another three physics questions?
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the answer to the first question is 5 m/s.\nF=ma; a=F/m=200N/40kg=5m/s\nthe second answer is 1m/s.\nF=ma; a=F/m=2N/2kg=1m/s\nthe third answer is 1:4.\nF=ma; m=F/a\n(1) M=f/a=F/a=m\n(2) M=f/a=4F/a=4m
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1. potential\n2. kinetic\n3. it left at 0\n4. friction\n5. the greater the mass, the less  distance\n\nYour source of errors and improvements must come from your lab
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1.  Biot savart law\n2.  Momentum equation\n3.  \n4.  Equation for a mangnetic field\n5.  Equation for curvature of the universe\n6.\n7.\n8.  The Einstein Tensor equation\n9.
804155
a) T = 0.5 sec\n\nb) freq = 1/T = 2 Hz\n\nc) w = 2pi*freq = 4 pi\n\nd) w^2 = k/m\nk = m*w^2 = 0.5*16 pi^2 = 8*pi^2\n\ne)  V_max = A*2*pi*f = 0.35*2*pi*2 = 1.4 pi m/sec\n\nf) A_max = A*(2*pi*f)^2 = 0.35*(2*pi*2)^2 = 5.6*pi^2 m/sec^2
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E1=M*g*h, M=(your mb)=32kg, - pot energy of the system at rest,\nh=2m and g=9.8m/s^2.\nE2=kinetic of M=32kg\nE3=kinetic of m=20kg\nE4=kinetic of pulley u=5kg\nE5=potential energy of m at the height h, E5=m*g*h\nThus energy conservation: E1=E2+E3+E4+E5.\nE2+E3=(M+m)*V^2/2, V to be found.\nE4=I*W^2/2, angular speed being W=V/r, r=0.1m, and I=u*r^2/2 for your pulley.\nHence 2*M*g*h=(M+m)*V^2+(u*r^2/2)*(V*r)^2+2*m*g*h,\n(M+m+u/2)*V^2=2*(M-m)*g*h, - does not depend on r.\nV=2.08m/s, don’t believe & check it! – good picture gal – a piece of art.
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cmon...dont u know basic physics...just calculate the resultant of these 2 forces\nanswer= [(132)^2 + (181)^2]^1/2\nif this makes sense!!
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a. V ice = pi(0.12^2-0.10^2)*3 = 0.041 m^3\nb. V ice = pi[(0.10 + x)^2 - 0.10^2]*3 = pi(0.10^2 + 0.20x + x^2 - 0.10^2)*3 = 3pi(0.20x + x^2)\ndV/dt = pi[0.60 + 6(0.02)]dx/dt = 0.000002pi\n= [0.60 + 6(0.02)]dx/dt = 0.000002\ndx/dt = 0.000002/[.60 + 6(.02)]
1086924
The three economic questions? \n\nSheesh...do your own homework!
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I do not know you that well yet...............\n\n\nHow about a physics question or something challenging?
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a)F = MV=  m1v1 +/-  m2v2  \nusing these we get,  95*4.1 + 85*5.5 (since same direction)=857kgm/s = (95+85)* V. so, V = 4.76m/s\nb) MV = M (6) = (9500+m)*6 = 9500*16 + m*0. \nso, (9500+m) = 9500*16/6= 25333.33 kg\nso, m = 25333.33-9500 =15833.33kg\nc)work done against friction= kinetic energy of bullet\nW = FScos(tetha)  since its horizontal, angle is zero, \nand cos0 = 1, so W = FS\nwe know, coeff of friction = tangent of angle between, the frictional force  and normal reaction...\nso, the \nF = normal reaction=(1.1 +0.015)* 9.81=10.938N\nfrictional force= 0.25 *10.938 = 2.7345N\ndistance is 9.5m...\nso W = 2.7345*9.5 = 25.97 J\nthe kinetic energy of bullet has been transferred to the block...\nso the kinetic energy of bullet is 25.97J\n1/2*(0.015)* v^2 = 25.97\nso, v^2 = 25.97*2/0.015=3463.74\nso, v = 58.85m/s\nhope this helps...:)
723260
Question 1:\n\nTorque is a force applied at a distance,\nT = F * r\nWhere T is the torque, F is the force, and r is the radius the force is applied away from the center or rotation.\n\nT = 40 Newtons * .15 meters = 6 Newton meters\n\nA torque causes an angular acceration,\nT = alpha * I\nwhere T is the torque,  alpha is the angular acceleration, and I is the rotational inertia ("moment of inertia").\n\nAn angular acceleration relates to a linear acceleration,\nalpha = a / R\nwhere alpha is the angular acceleration, a is the linear acceleration, and R is the radius of rotation (the wheel's radius in this case).\n\nThe linear acceleration can be found by finding the change in linear velocity with respect to time,\na = delta v / delta t\na = 2.25 m/s / 3 s = .75 m/s^2\n\nalpha = .75 m/s^2 / .5 meters = 1.5 radians per second^2\n\nT = alpha * I \n--> I = T / alpha\nI = 6 Nm / 1.5 rad/s^2 = 4 kg m^2\n \nDo your own homework next time (like on question 2).\nhttp://sln.fi.edu/qa97/spotlight3/spotlight3.html
