1. A force of 10 N acts on a particle of mass 0.4 kg. Find the acceleration of the particle. 2. A body of mass 1 kg is kept at rest. A constant force of 6.0 N starts acting on it. Find the time taken by the body to move through a distance of 12 m. 3. The velocity of a particle of mass 150 g changes from 8 m/s to 12 m/s in two seconds. Assuming that a constant force acts on it, find the magnitude of the force. 4. A force of 4.0 N acts on a body of mass 2.0 kg for 4.0 s. Assuming the body to be initially at rest, find (a) its velocity when the force stops acting, (b) the distance covered in 10 s after the force starts acting 5. A coin of mass 20 g is pushed on a table. The coin starts moving at a speed of 25 cm/s and stops in 5 seconds. Find the force of friction exerted by the table on the coin. 6. When a body is dropped from a height, it falls with an acceleration of 10 m/s. If its mass is 250 g, how much force is exerted on it downwards? Who exerts this force on the body?

1. A force of 10 N acts on a particle of mass 0.4 kg. Find the acceler

1.	1. A force of 10 N acts on a particle of mass 0.4 kg. Find the acceleration of the particle. 2. A body of mass 1 kg is kept at rest. A constant force of 6.0 N starts acting on it. Find the time taken by the body to move through a distance of 12 m. 3. The velocity of a particle of mass 150 g changes from 8 m/s to 12 m/s in two seconds. Assuming that a constant force acts on it, find the magnitude of the force. 4. A force of 4.0 N acts on a body of mass 2.0 kg for 4.0 s. Assuming the body to be initially at rest, find (a) its velocity when the force stops acting, (b) the distance covered in 10 s after the force starts acting 5. A coin of mass 20 g is pushed on a table. The coin starts moving at a speed of 25 cm/s and stops in 5 seconds. Find the force of friction exerted by the table on the coin. 6. When a body is dropped from a height, it falls with an acceleration of 10 m/s. If its mass is 250 g, how much force is exerted on it downwards? Who exerts this force on the body?
Answer» 1. F=M×A,=10=0.4 ×AAcceleration is 25m/S22. M=F/A,= 1=6/AA=6m/S2using s=ut+0.5at2,ut=0,0.5×6×t2=12which gives t =2 seconds3 F =(mv-mu)/t150g =0.15kgF=0.15(12-8)/2=0.3 N4. a F=M×A,A=4/22m/s2using,v=u+at,u=0v=2×48m/sb. using s=ut+0.5at2=0.5×2×10×10100m5. using v=u+at,force of friction is negetive as it is exerted in opposite direction0=0.25+(-a×5)a is -0.5m/s2F=M×A20g is 0.02kg0.02×(-0.5)-0.01N6. 250G is 0.25kgF=M×A=0.25×102.5NThis force which is exerted is the gravitational force. It is exerted by the surface of the Earth due to the universal LAW of gravitation proposed by Sir Isaac Newton.I HOPE THIS WILL HELP YOUPLEASE mark it the brainliest

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