Physics, 27.02.2020 04:25 jdkrisdaimcc11

A horizontal tension force pulls to the right on block A. Block A slides to the right on a horizontal surface that is not frictionless. Block B is stacked on top of block A and the two blocks move together as a unit (block B does not slide on top of block A). Block A is 6 kg and block B is 4 kg. While a 10 N tension force is applied, the blocks slide to the right at constant speed. What is the magnitude of the horizontal force that block A exerts on block B?a. 0Nb. 2Nc. 4Nd. 6Ne. 10NThe tension force is reduced to 5 N and the blocks continue sliding to the right, but they are now slowing down. Which direction is the horizontal force that block B exerts on block A?a. to the left. b. to the right. c. block A exerts no horizontal force on block B. d. the rate at which the blocks are slowing down is needed to determine e. the direction of the horizontal force depends on the speed at which the blocks slide

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force is based upon mass and acceleration.the statement is not always true.

it can be explained from newton's second law of motion.

as per newton's second law of motion the rate of change of momentum is directly proportional to the applied force and takes place along the direction of force.

mathematically it can be written as -

$f\alpha \frac{dp}{dt}$

⇒ $f= k \frac{dp}{dt}$

here k is the proportionality constant whose value is 1.

$f=\frac{dp}{dt}$

$f= \frac{d[mv]}{dt}$

$f= m\frac{dv}{dt} +v\frac{dm}{dt}$

$f=ma+v\frac{dm}{dt}$ [a is the acceleration=dv/dt]

hence we see that force not only depends on mass and acceleration but also on velocity and rate of change of mass. force f will depend on mass and acceleration only when mass is constant.

hence the statement that force is based upon mass and acceleration is false.

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The particles of the gas inside the ball heats up, bouncing off each other and spreading farther apart.hope my answer has you : )

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Answer(b) both involve the use of levers;

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The figure is missing but we can answer anyway.the correct answer is "releasing the string". in fact, when the person is holding the bow and the arrow, the kinetic energy of the arrow is still zero (because the arrow is not moving). the moment the arrow is released, it acquires a velocity v, so its kinetic energy will become different from zero: $k= \frac{1}{2}mv^2$ where m is the mass of the arrow and v its velocity. all the other options, instead, do not increase the kinetic energy of the system: - pulling farther back on the string : this will just increase the elastic potential energy of the system- holding the arrow higher: this will increase the gravitational potential energy of the system (which is given by u=mgh, where h is the height above the ground)- decreasing the mass of the arrow: this will decrease the gravitational potential energy of the system (which is given by u=mgh, where m is the mass)

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A horizontal tension force pulls to the right on block A. Block A slides to the right on a horizonta...

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