If an automobile of mass m rounds a curve, then its inward vector component of acceleration a N N is caused by the frictional force F of the road. Thus, it follows from the vector form of Newton’s second law [Equation (19)] that the frictional force and the normal component of acceleration are related by the equation F = m a N N . Thus, F = m k d s d t 2 Use this result to find the magnitude of the frictional force in newtons exerted by road on a 500 kg go-cart driven at a speed of 10km/h around a circular track or radius 15m. [ N o t e : 1 N = 1 kg . m / s 2 . ]
If an automobile of mass m rounds a curve, then its inward vector component of acceleration a N N is caused by the frictional force F of the road. Thus, it follows from the vector form of Newton’s second law [Equation (19)] that the frictional force and the normal component of acceleration are related by the equation F = m a N N . Thus, F = m k d s d t 2 Use this result to find the magnitude of the frictional force in newtons exerted by road on a 500 kg go-cart driven at a speed of 10km/h around a circular track or radius 15m. [ N o t e : 1 N = 1 kg . m / s 2 . ]
If an automobile of mass m rounds a curve, then its inward vector component of acceleration
a
N
N
is caused by the frictional force F of the road. Thus, it follows from the vector form of Newton’s second law [Equation (19)] that the frictional force and the normal component of acceleration are related by the equation
F
=
m
a
N
N
.
Thus,
F
=
m
k
d
s
d
t
2
Use this result to find the magnitude of the frictional force in newtons exerted by road on a 500 kg go-cart driven at a speed of 10km/h around a circular track or radius 15m.
[
N
o
t
e
:
1
N
=
1
kg
.
m
/
s
2
.
]
Quantities that have magnitude and direction but not position. Some examples of vectors are velocity, displacement, acceleration, and force. They are sometimes called Euclidean or spatial vectors.
For each given function f(x) find f'(x) using the rules learned in section 9.5.
1. f(x)=x32
32x
2. f(x)=7x+13
3. f(x) =
x4
4. f(x) = √√x³
5. f(x) = 3x²+
3
x2
Find:
lim x →-6 f (x)
limx-4 f (x)
lim x-1 f (x)
lim x →4 f (x)
(-6,3) •
(-1,5)
-8
-7
(-6,-2)
4+
(4,5)
(4,2) •
(-1,1)
-6
College Algebra with Modeling & Visualization (5th Edition)
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