ajayb

Resistance R = r1* l/a1 = r2*l/a2 where r1 & r2 are resistivies of two wires. So a1/a2 = r1/r2

13 years ago

The charged particle on the incline stays in the middle because the electrstatic force of repulsion balances the gravitational force acting along the incline downwards. If the distance between the charged particles is x: k Q^2/x^2 = mg*(sin21deg) here k =...

13 years ago

Since the post-lunch momentum is zero, the momentum of larger fish should be equal to the momentum of smaller fish in terms of magnitude but opposite in direction. Speed of the smaller fish v2: v2 = (m1*v1)/m2 = 5*1/1 = 5m/s

13 years ago

a) Stationary observer on the train: As the observer himself is having speed that of the train, Indiana Jones would appear to be running at 10 km/h. b) An observer on the train running in opp. direct to the train at speed of 10km/h: Relative to such an ob...

13 years ago

Momentum of the 'two boys' system shall be conserved (because there is no external force in the horizontal direction) So, m1v1 = m2v2 m1 = 40*10/8 = 50 Kg

13 years ago

Momentum = mass * velocity You know two parameters of the above equation - so you can find the third.

13 years ago

You are solving the problem correctly. Having found the acceleration of each block to be a= 2.4 mtr/sec sq, you need to feed its value in the second equation: T = m1a + m1g sin(theta1) and get the answer as T = 24.0 Nt.

13 years ago

Assume the particle's acceleration is a cm/sec^2 and that it moves a distance of S(t)cms in total time of t secs. Also it covers S(t-1) cms. in (t-1) secs. Since it covers 6cm in the first second => 6 = 0.1 + (1/2).a.1^2 => a=12cm/sec^2. Now, S(t)-S(t-1)=...

13 years ago

X distance moved in first 2 secs. => x=0.2+(1/2).a.2^2= 2a ...(1) velocity at the end of 2 secs.(say v) => v = 0+a.2 = 2a ....(2) y distance covered in the next 2 secs. => y = v.t +(1/2).a.t^2 or y = 2a.2 + (1/2).a.2^2 = 6a ...(2) From, (1) & (2) y = 3x ....

13 years ago

The momentum of the book is m1*v1 in +ve x-direction. Since there is no external force in the horizontal direction, the linear momentum of the 'book and you' system will be conserved. Therefore you would have the same magnitude of momentum in the opposite...

13 years ago

If the amplitude is A, you are required to find the velocity of the mass when it is at a distance A/2 from the equilibrium position (i.e. midway between the extreme and equilibrium positions) use the expression: v = w*sqrt(A^2 - x^2) where v - velocity at...

13 years ago

a) linear speed v = w*r (where w is the angular speed in radians/sec and r is the radial distance from the center) Here, w = 2990 rpm = 49.83 rev.per sec. = 49.83*2*pi rad/sec =313.11 rad/sec & r = 0.59/2 = 0.295 m So, v = 313.11*0.295 = 92.4m/s b) w(fina...

13 years ago

First compute mu(static): When the box just begins to move, the applied force equals mu(s)*m*g So mu(s)= F/m*g = 26.9/(9.1*9.8)= 0.30 Now when the box gets into motion, mu(k) comes in picture and frictional force f = mu(k)*m*g So F - mu(k)*m*g = m*a mu(k)...

13 years ago

Speed(of first wave) = freq*wavelength (m/s) Speed of the second wave is also the same. Since its wavelength is given, find the frequency. Reciprocal of the frequency is the time period in secs.

13 years ago

Consider 'Free body diagram' of one of the beams (say left one) i.e. isolate it and see the forces acting on it. Let's say end B is on the ground and end A is hinged to the other beam of the frame. Which are the forces acting on it? 1.Normal reaction N (v...

13 years ago

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

13 years ago

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.

13 years ago

Because Torque is a cross product of radius vector(r) and Force vector(F). If r =Ax+By+Cz & F = ax+by+cz (where x,y and z are unit vectors in x,y and z directions) then x-component of torque Tx is : Tx = (Bc-Cb) = 3*0-3*4 = -12 Therefore x-comp. of the To...

13 years ago

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 =1...

13 years ago

How the spring is connected to the block?

13 years ago

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

13 years ago

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

13 years ago

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

13 years ago

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 = 79.3...

13 years ago

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)/...

13 years ago

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

13 years ago

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.

13 years ago

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

13 years ago

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...

13 years ago

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

13 years ago

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 pia...

13 years ago

correct option - (C)

13 years ago

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

13 years ago

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 -...

13 years ago

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

13 years ago

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.

13 years ago

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

13 years ago

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

13 years ago

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.

13 years ago

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

13 years ago

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

13 years ago

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

13 years ago

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

13 years ago

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...

13 years ago

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

13 years ago

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

13 years ago

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

13 years ago

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

13 years ago

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 =...

13 years ago

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

13 years ago

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 (...

13 years ago

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

13 years ago

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

13 years ago

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...

13 years ago

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 (u...

13 years ago

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

13 years ago

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

13 years ago

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)

13 years ago

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)

13 years ago

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

13 years ago

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)

13 years ago

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...

13 years ago

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

13 years ago

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

13 years ago

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

13 years ago

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

13 years ago

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.

13 years ago

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 origin...

13 years ago

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 =...

13 years ago

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

13 years ago

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

13 years ago

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.

13 years ago

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 =...

13 years ago

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...

13 years ago

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

13 years ago

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

13 years ago

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

13 years ago

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

13 years ago

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

13 years ago

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

13 years ago

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

13 years ago

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

13 years ago

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.

13 years ago

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.

13 years ago

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

13 years ago

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 dir...

13 years ago

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

13 years ago

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.

13 years ago

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

13 years ago

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-seco...

13 years ago

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

13 years ago

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...

13 years ago

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

13 years ago

You've to consider the motion in vertical and horizontal directions separately. 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...

13 years ago

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

13 years ago

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.

13 years ago