j Mag. of av. acc.=sqrt(Ax^2+Ay^2)

*July 1, 2012*

**science**

Yes the bullet will hit the monkey as both fall with downward accleration of g m/s^2. Hint: The bullet will hit the monkey at a ht. S above the ground given by: S = V*Sin(theta)*t - (1/2)*g*t^2 {V-initial velocity at an angle theta with horizontal} During this time t, the ...

*July 1, 2012*

**Engineering Physics**

Since the platform is 3m above ground(vertical height), the gravitational PE gained would be m*g*h - no need to multiply by Sin 30

*July 1, 2012*

**physics**

The truck is moving at constant velocity - means zero acceleration - that can happen only when the net force acting on the truck is zero.The frictional force opposes the motion and its magnitude is same as that produced by the engine for forward motion (under constant velocity...

*June 30, 2012*

**physics**

1. The apparent frequecy should be 20kHz i.e.lower than the actual freq. of 21kHz. Therefore the sound source should move away from you. If the rider speed is U - F' = v*F/(v+U) 20*10^3= 330*21*10^3/(330+U) 330+U= 330*21/20 U = 330*[21/20 -1]= 330*1/20 = 16.5m/s (min....

*June 30, 2012*

**dynamic**

How much is the acceleration? Not stated in the problem! anyway, you can use the formula: V^2 = U^2 + 2*a*s here a = accleration, U=0 and s = 2-1= 1m

*June 30, 2012*

**Physics**

Solution to a similar problem has been posted earlier which please check. B) The total mechanical energy = KE + PE of the object at the surface of Eris will remain constant and will be equal to PE at the max. height h (above the surface). The object's KE will be zero at ...

*June 29, 2012*

**Physics**

Can't be solved without the diagram showing the incline gradient

*June 29, 2012*

**Physics**

use T^2 = 4*pi^2*r^2/GM T and r are given, G -Universal constant of gravitation. Compute M - mass of star

*June 29, 2012*

**Physics**

use the law: T^2 proportional to r^3

*June 29, 2012*

**Physics**

Since the box is moving through sea water at a constant speed, the net force on it must be zero. The forces are: i) W= mg its weight downwards ii) Fb - buoyant force upwards iii) Ff - frictional force offered by water acting upwards. now W = mg = V*Db*g (Db-density of box) and...

*June 29, 2012*

**Physics**

tension in the string = m*V^2/r max tension is 70N so Vmax = sqrt(70*r/m)

*June 29, 2012*

**physics**

I suppose you have to find time for the bomb to strike the ground. h = (1/2)*g*t^2 h = 3000ft g = 32 ft/s^2 find out t The data - jet's horizontal velocity and mass of the bomb - is irrelevant for the above computation of time

*June 29, 2012*

**Physics**

a) s1= vt - for motorist s2= (1/2)*a*t^2 - for police officer when the cop cathes up the motorist: s1 = s2 So vt = (1/2)*a*t^2 15*t= (1/2)*3*t^2 t = 15*2/3 = 10 sec b) v = u + a*t = 0 + 3*10 = 30 m/s c) distance s1=s2= 15*10 = 150m

*June 29, 2012*

**physics**

W2^2 = W1^2 + 2*alpha*theta 13.5^2 = 22.0^2 +2*alpha*13.8 alpha = (13.5^2 - 22.0^2)/2*13.8 = - 10.9 rad/s^2 So, magnitude of angular acceleration = 10.9 rad/s^2

*June 28, 2012*

**Physics**

A) Escape vel. = Ue = sqrt(2GM/R) ...(1) also g = GM/R^2 So GM/R = gR .....(2) From(1) & (2) Ue = sqrt(2gR) = sqrt(2*0.77*1200,000) = 1360 m/s B) u = Ue/2 = (1/2)*sqrt(2GM/R) Now KEi+PEi = PEf+KEf (conservation of mechanical energy) or (1/2)mu^2 - GMm/R = -GMm/(R+h)+0 or GM/4R...

*June 28, 2012*

**Physics**

The total load is 980+230 = 1210N but how it shall be shared by the supporting poles would depend on the location of the painter on the board. The question can not be answered in absence of this information/diagram.

*June 27, 2012*

**Physics**

F= m*a = m*(v-u)/t v is the velocity after time t(5.5s) u is the initial velocity = 0 Plug in the values to find the av. net force F

*June 26, 2012*

**Physics**

F = dP/dt = 0.70(6.9-0)/0.15 = 32.2 N

*June 26, 2012*

**physics**

Resultant force=160-40-60=60 N to the left.

*June 26, 2012*

**Physics**

g(n) = G*Mn/Rn^2 & g(e)= G*Me/Re^2 Mn=d*(4/3)*pi*Rn^3 (Mn->new planet's mass) Me=d*(4/3)*pi*Re^3 (Me-> Earth's mass) So, g(n)/g(e) = Rn/Re .....(1) Now, Mn/Me = 2 (given) So Rn^3/Re^3 = 2 or Rn/Re = 2^1/3 ......(2) From (1) & (2) g(n)/g(e)= Rn/Re = 2^1/3 So, if ...

*June 26, 2012*

**physics**

use the formula: S = ut +(1/2)*a*t^2 S - distance covered =1.25m u - initial velocity = 75m/s^2 a - acceleration = 25m/s^2 t - time taken to cover the distance

*June 26, 2012*

**Physics.**

Try E = mc^2 Take mass of electron and positron to be 9.1x10^-31Kg and 3x10^8m/s as speed of light.

