Answer:
An Empty Tank Is Slowly Filled With Air. Water Is Then Added To The Tank, Decreasing The Volume For The Air In The Tank. The Temperature Remains Constant
Explanation:
The coefficient of friction must be 0.196
Explanation:
For a car moving on a circular track, the frictional force provides the centripetal force needed to keep the car in circular motion. Therefore, we can write:
where the term on the left is the frictional force acting between the tires of the car and the road, while the term on the right is the centripetal force. The various terms are:
is the coefficient of friction between the tires and the road
m is the mass of the car
is the acceleration of gravity
v is the speed of the car
r is the radius of the curve
In this problem,
r = 750 m is the radius
is the speed
And solving for , we find the coefficient of friction required to keep the car in circular motion:
Learn more about circular motion:
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If an object is thrown in an upward direction from the top of a building 1.60 x 102 ft. high at an initial velocity of 21.82 mi/h, what is its final velocity when it hits the ground? (Disregard wind resistance. Round answer to nearest whole number and do not reflect negative direction in your answer.)
this question is troubling me i guessed 96 ft/s
can someone help me out and explain it thanks so much!!!!!!
Accordingly Newton's findings, astronomy and physics have industrialized
hugely over the period. Scientists now recognize that every object in the
world has a force that draws each other and the power of the force hinge on the
mass of the object. Also, Newton's Laws of Motion offer individuals a
better understanding of what is likely concerning movement. This is very helpful,
particularly in mechanics and space travel. Generally, Newton had a
huge and permanent impact on science.
Answer:
Slope = 2 m / 10 m = 1/5
For every 5 m of effort the object will be raised 1 m
W = work done on object = M g h increase in PE of object
E S = W where E is effort and S the distance thru which the effort acts
E S = M g H
E = 100 kg * 9.8 m/s^2 * 2 m / 10 m = 196 kg m / s^2 = 196 N
Check: total work = 2 * 9.8 * 100 = 1960 J
Force Needed = 1960 J / 2 m = 980 Newtons
Mechanical advantage = 980 / 196 = 5 as one would expect since the object is raised 1 m for every 5 m of force input
