Which applied force will allow a 7.65 kg block of ice to begin sliding on a sheet of ice? The block of ice has a kinetic coeffic
ient of friction of 0.030 and a static coefficient of friction of 0.10.
2 answers:
Answer:
force for start moving is 7.49 N
force for moving constant velocity 2.25 N
Explanation:
given data
mass = 7.65 kg
kinetic coefficient of friction = 0.030
static coefficient of friction = 0.10
solution
we get here first weight of block of ice that is
weight of block of ice = mass × g
weight of block of ice = 7.65 × 9.8 = 74.97 N
so here Ff = Fa
so for force for start moving is
Fa = weight × static coefficient of friction
Fa = 74.97 × 0.10
Fa = 7.49 N
and
force for moving constant velocity is
Fa = weight × kinetic coefficient of friction
Fa = 74.97 × 0.030
Fa = 2.25 N
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ANSWER
EXPLANATION
Parameters given:
Mass of the student, M = 70 kg
Mass of the textbook, m = 1 kg
Distance, r = 1 m
To find the gravitational force acting between the student and the textbook, apply the formula for gravitational force:
where G = gravitational constant
Therefore, the gravitational force acting between the student and the textbook is:
That is the answer.
Answer:
R = 73.25 m
Explanation:
We have,
Initial speed of the ball is 27 m/s
It is projected at an angle of 40 degrees
The maximum range of the ball is given by :
Plugging all the values we get :
So, the maximum range of the ball is 73.25 m