In pure translation or in roll without slip?
In pure translation, just use Fnet = m*a. Fnet = 7.5 Newtons
In roll without slip it has effectively more inertia than 5 kilograms. The effective inertia can be found as:
m_effective = m + I/r^2
For a uniform solid sphere, I = 2/5*m*r^2
So m_effective = 7/5*m
To make a 5 kg bowling ball accelerate at 1.5 m/s^2 in roll without slip, it requires Fnet = 10.5 Newtons, and this ASSUMES that all forces causing the net force on it have lines of action through its center of mass (most common case).
i hope this is what ur looking for
Answer:
-11/30, -1/6, 2/5
Step-by-step explanation:
first, make the question in the same denominator
= 12/30, -5/30, -11/30
= -11/30, -5/30, 12/30
= -11/30, -1/6, 2/5
F(x)= 2x-3
f(0)=2•0-3=0-3=-3 ✔️
f(7)=2•7-3=14-3=11 ✔️
F(x) is 25-x^
But x is being squared, cubed or what. What is the exponent of x.
a^2 + b^2 = c^2
a = sqrt(c^2 - b^2) = sqrt (20^2 - 16^2)
= 12
sin
= a/c = 12/20
= 3/5
cos = b/c = 16/20
= 4/5
tan = a/b = 12/16
= 3/4
csc
= c/a = 20/12
= 5/3
= 1 2/3
sec = c/b = 20/16
= 5/4
= 1 1/4
