Answer:
a = F-ff/m
Explanation:
According to Newton's second law of motion which states that "the rate of change in momentum of a body is directly proportional to the applied force F and acts in the direction of the force.
Mathematically;
F = ma
Since two forces acts on the cart i.e the moving force F and the frictional force Ff , we will take the sum of the forces.
∑F = ma where
m is the mass of the cart
a is its acceleration
∑F = F+(-ff )(since frictional force is an opposing force)
F - ff = ma
Dividing both sides by mass m
a = F-ff/m
Solution :
Given
Diameter of the roulette ball = 30 cm
The speed ball spun at the beginning = 150 rpm
The speed of the ball during a period of 5 seconds = 60 rpm
Therefore, change of speed in 5 seconds = 150 - 60
= 90 rpm
Therefore,
90 revolutions in 1 minute
or In 1 minute the ball revolves 90 times
i.e. 1 min = 90 rev
60 sec = 90 rev
1 sec = 90/ 60 rec
5 sec = 
= 75 rev
Therefore, the ball made 75 revolutions during the 5 seconds.
Answer:
The images output from your new color laser printer seem to be a little too blue. to fix this problem we need to calibrate the printer.
Explanation:
This can be done by opening the toolbox, clicking in the device setting folder their you get print quality page click on it. Under the print quality option click on the calibrate next to calibrate now. Then click OK unless when the 'your request has been sent to the device' appears on the screen. When the calibration ends again try printing. calibrating is useful for managing the proper alignment of the inkjet cartridge nozzle to the paper and each other, without proper calibration the print quality deteriorates.
Answer:
Explanation:
If friction is neglected, the wheel cannot roll and can only slide frictionlessly and will have the same velocity at the bottom of the ramp as if it had been in free fall as it has converted the same amount of potential energy.
mgh = ½mv²
v = √(2gh) = √(2(9.81)(2.00)) = 6.26418... = 6.26 m/s
However if we do not ignore all friction and the wheel rolls without slipping down the slope, the potential energy becomes linear and rotational kinetic energy
mgh = ½mv² + ½Iω²
mgh = ½mv² + ½(½mR²)(v/R)²
2gh = v² + ½v²
2gh = 3v²/2
v = √(4gh/3) =√(4(9.81)(2.00)/3) = 5.11468... = 5.11 m/s
