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ajayb
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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.
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
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
I just answered your questions in the previous post. You are welome to ask again in case of any doubt.
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
Remember Newton's third law of motion. The reactionary force on the car would be of the same magnitude i.e. 1580N.
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
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
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.
You have been given the linear expansion coefficient - find increase in radius R and then find the disk's MI.
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
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
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.
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.
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
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
That is correct. This is a case of inelastic collision where linear momentum is conserved because there is no external force. So, Pi = Pf
E=V/d = 60/.03 = 2000 v/m W = qV = 5*12.45 = 62.25 J
h+g*t^2/2 t=sqrt(2h/g) =sqrt(2*51/9.8) = 3.2 sec
U^2 = 2*a*s s = U^2/2*a = (16.66)^2/(2*30*9.8) = 0.47m or 47 cm
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
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
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
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 =
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.
Work reqd.(W)= gain in potential energy = m*g*h Solve it to get W in joules
F=Pressure*area = 2*10^5(N/m^2)*pi*1.5*10^-6(m^2) = 3*pi*10^-1 N
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 =
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
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.
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
Work required = Gain in PE = m*g*h = 1460*9.8*14Sin12.5 J (h-vertical height gained)
(1/2)mV^2 = (1/2)kx^2 V = x*sqrt(k/m) = 0.6*sqrt(650/0.1)m/s
KE gained = PE lost (1/2)mv^2 = mgh v = sqrt(2gh)= sqrt(2*9.8*0.54)m/s
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
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)
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
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)
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)
Can't be solved without the diagram showing the incline gradient
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
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
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
tension in the string = m*V^2/r max tension is 70N so Vmax = sqrt(70*r/m)
use the law: T^2 proportional to r^3
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
use T^2 = 4*pi^2*r^2/GM T and r are given, G -Universal constant of gravitation. Compute M - mass of star
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 =
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
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
