Raindrops of raidus 1mm and mass 4 mg are falling with a speed of 30 m/s on the head of a bald person. The drops splash on the head and come to rest. Assuming equivaletly that the drops cover a distance equal to their radii on the head, estimate the force exerted by the each drop on the head.
Raindrops of raidus 1mm and mass 4 mg are falling with a speed of 30 m/s on the head of a bald person. The drops splash on the head and come to rest. Assuming equivaletly that the drops cover a distance equal to their radii on the head, estimate the force exerted by the each drop on the head.
Text Solution
Verified by Experts
Given, `r=1 mm=10^-3 m,
`m=4 mg = 4xx 10^-6 kg`
`S= 10^-3 m,`
`V=0, u=30 m/s`
We knw, `a= (v^2-u^2)/(2xx10^-3)= (-30xx30)/(2xx10^-3)
` =-4.5xx10^5 ms^-2` (decelerating)
Taking magnitude only, ltBrgt deceleration = `4.5xx10 m/s^2`
`:. force, F=4xx10^-6xx4.5xx10^5`
`=1.8N`.
`m=4 mg = 4xx 10^-6 kg`
`S= 10^-3 m,`
`V=0, u=30 m/s`
We knw, `a= (v^2-u^2)/(2xx10^-3)= (-30xx30)/(2xx10^-3)
` =-4.5xx10^5 ms^-2` (decelerating)
Taking magnitude only, ltBrgt deceleration = `4.5xx10 m/s^2`
`:. force, F=4xx10^-6xx4.5xx10^5`
`=1.8N`.
Similar Questions
Explore conceptually related problems
A raindrop of mass 1.00 g falling from a height of 1m hits the ground with a speed of 50 ms^(-1) Calculate (a) the loss of PE of the drop (b) the gain in KE of the drop ( c ) Is the gain in KE equal to loss of PE ? If not why ? Take, g =10 ms^(-2)
A student performs an experiment to determine how the range of a ball depends on the velocity with which it is projected. The "range" is the distance between the points where the ball lends and from where it was projected, assuming it lands at the same height from which it was projected. It each trial, the student uses the same baseball, and launches it at the same angle. Table shows the experimental results. |{:("Trail","Launch speed" (m//s),"Range"(m)),(1,10,8),(2,20,31.8),(3,30,70.7),(4,40,122.5):}| Based on this data, the student then hypothesizes that the range, R, depends on the initial speed v_(0) according to the following equation : R=Cv_(0)^(n) , where C is a constant and n is another constant. The student speculates that the constant C depends on :- (i) The angle at which the ball was launched (ii) The ball's mass (iii) The ball's diameter If we neglect air resistance, then C actually depends on :-
Two balls of equal mass have a head-on collision with speed 6 m//s . If the coefficient of restitution is (1)/(3) , find the speed of each ball after impact in m//s
It is a common observation that rain clouds can be at about a kilometer altitude above the ground. (a) If a rain drop falls from such a height freely under gravity, what will be its speed? Also calculate in km/h (g = 10 m//s^(2) ). (b) A typical rain drop is about 4 mm diameter. Momentum is mass x speed in magnitude. Estimate its momentum when it hits ground. (c) Estimate the time required to flatten the drop. (d) Rate of change of momentum is force. Estimate how much force such a drop would exert on you. (e) Estimate the order of magnitude force on umbrella. Typical lateral separation between two rain drops is 5 cm. (Assume that umbrella is circular and has a diameter of 1 m and cloth is not pierced through.)
Statement I: If a sphere of mass m moving with speed u undergoes a perfectly elastic head-on collision with another sphere of heavier mass M at rest ( M gt m ), then direction of velocity of sphere of mass m is reversed due to collision (no external force acts on system of two spheres). Statement II: During a collision of spheres of unequal masses, the heavier mass exerts more force on the lighter mass in comparison to the force which lighter mass exerts on the heavier one,
A circus wishes to develop a new clown act. Fig. (1) shows a diagram of the proposed setup. A clown will be shot out of a cannot with velocity v_(0) at a trajectory that makes an angle theta=45^(@) with the ground. At this angile, the clown will travell a maximum horizontal distance. The cannot will accelerate the clown by applying a constant force of 10, 000N over a very short time of 0.24s . The height above the ground at which the clown begins his trajectory is 10m . A large hoop is to be suspended from the celling by a massless cable at just the right place so that the clown will be able to dive through it when he reaches a maximum height above the ground. After passing through the hoop he will then continue on his trajectory until arriving at the safety net. Fig (2) shows a graph of the vertical component of the clown's velocity as a function of time between the cannon and the hoop. Since the velocity depends on the mass of the particular clown performing the act, the graph shows data for serveral different masses. If a clown holds on to hoop instead of passing through it, what is the position of the cable so that he doesn't hit his head on the ceiling as he swings upward?
