8 spherical rain drops of equal size are falling vertically through air with uniform speed of 1m/s. What would be the uniform speed if these drops were to combine to form one large spherical drop?
Falling with uniform speed means the drops have achieved terminal speed.Let us find the radius [math]R[/math] of the bigger drop in terms of the radius [math]r[/math] of the smaller drop. We use conservation of volume to arrive at the result.[math]8.\frac 43 \pi r^3 = \frac 43 \pi R^3 \Rightarrow R = 2r[/math]Now terminal speed is directly proportional to the square of the radius. Thus[math]\frac{v_1}{v_2} = \frac{r^2}{R^2}=\frac{r^2}{(2r)^2} \Rightarrow v_2 = 4v_1[/math]
A mercury drop of radius 1cm is broken into 10^6 drops of equal size. What is the work done? (given T= 35 x 10?
1. calculate change in surface area for breaking of the drop. (volume is constant) 2. surface energy=suface-tension per unit area. 3.nb:heat change calculation data nt given. sorry,(if i give ans, ur braing be unused, plz use brain)
Two objects with a mass of 1 kg and 2 kg are dropped at the same time and height and both have the same shape. Which one reaches the ground earlier?
The answer depends if there exist an air resistance or not.If the objects are dropped in vacuumIn this case both objects will reach the surface at the same time. Near earth surface acceleration due to gravity is same for all objects irrespective of their mass. The force of gravity is greater on the heavier object than on the lighter object. This means that an object with twice the mass will be pulled toward the earth with twice the force. But the acceleration is equal to the force divided by the mass. This means that an object that is twice the mass of another object will be accelerated twice as slowly as the lighter object given the same force.If there is air resistanceIn this case there is a resistance force in addition to force of gravity acting opposite to the direction of motion. Let the object of 1kg be A and object of 2kg be B. Let the air exert a constant resistance force f (in upward direction) which is same in both cases.Acceleration of 1 kg object,[math]a_A = (F-f)/m_A = (m_Ag - f[/math][math])/m_A = g - f/m_A[/math]Similarly acceleration of 2 kg object,[math]a_B = g - f/m_B[/math]Now, [math]m_B > m_A[/math]Hence, [math]a_B > a_A[/math]This shows that object B (2 kg) will reach the surface earlier than A (1 kg).
Help me with this physics problem it's urgent?
If the area of the liquid surface has to be increased work has to be done against the force of surface tension. The work done to form a film is stored as potential energy in the surface and the amount of this energy per unit area of this surface under isothermal condition is the "intrinsic surface energy" or free surface energy density. work done in small displacement dX is dW = f x dX therefore, W=2TLX=TA therefore T = W/A the problem R=1cm = 10^-2 M = 4/3 (pi) R^3 p where p is the surface density This is broken into 10^6 small drops, each of radius r and mass m M = 10^6 m or 4/3 (pi) R^3 p = 10^6 . 4/3 (pi) r^3 p R = 100r r = R/100=10^-2/100 = 1 x 10^-4 energy spent in spraying droplets = TA = T(10^6 x 4(pi) r^2 - 4 (pi) R^2) = 4(pi) T (10^6 r^2 - R ^-3) =4.356 x 10^-3. surface tension of mearcury = T = 35 x 10^-3