what type of simple machine is shown in the diagram?write the length and height of slope in it. Please help ASAP!!!
2 answers:
You might be interested in
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:
change in height is 1.664 mm
Explanation:
Given data
drops = 3.00 mm
diameter = 5.00 cm = 0.05 mm
decrease = 350 cm^3
temperature = 95°C to 44.0°C
to find out
the decrease in millimeters in level
solution
we will calculate here change in volume so we can find how much level is decrease
change in volume = β v change in temp ...............1
here change in volume = area× height
so = /4 × d² h
so we can say change in volume = /4 × d² × change in height .......2
so from equation 1 and 2 we calculate change in height
( β(w) -β(g) )× v× change in temp = /4 × d² × change in height
change in height = 4 × ( β(w) -β(g) ) v× change in temp / /4 × d²
put all value here
change in height = 4 × ( 210 - 27 )(350 ) × (95-44) /
/4 × 0.05²
change in height is 1.664 mm