Physics-
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Question

When jet of liquid strikes a fixed or moving surface, it exerts thrust on it due to rate of change of momentum. F = open parentheses rho A v subscript 0 end subscript close parentheses v subscript 0 end subscript minus open parentheses rho A v subscript 0 end subscript close parentheses v subscript 0 end subscript c o s invisible function application theta equals rho A v subscript 0 end subscript superscript 2 end superscript left square bracket 1 minus c o s invisible function application theta right square bracket

If surface is free and starts moving due to thrust of liquid, then at any instant, the above equation gets modified based on relative change of momentum with respect to surface. Let any instant the velocity of surface is u, then above equation becomes – F equals rho A open parentheses v subscript 0 end subscript minus u close parentheses to the power of 2 end exponent left square bracket 1 minus c o s invisible function application theta right square bracket

Based on above concept, in the below given figure, if the cart is frictionless and free to move in horizontal direction, then answer the following :Given cross–section area of jet equals 2 cross times 10 to the power of negative 4 end exponent m to the power of 2 end exponent velocity of jet V subscript 0 end subscript = 10 m/s, density of liquid = 1000 kg/ m to the power of 3 end exponent , Mass of cart M = 10 kg: In the above problem, what is the acceleration of cart at this instant –

  1. 1.6 text end text m divided by s to the power of 2 end exponent    
  2. 1 text end text m divided by s to the power of 2 end exponent    
  3. 0.64 text end text m divided by s to the power of 2 end exponent    
  4. 0.16 text end text m divided by s to the power of 2 end exponent    

The correct answer is: 0.16 text end text m divided by s to the power of 2 end exponent

Related Questions to study

General
physics-

When jet of liquid strikes a fixed or moving surface, it exerts thrust on it due to rate of change of momentum. F = open parentheses rho A v subscript 0 end subscript close parentheses v subscript 0 end subscript minus open parentheses rho A v subscript 0 end subscript close parentheses v subscript 0 end subscript c o s invisible function application theta equals rho A v subscript 0 end subscript superscript 2 end superscript left square bracket 1 minus c o s invisible function application theta right square bracket

If surface is free and starts moving due to thrust of liquid, then at any instant, the above equation gets modified based on relative change of momentum with respect to surface. Let any instant the velocity of surface is u, then above equation becomes – F equals rho A open parentheses v subscript 0 end subscript minus u close parentheses to the power of 2 end exponent left square bracket 1 minus c o s invisible function application theta right square bracket

Based on above concept, in the below given figure, if the cart is frictionless and free to move in horizontal direction, then answer the following :Given cross–section area of jet equals 2 cross times 10 to the power of negative 4 end exponent m to the power of 2 end exponent velocity of jet V subscript 0 end subscript = 10 m/s, density of liquid = 1000 kg/ m to the power of 3 end exponent , Mass of cart M = 10 kg: Velocity of cart at t = 10 s. is equal to:

When jet of liquid strikes a fixed or moving surface, it exerts thrust on it due to rate of change of momentum. F = open parentheses rho A v subscript 0 end subscript close parentheses v subscript 0 end subscript minus open parentheses rho A v subscript 0 end subscript close parentheses v subscript 0 end subscript c o s invisible function application theta equals rho A v subscript 0 end subscript superscript 2 end superscript left square bracket 1 minus c o s invisible function application theta right square bracket

If surface is free and starts moving due to thrust of liquid, then at any instant, the above equation gets modified based on relative change of momentum with respect to surface. Let any instant the velocity of surface is u, then above equation becomes – F equals rho A open parentheses v subscript 0 end subscript minus u close parentheses to the power of 2 end exponent left square bracket 1 minus c o s invisible function application theta right square bracket

Based on above concept, in the below given figure, if the cart is frictionless and free to move in horizontal direction, then answer the following :Given cross–section area of jet equals 2 cross times 10 to the power of negative 4 end exponent m to the power of 2 end exponent velocity of jet V subscript 0 end subscript = 10 m/s, density of liquid = 1000 kg/ m to the power of 3 end exponent , Mass of cart M = 10 kg: Velocity of cart at t = 10 s. is equal to:

physics-General
General
physics-

When jet of liquid strikes a fixed or moving surface, it exerts thrust on it due to rate of change of momentum. F = open parentheses rho A v subscript 0 end subscript close parentheses v subscript 0 end subscript minus open parentheses rho A v subscript 0 end subscript close parentheses v subscript 0 end subscript c o s invisible function application theta equals rho A v subscript 0 end subscript superscript 2 end superscript left square bracket 1 minus c o s invisible function application theta right square bracket