Two stars each of one solar mass (= 2 xx 10^(30) kg) are approaching each other for a head on collision. When they are a distance 10^9 km, their speeds are negligible. What is the speed with which they collide ? The radius of each star is 10^4 km. Assume the stars to remain undistorted until they collide. (Use the known value of G).
When an object moves through a fluid, as when a ball falls through air or a glass sphere falls through water te fluid exerts a viscous foce F on the object this force tends to slow the object for a small sphere of radius r moving is given by stoke's law, F_(w)=6pietarv . in this formula eta in the coefficient of viscosity of the fluid which is the proportionality constant that determines how much tangential force is required to move a fluid layer at a constant speed v, when the layer has an area A and is located a perpendicular distance z from and immobile surface. the magnitude of the force is given by F=etaAv//z . For a viscous fluid to move from location 2 to location 1 along 2 must exceed that at location 1, poiseuilles's law given the volumes flow rate Q that results from such a pressure difference P_(2)-P_(1) . The flow rate of expressed by the formula Q=(piR^(4)(P_(2)-P_(1)))/(8etaL) poiseuille's law remains valid as long as the fluid flow is laminar. For a sfficiently high speed however the flow becomes turbulent flow is laminar as long as the reynolds number is less than approximately 2000. This number is given by the formula R_(e)=(2overline(v)rhoR)/(eta) In which overline(v) is the average speed rho is the density eta is the coefficient of viscosity of the fluid and R is the radius of the pipe. Take the density of water to be rho=1000kg//m^(3) Q. What is the viscous force on a glass sphere of radius r=1mm falling through water (eta=1xx10^(-3)Pa-s) when the sphere has speed of 3m/s?
When an object moves through a fluid, as when a ball falls through air or a glass sphere falls through water te fluid exerts a viscous foce F on the object this force tends to slow the object for a small sphere of radius r moving is given by stoke's law, F_(w)=6pietarv . in this formula eta in the coefficient of viscosity of the fluid which is the proportionality constant that determines how much tangential force is required to move a fluid layer at a constant speed v, when the layer has an area A and is located a perpendicular distance z from and immobile surface. the magnitude of the force is given by F=etaAv//z . For a viscous fluid to move from location 2 to location 1 along 2 must exceed that at location 1, poiseuilles's law given the volumes flow rate Q that results from such a pressure difference P_(2)-P_(1) . The flow rate of expressed by the formula Q=(piR^(4)(P_(2)-P_(1)))/(8etaL) poiseuille's law remains valid as long as the fluid flow is laminar. For a sfficiently high speed however the flow becomes turbulent flow is laminar as long as the reynolds number is less than approximately 2000. This number is given by the formula R_(e)=(2overline(v)rhoR)/(eta) In which overline(v) is the average speed rho is the density eta is the coefficient of viscosity of the fluid and R is the radius of the pipe. Take the density of water to be rho=1000kg//m^(3) Q. If the sphere in previous question has mass of 1xx10^(-5)kg what is its terminal velocity when falling through water? (eta=1xx10^(-3)Pa-s)
Recommended Questions
- Raindrops of raidus 1mm and mass 4 mg are falling with a speed of 30 ...
Text Solution
|
- Raindrops of raidus 1mm and mass 4 mg are falling with a speed of 30 ...
Text Solution
|
- A dics is rotating in a room. A boy standing near the rim of the disc ...
Text Solution
|
- Two small spheres of equal mass, and heading towards each other with e...
Text Solution
|
- सड़क पर खड़े एक व्यक्ति को वर्षा से बचने के लिए छाते को ऊर्ध्वाधर से 30 ...
Text Solution
|
- 1 mm त्रिज्या तथा 4 mg द्रव्यमान की वर्षा की बूँदें एक गंजे व्यक्ति के...
Text Solution
|
- एक व्यक्ति एक सिक्के को 3 बार उछालता है। एक शीर्ष आने की,
Text Solution
|
- A person walking at 4 m/s finds rain drops falling slantwise into his ...
Text Solution
|
- If velocity of flow is 4 m/s, then velocity head is
Text Solution
|