How we interact to a human person or people in a good way
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Answer:
The constant force with least magnitude that must be applied to the board in order to pull the board out from under the box is
Explanation:
The Newton’s second law states that the net force on an object is the product of mass of the object and final acceleration of the object. The expression of newton’s second law is,
Here, is the sum of all the forces on the object, mm is mass of the object, and aa is the acceleration of the object.
The expression for static friction over a horizontal surface is,
Here, is the coefficient of static friction, mm is mass of the object, and
is the acceleration due to gravity.
Use the expression of static friction and solve for maximum static friction for box of mass
Substitute for in the expression of maximum static friction
Use the Newton’s second law for small box and solve for minimum acceleration aa to pull the box out.
Substitute for , [/tex]{m_1}[/tex] for in the equation .
Substitute for
in the equation
Rearrange for a.
The minimum acceleration of the system of two masses at which box starts sliding can be calculated by equating the pseudo force on the mass with the maximum static friction force.
The pseudo force acts on in the direction opposite to the motion of the board and the static friction force on this mass acts in the direction opposite to the pseudo force. If these two forces are cancelled each other (balanced), then the box starts sliding.
Use the Newton’s second law for the system of box and the board.
Substitute for for in the equation .
Substitute for in the above equation .
The constant force with least magnitude that must be applied to the board in order to pull the board out from under the box is
There is no friction between the board and the surface. So, the force required to accelerate the system with the minimum acceleration to slide the box over the board is equal to total mass of the board and box multiplied by the acceleration of the system.
Answer:
(a) 1500 m
(b) 2827.43m
Explanation:
Given that the time for one cycle of the swing is 1 s
The radius of the swing, R= 3.0m
The angle covered, \theta, by each swing is a quarter of the circle. i.e.
Speed of running of women =5 m/s.
Time of running = 5 minutes= 5 x 60 secondes= 300 s.
(a) As distance= (speed) x (time)
So, the required distance= 5 x 300 m= 1500 m.
(b) As there are two swings in one cycle, so, the distance covered in one swing is the length of the circular shown in the figure.
As arc length, , where \theta is the angle, in radian, subtended by the arc at the center, and R is the radius of curvature of the arc.
So, the distance covered by the child in 1 swing .
In 1 cycle, there are 2 swings, so distance covered in 1 cycle .
Now, in 1 second there is 1 cycle, so in 5 minutes there will be 300 cycles.
So, the total distance covered by the child
.
