A fuse (a special cord, burning with a low rate) is used to iginite a bomb. The flame propagates with constant velocity v=0.8 cm/sec. what is the length of the fuse so that the man could run off to a safe distance s=120m, while the flame will not reach the bomb ? the velocity of person `v_(1)=4 m//sec`

A whistle emitting a sound of frequency 440 H z is tied to string of 1.5 m length and rotated with an angular velocity of 20 rad//sec in the horizontal plane. Then the range of frequencies heard by an observer stationed at a large distance from the whistle will (v=330 m//s)

A whistle emitting a sound of frequency 440 H z is tied to string of 1.5 m length and rotated with an angular velocity of 20 rad//sec in the horizontal plane. Then the range of frequencies heard by an observer stationed at a large distance from the whistle will (v=330 m//s)

Choose the correct option: A trolley is moving horizontally with a constant velocity of v m//s w.r.t. earth. A man starts running from one end of the trolley with velocity 1.5v m//s w.r.t. to trolley. After reaching the opposite end, the man return back and continues running with a velocity of 1.5 v m//s w.r.t. the trolley in the backward direction. If the length of the trolley is L then the displacement of the man with respect to earth during the process will be :-

A man holds his umbrella vertically upward while walking due west with a constant velocity of magnitude V_(m) = 1.5 m/sec in rain. To protect himself from rain, he has to rotate his umbrella through an angle phi = 30^(@) when he stops walking. Find the velocity of the rain.

A man holds his umbrella vertically upward while walking due west with a constant velocity of magnitude V_(m) = 1.5 m/sec in rain. To protect himself from rain, he has to rotate his umbrella through an angle phi = 30^(@) when he stops walking. Find the velocity of the rain.

Suppose that a man (60 kg) is standing in an elevator. The elevator accelerates upward from rest at 1m//s^(2) for 2 sec then for further 10 sec it moves with the constant velocity and then decelerates at the same rate for 2 sec (a) For the whole motion, how much work is done by the normal force on the man by the elevator floor? (b) By man's weight (c) what average power is delivered by the normal force for the whole motion (d) What instantaneous power is delivered by the normal force at 7 sec ? (e) at 13 sec ?

Direction : Resistive force proportional to object velocity At low speeds, the resistive force acting on an object that is moving a viscous medium is effectively modeleld as being proportional to the object velocity. The mathematical representation of the resistive force can be expressed as R = -bv Where v is the velocity of the object and b is a positive constant that depends onthe properties of the medium and on the shape and dimensions of the object. The negative sign represents the fact that the resistance froce is opposite to the velocity. Consider a sphere of mass m released frm rest in a liquid. Assuming that the only forces acting on the spheres are the resistive froce R and the weight mg, we can describe its motion using Newton's second law. though the buoyant force is also acting on the submerged object the force is constant and effect of this force be modeled by changing the apparent weight of the sphere by a constant froce, so we can ignore it here. Thus mg - bv = m (dv)/(dt) rArr (dv)/(dt) = g - (b)/(m) v Solving the equation v = (mg)/(b) (1- e^(-bt//m)) where e=2.71 is the base of the natural logarithm The acceleration becomes zero when the increasing resistive force eventually the weight. At this point, the object reaches its terminals speed v_(1) and then on it continues to move with zero acceleration mg - b_(T) =0 rArr m_(T) = (mg)/(b) Hence v = v_(T) (1-e^((vt)/(m))) In an experimental set-up four objects I,II,III,IV were released in same liquid. Using the data collected for the subsequent motions value of constant b were calculated. Respective data are shown in table. {:("Object",I,II,II,IV),("Mass (in kg.)",1,2,3,4),(underset("in (N-s)/m")("Constant b"),3.7,1.4,1.4,2.8):} If an object of mass 2 kg and constant b = 4 (N-s)/(m) has terminal speed v_(T) in a liquid then time required to reach 0.63 v_(T) from start of the motion is :

Direction : Resistive force proportional to object velocity At low speeds, the resistive force acting on an object that is moving a viscous medium is effectively modeleld as being proportional to the object velocity. The mathematical representation of the resistive force can be expressed as R = -bv Where v is the velocity of the object and b is a positive constant that depends onthe properties of the medium and on the shape and dimensions of the object. The negative sign represents the fact that the resistance froce is opposite to the velocity. Consider a sphere of mass m released frm rest in a liquid. Assuming that the only forces acting on the spheres are the resistive froce R and the weight mg, we can describe its motion using Newton's second law. though the buoyant force is also acting on the submerged object the force is constant and effect of this force be modeled by changing the apparent weight of the sphere by a constant froce, so we can ignore it here. Thus mg - bv = m (dv)/(dt) rArr (dv)/(dt) = g - (b)/(m) v Solving the equation v = (mg)/(b) (1- e^(-bt//m)) where e=2.71 is the base of the natural logarithm The acceleration becomes zero when the increasing resistive force eventually the weight. At this point, the object reaches its terminals speed v_(1) and then on it continues to move with zero acceleration mg - b_(T) =0 rArr m_(T) = (mg)/(b) Hence v = v_(T) (1-e^((vt)/(m))) In an experimental set-up four objects I,II,III,IV were released in same liquid. Using the data collected for the subsequent motions value of constant b were calculated. Respective data are shown in table. {:("Object",I,II,II,IV),("Mass (in kg.)",1,2,3,4),(underset("in (N-s)/m")("Constant b"),3.7,1.4,1.4,2.8):} If an object of mass 2 kg and constant b = 4 (N-s)/(m) has terminal speed v_(T) in a liquid then time required to reach 0.63 v_(T) from start of the motion is :