Area of small circle = PI x r^2
r = 1/2 the diameter ( 1/2 * 3 = 1.5)
3.14 x 1.5^2 = 7.065 square cm
there are 2 small circles so total area for them = 7.065 * 2 = 14.13 square cm
the area for the large circle = PI x r^2
the radius of the large circle is 3 cm
3.14 x 3^2 = 28.26 square cm
the area of the shaded area is: 28.26 - 14.13 = 14.13 square cm, which can round down to 14 square cm.
Assuming metric units, metre, kilogram and seconds
Best approach: draw a free body diagram and identify forces acting on the child, which are:
gravity, which can be decomposed into normal and parallel (to slide) components
N=mg(cos(theta)) [pressing on slide surface]
F=mg(sin(theta)) [pushing child downwards, also cause for acceleration]
m=mass of child (in kg)
g=acceleration due to gravity = 9.81 m/s^2
theta=angle with horizontal = 42 degrees
Similarly, kinetic friction is slowing down the child, pushing against F, and equal to
Fr=mu*N=mu*mg(cos(theta))
mu=coefficient of kinetic friction = 0.2
The net force pushing child downwards along slide is therefore
Fnet=F-Fr
=mg(sin(theta))-mu*mg(cos(theta))
=mg(sin(theta)-mu*cos(theta)) [ assuming sin(theta)> mu*cos(theta) ]
From Newton's second law,
F=ma, or
a=F/m
=mg(sin(theta)-mu*cos(theta)) / m
= g(sin(theta)-mu*cos(theta)) [ m/s^2]
In case imperial units are used, g is approximately 32.2 feet/s^2.
and the answer will be in the same units [ft/s^2] since sin, cos and mu are pure numbers.
Answer:
since M lies between point A and B , we came to know that,
M(4,6) =(x,y)
A(-2,-1)=(x1 ,y1)
B( _, _ )=(x2,y2)
Now using mid point formula,
x=x1×x2÷2 y=y1+y2÷2
so, point B is (10,13)
What would you like help with?
Answer:
transversal line is line A. parallel lines are B and C. corresponding to angle A is angle D. and alternate angle to E is G.
