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.
2x+4= 3x-6+2
2x+4=3x-4
2x-3x+4=3x-3x-4
-x+4=-4
-x+4-4= -4-4
-x=-8
Need to multiply by -1 for x to be positive and also for -8
-1(-x)= -1(-8)
x=8
Answer:
Distributive Property
Step-by-step explanation:
You are distributing 3 to x and -4
Answer with Step-by-step explanation:
we are given a event of rolling as die several times:
Outcome: [1 2 3 4 5 6]
Number of times outcome occurred: [10 6 4 8 6 6]
We have to find the experimental probability of rolling a 1 or a 5
number of times 1 or 5 came up= 10+6=16
and total number of outcomes= 10+6+4+8+6+6= 40
P( rolling a 1 or a 5)=
number of times 1 or 5 came up/ total number of outcomes
= 16/40
= 2/5
Hence, the experimental probability of rolling a 1 or a 5 is:
2/5