If surface is free and starts moving due to thrust of liquid, then at any instant, the above equation gets modified based on relative change of momentum with respect to surface. Let any instant the velocity of surface is u, then above equation becomes – F equals rho A open parentheses v subscript 0 end subscript minus u close parentheses to the power of 2 end exponent left square bracket 1 minus c o s invisible function application theta right square bracket

Based on above concept, in the below given figure, if the cart is frictionless and free to move in horizontal direction, then answer the following :Given cross–section area of jet equals 2 cross times 10 to the power of negative 4 end exponent m to the power of 2 end exponent velocity of jet V subscript 0 end subscript = 10 m/s, density of liquid = 1000 kg/ m to the power of 3 end exponent , Mass of cart M = 10 kg: Initially (t = 0) the force on the cart is equal to :

When jet of liquid strikes a fixed or moving surface, it exerts thrust on it due to rate of change of momentum. F = open parentheses rho A v subscript 0 end subscript close parentheses v subscript 0 end subscript minus open parentheses rho A v subscript 0 end subscript close parentheses v subscript 0 end subscript c o s invisible function application theta equals rho A v subscript 0 end subscript superscript 2 end superscript left square bracket 1 minus c o s invisible function application theta right square bracket

If surface is free and starts moving due to thrust of liquid, then at any instant, the above equation gets modified based on relative change of momentum with respect to surface. Let any instant the velocity of surface is u, then above equation becomes – F equals rho A open parentheses v subscript 0 end subscript minus u close parentheses to the power of 2 end exponent left square bracket 1 minus c o s invisible function application theta right square bracket

Based on above concept, in the below given figure, if the cart is frictionless and free to move in horizontal direction, then answer the following :Given cross–section area of jet equals 2 cross times 10 to the power of negative 4 end exponent m to the power of 2 end exponent velocity of jet V subscript 0 end subscript = 10 m/s, density of liquid = 1000 kg/ m to the power of 3 end exponent , Mass of cart M = 10 kg: Initially (t = 0) the force on the cart is equal to :

physics-General
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Statement–I : A block is immersed in a liquid inside a beaker, which is falling freely. Buoyant force acting on block is zero

Statement–II : In case of freely falling liquid there is no pressure difference between any two points

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Calculate the force ' F 'that is applied horizontally at the axle of the wheel which is necessary to raise the wheel over the obstacle of height 4m Radius of the wheel is 1m and its mass is 10 kg open parentheses g equals 10 m s to the power of negative 2 end exponent close parentheses

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A wheel of radius ' r 'and mass 'm' stands in front of a step of height ' h' The least horizontal force which should be applied to the axle of the wheel to allow it to raise onto the step is

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A cubical block of side ‘L’ rests on a rough horizontal surface with coefficient of friction A horizontal force F is applied on them' ' block as shown If the coefficient of friction is sufficiently high so that the block does not slide before toppling, the minimum force required to topple the block is

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physics-General
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If the time period of a pendulum is 1 sec, then what is the length of the pendulum at point of l minus T and l minus T to the power of 2 end exponent intersection of graph

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The potential energy of a particle of mass 1kg in motion along x-axis is given by U equals 4 left parenthesis 1 minus c o s invisible function application 2 x right parenthesis joule, where  ' x' is in meter The period of small oscillations in second is

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For the given figure, calculate zero correction.

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If ABCD is a square as shown in the figure then which of the following are adjacent sides ?

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Pick up the correct statement/statements: 1. gas A will tend to lie at the bottom. 2. the number of atoms of various gases A, Band Care same. 3. the gases will diffuse to form homogeneous mixture, 4. average kinetic energy of each gas is same,

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chemistry-General
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Which of the above gases cannot be liquefied at 100 K and 50 - atm?

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V subscript c end subscript equals 3 b comma P subscript c end subscript equals fraction numerator a over denominator 27 b to the power of 2 end exponent end fraction comma T subscript c end subscript equals fraction numerator 8 a over denominator 27 R b end fraction

Which of the above gases cannot be liquefied at 100 K and 50 - atm?

chemistry-General
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chemistry-General
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