*June 26, 2012*

**physics**

A) position: x(2)= (2*2-3)^2= 1m v = dx/dt = 2(2t-3)*2 so v(2)= 4 m/s a = dv/dt = 8 m/s^2 B) v at origin: x=0=(2t-3)^2 so t=3/2 at t=3/2, v = 4(2*3/2-3)= 0

*June 26, 2012*

**phsc**

If the external force on the satellite is zero, the linear momentum will be conserved. This would be the case for a satellite out in the space where the gravitational pull of earth is nearly zero. In such a situation the fragments will fly off in all the directions obeying the...

*June 26, 2012*

**phsc**

The larger contact time means longer duration of the impulse imparted to the golf ball. (Impulse I = F*t) Largere impulse causes larger change in the momentum of the ball. And the distance covered is directly proportional to the momentum imparted.

*June 26, 2012*

**phsc**

Since momentum in each case is the same, the damage will be identical. Both the options are equally bad.

*June 26, 2012*

**Physics: Theory**

correct option - (C)

*June 26, 2012*

**Physics**

1) r = m*v/(q*B) r(a)= m(a)*2v/(q(a)*B).....(1) r(p)= m(p)*v/(q(p)*B).....(2) r(a)/r(p)= m(a)*2v*q(p)/[m(p)*v*q(a)] = 4m(p)*2*q(p)/[m(p)*2q(p)] = 4*2/2 = 4 2) Ma*Va/[Qa*B] = Mp*Vp/[Qp*B] So, Va = 4.4*2/4 = 2.2m/s

*June 25, 2012*

**physics**

a)As the acceleration is zero, the force exerted by the man equal the comp. of gravitational force on the body along the incline. F = mg*Sin22 b)Wm = F.d = - 3.8*F ( -ve because F and d are in opposite directions) c) Wg = +3.8*mg*Sin22 d) Net force on piano = 0. So Wnet = 0

*June 25, 2012*

**physics**

What is the unit of work given in the problem? Is it Joules? W = mgh should give you the height h.

*June 25, 2012*

**Physics**

The moments of the gravitational forces should balance each other to prevent plank's rotation(about the wedge). a)Taking moments about the wedge position: 40g*X = 60g*10 X = 15m from the wedge b) You can't balance with a 10Kg mass. 40g*10 = 10g*X X= 40m - beyond the ...

*June 25, 2012*

**Physics**

kx = mg k = mg/x = 2*10/0.25 = 80 N/m

*June 25, 2012*

**Physics**

Try G*Ms*Mu/R^2 = Mu*Vu^2/R here Ms = mass of Sun Mu = mass of Uranus Vu = Speed of Uranus R =Orbital radius Also 2*pi*R = Vu*T will give value of Vu which when plugged in the first equation will give R.

*June 25, 2012*

**Physics**

First find the orbital speed(V)by equating force of attraction (between the two bodies) with the centrifugal force. Having found the speed you can find the period T= 2*pi*R/V

*June 25, 2012*

**Physics**

S=(1/2)gt^2 since u=0 4000=(1/2)*10*t^2 t=sqrt(800)secs - time to reach ground So hor. distance covered= 80*sqrt(800) = 2263m

*June 22, 2012*

**physics**

The force required to move the crate at constant speed should just be greater than the frictional force. F (ext.)= 180 N Work done = F*s = 180*5.5 = 990 J

*June 21, 2012*

**Physics(Please help)**

W = p*dV = p(Vf - Vi)--for isbaric p is constant. Plug in values of W,p and Vi to find Vf

*June 21, 2012*

**Physics**

let q1= -2C, q2= +5C, q3= -4C also r13=10+4=14cm=0.14m & r23= 4cm = 0.04m a) Force on q3 due to q2: F32= k*q3*q2/(r23)^2 towards left b)Force on q3 due to q1: F31= k*q3*q1/(r13)^2 towards right c) Net force on q3 = F32-F31

*June 21, 2012*

**Physics**

Assume the string makes an angle theta with the horizontal and tension in the string is T. T*Cos theta = m*v^2/r (centrifugal force) and T*Sin theta = m*g By dividing: Tan theta = g*r/v^2 = 9.8*2.2/7.2^2 = 0.41 So Theta = 22.5 deg Now T=m*g/Sin22.5 = 79.3N

*June 21, 2012*

**Physics**

Assume the rope makes an angle theta with horizontal and tension in the rope is T. Let the normal reaction at the pole and wall contact point is N (in horizontal dir.) T*Cos theta = N ....(1) T*Sin theta + mu*N = Mg ........(2) So, Tan theta = (M*g-mu*N)/N or, Tan theta = M*g/...

*June 21, 2012*

**Physics**

How much is the ramp's inclination with the horizontal?

*June 19, 2012*

**Physics**

v^2 = u^2 - 2*a*s here, v = 0 (final speed) u = 24 m/s a = 0.065 m/s^2 Solve above to obtain s - distance covered while slowing down.

*June 19, 2012*

**Physics**

Since the skier is pulled at a constant speed, the pulling force (for one skier)is given by: F = m*g[sin theta + mu_k*cos theta] = 80*10[sin25+0.15*cos25] = 446.8N (along the incline upwards) Work done by F in moving distance of 230m: W=F*x = 446.8*230 =102,776 Joules Power ...

*June 18, 2012*