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There are three objects; 5kg, 9kg, and a cable of negligible mass.\n\nWhen the three move, they must have the SAME acceleration.\n\nTherefore, consider the three as a single object of mass 5 + 9 = 14 kg, with acceleration ‘a’.\n\nThe only force that is pulling the three is the force given by the 9kg object. (The gravitational pull is only on this 9kg.)\n\nThis force is 9 x 9.8 N = F.\n\nSince F = M a,   9 x 9.8 = 14 x a.\n\nFrom this a = 6.3 m/s^2.\n\nAll the three objects are moving with an acceleration 6.3 m/s^2.\n\nIf the body 5kg is moving with an acceleration of 6.3 m/s^2, then what is the force acting on it?\n\nWe know that f = ma; hence, f = 5 x 6.3 = 31.5 Newton.\n\nSince the tension of the rope is the pulling force, tension =31.5 Newton.\n\nThe tension can be calculated using the forces acting on the other end of the cable also.\n\nThe downward force on 9kg is 9 x 9.8 N; the tension is the upward force T. \n\nThe net force acting on 9kg is (9 x 9.8 - 9 x 6.3) = 9 x 3.5 = 31.5 N = T.
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1.because of the void\n2.the air inside is geting hot and the volume expand so the baloon BUM\n3.p=5/2=2.5\n   p=10/2=5   it doubles\n   p=5/1=5 it doubles\n\nhere force is mesured in newtons and area in meters
705574
1) *more info?\n2) microscopy\n3) electron
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you need to identify that there're two parts of the force, one acting horizontally, the other acting vertically.\nSo:\n1) the vertical part is:\n70sin30\n2) tje horizontal part is:\n70cos30
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(a) Displacement is given by the equation:\n\nx = 6 cos[3π(t+(1/9))], so at t = 2 sec\n\nx = 6 cos[3π(2+(1/9))] = 3 m\n\n(b) Velocity is v = dx/dt = -18πsin[3π(t+(1/9))], so at t = 2 sec\n\nv = -18πsin[3π(2+(1/9))] = -48.97 m/s (in the negative x direction)\n\n(c) Acceleration is a = dv/dt = -54π^2cos[3π(t+(1/9))]\n\nso at t = 2 sec, a =  -266.48 m/s^2\n\n(d) Phase of the motion is θ = 3π(t+(1/9)), for t = 2\n\nθ = 19.89 radians\n\n(e) The frequency = ω/2π = 3π/2π = 1.5 revolutions per sec\n\n(f) Period = 2π/ω = 2/3 sec
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1) The gravitational potential energy of water at the top of a waterfall is first converted to kinetic energy (falling) and then converted to both acoustic (sound) and thermal (temperature) energy when it hits the pool of water at the bottom. The temperature of the water in the pool will be slightly higher than the temperature of the water at the top of the waterfall.\n\n2) The thing to remember for this problem is that the energy in a system before some action = the energy in that system after some action:\n\nP0 + K0 + Th0 = P1 + K1 + Th1\nWhere P is gravitational potential energy (mgh), K is kinetic energy (0.5mv^2), and Th is thermal energy (leftover).\n\n22*9.81*4.5 + 0 + 0 = 0 + 0.5*22*2.5^2 + Th1\nSolve for Th1:\nTh1 = 22*9.81*4.5 - 0.5*22*2.5^2 =  902.44 Joules of thermal energy were generated due to friction.\n\n3) Split this problem into two cases, the first 10m and the second 10m. Find the velocity of the crate after the first 10m:\na = F/m = 200/70 = 2.86 m/s^2\nv = sqrt(2*a*d) = sqrt(2*2.86*10) = 7.56 m/s\n\nFor the second 10m, the force of friction on the crate is:\nFf = 0.3*70*9.81 = 206 N\n\nThe net force on the crate is 6N in the direction opposite of movement. The acceleration for the second 10m is:\na = F/m = 6/70 = 0.086 m/s^2\nvf = sqrt(vi^2 + 2*a*d) = sqrt(7.56^2 + 2*-0.086*10) = 7.45 m/s\n\n5) Since watts (W) are measured in J/s and joules (J) are measured in N*m, a watt is the same as N*m/s. Convert lb to N and ft to m to go from horsepower to watts:\n\n550 ft*lb/s * 0.3048 m/ft * 4.45 N/lb = 746 N*m/x = 746 W\n\nPS - You have in your problem description that horsepower is equal to 764 W but that should be 746 W.\n\nI hope this helps!