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Wavelength of the wave = 343/110 = 3.118 mLet the distance of the observer from B be 'b' and A be 'a' respectively. then we have a = sq rt[b^2 + 5.7^2] = b[1 + (1/2)*(5.7/b)^2] neglecting higher powers, or a = b + [0.5*(5.7^2)]/b or a-b = 16.245/b = 3.118/2 for least value of b. or b = 16.245...

Neglecting force due to air friction, we have a) 300 = 0.5*9.81*t^2 or t = sq rt[300/(0.5*9.81)] = 7.82 s b) horizontal distance traveled = 60*7.82 = 469 m c) Horizontal component, Vx = 60 m/s and vertical component, Vy is given by Vy = 9.81*7.82 = 77 m/s d) Helicopter has also traveled the same...

The energy of the spring at the end of the impact = 0.5*(0.650/0.0025)*(.15^2) = 2.925 J = 0.5 (M+m) v^2, where M is the mass of block, m is the mass of bullet and v is the velocity of block + bullet just after impact. So we have v^2 = 2.925/[0.5*(0.992+0.008)] = 5.85 or v = 2.42 m/s^2 Let the...

The energy of the spring at the end of the impact = 0.5*(0.650/0.0025)*(.15^2) = 2.925 J = 0.5 (M+m) v^2, where M is the mass of block, m is the mass of bullet and v is the velocity of block + bullet just after impact. So we have v^2 = 2.925/[0.5*(0.992+0.008)] = 5.85 or v = 2.42 m/s^2 Let the...

The answer is so obvious. And it is D)