Explanation:
Below is an attachment containing the solution.
A 7.5kg block is sliding down a wall with constant velocity. The coefficient of static friction between the block and the wall
1 answer:
Ok, a couple of things have to be accounted for here. First, since the block is moving relative to the wall we have to use the kinetic coefficient of friction, 0.40. The second consideration is that since the block is moving at a constant velocity, the acceleration is zero. This means, by Newton's second Law, that the net force is zero. So the force of gravity must be equal to the friction force of the wall resisting its fall. This friction force is the product of the normal force (which we are seeking) and the kinetic coefficient of friction. We can then set these two forces equal:
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Answer:
The answer to the question is;
The rate at which the distance between the two cars is changing is equal to 14.4 ft/sec.
Explanation:
We note that the distance traveled by each car after 4 seconds is
Car A = 19 ft in the west direction.
Car B = 26 ft in the north direction
The distance between the two cars is given by the length of the hypotenuse side of a right angled triangle with the north being the y coordinate and the west being the x coordinate.
Therefore, let the distance between the two cars be s
we have
s² = x² + y²
= (19 ft)² + (26 ft)² = 1037 ft²
s = = 32.202 ft.
The rate of change of the distance from their location 4 seconds after they commenced their journeys is given by;
Since s² = x² + y² we have
→ which gives
We note that the speeds of the cars were given as
Car B moving north = 12 ft/sec, which is the y direction and
Car A moving west = 8 ft/sec which is the x direction.
Therefore
= 12 ft/sec and
= 8 ft/sec
becomes
= 464 ft²/sec
= 14.409 ft/sec ≈ 14.4 ft/sec to one place of decimal.
Answer:
Explanation:
The solution of the question has been put in attachment.
