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Rainbow [258]
1 year ago
14

Determine whether the statement is true or false. If f '(c) = 0, then f has a local maximum or minimum at c.

Physics
1 answer:
ankoles [38]1 year ago
4 0

It is False that, If f '(c) = 0, then f has a local maximum or minimum at c.

Local maximum and minimum points are very distinctive on the graph of a function and are thus, helpful in grasping the shape of the graph. Either a local minimum or a local maximum can be considered a local extremum.

A counterexample can be used for:

f(x) = x³

f'(x) = 2 × x²

and,

f'(0) = 2×0² = 0

However, the assertion is false because x = 0 is actually an inflection point rather than a maximum or minimum in f(x) = x³.

Know more about f(x) here: brainly.com/question/28058540

#SPJ4

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Two metal disks are welded together and are mounted on a frictionless axis through their common centers. One disk has a radius R
Airida [17]

The Inertia is 22. 488 kg. m² and the speed just before it hits the ground is 6. 4 m/s

<h3>How to determine the inertia</h3>

Using the formula:

I = 1/2 M₁R₁² + 1/2 M₂R₂²

Where I = Inertia

I = 1/2 * 0.810* (2. 60)² + 1/2 * 1. 58 * (5)²

I = 1/2 * 5. 476 + 1/2 * 39. 5

I = 2. 738 + 19. 75

I = 22. 488 kg. m²

To determine the block's speed, use the formula

v = \sqrt{2gh}

v = \sqrt{2* 10 * 2. 10}

v = \sqrt{42}

v = 6. 4 m/s

Therefore, the Inertia is 22. 488 kg. m² and the speed just before it hits the ground is 6. 4 m/s

Learn more about law of inertia here:

brainly.com/question/10454047

#SPJ1

7 0
2 years ago
How do I do these? My teacher didn’t show us how.
melisa1 [442]

Explanation:

Displacement is simply the change in position.  So in the first part of problem 1, looking at the graph between 0 s and 2 s, the position changes from 0 m to -4 m.  So the displacement is:

Δx =  -4 m − 0 m

Δx = -4 m

Between 2 s and 4 s, the position stays at -4 m.  The displacement is:

Δx = -4 m − (-4 m)

Δx = 0 m

Finally, between 4 s and 6 s, the position goes from -4 m to 6 m.  The displacement is:

Δx = 6 m − (-4 m)

Δx = 10 m

The net displacement is the change in position from 0 s to 6 s:

Δx = 6 m − 0 m

Δx = 6 m

In the second part of problem 1, we have a velocity vs time graph.

Car 1 starts with 0 velocity and ends with a velocity of 6 m/s, so it is accelerating and constantly moving to the right.

Car 2 starts with a velocity of -6 m/s and ends with a velocity of 6 m/s.  It is also accelerating, but first it is moving to the left, comes to a stop at t = 3 s, then moves to the right.

Car 3 starts with a velocity of 2 m/s and ends with a velocity of 2 m/s.  So it is moving constantly to the right, but never speeds up or slows down.

We want to know when two of the cars meet.  Unfortunately, this isn't as easy as looking for where the lines cross on the graph.  We need to calculate their displacements.  We can do this by finding the area under the graph (assuming all the cars start from the same point).

Let's start with Car 2.  Half of the area is below the x-axis, and half is above.  Without doing calculations, we can say the total displacement for this car is 0.  This means it ends back up where it started, and that it never meets either of the other cars, both of which have positive displacements.

So we know Car 1 and Car 3 meet, we just have to find where and when.  For Car 1, the area under the curve is a triangle.  So its displacement is:

Δx = ½ t v(t)

where t is the time and v(t) is the velocity of Car 1 at that time.  Since the line has a slope of 1 and y intercept of 0, we know v(t) = t.  So:

Δx = ½ t²

Now look at Car 3.  The area under the curve is a rectangle.  So its displacement is:

Δx = 2t

When the two cars have the same displacement:

½ t² = 2t

t² = 4t

t² − 4t = 0

t (t − 4) = 0

t = 0, 4

t = 0 refers to the time when both cars are at the starting point, so t = 4 is the answer we're looking for.  Where are the cars at this time?  Simply plug in t = 4 into either of the equations we found:

Δx = 8

So Cars 1 and 3 meet at 4 s and 8 m.

7 0
3 years ago
A sled is slowing down at the bottom of a snowy hill.
Kitty [74]

Answer:

A sled and its rider are moving at a speed of along a horizontal stretch of snow, as Figure 4.24a illustrates. The snow exerts a kinetic frictional force on the runners of the sled, so the sled slows down and eventually comes to a stop. The coefficient of kinetic friction is 0.050. What is the displacement x of the sled?

5 0
2 years ago
A prismatic bar AB of length L and solid circular cross section (diameter d) is loaded by a distributed torque of constant inten
Lyrx [107]

Answer:

a) the maximum shear stress τ_{max} the bar is 16T_{max} /πd³

b) the angle of twist between the ends of the bar is 16tL² / πGd⁴  

Explanation:

Given the data in the question, as illustrated in the image below;

d is the diameter of the prismatic bar of length AB

t is the intensity of distributed torque

(a) Determine the maximum shear stress tmax in the bar

Maximum Applied torque  T_max = tL

we know that;

shear stress τ = 16T/πd³

where d is the diameter

so

τ_{max} = 16T_{max} /πd³

Therefore, the maximum shear stress τ_{max} the bar is 16T_{max} /πd³

(b) Determine the angle of twist between the ends of the bar.

let theta (\theta) be the angle of twist

polar moment of inertia I_p} = πd⁴/32

now from the second image;

lets length dx which is at distance of "x" from "B"

Torque distance x

T(x) = tx

Elemental angle twist = d\theta = T(x)dx / GI_{p}

so

d\theta = tx.dx / G(πd⁴/32)

d\theta = 32tx.dx / πGd⁴

so total angle of twist \theta will be;

\theta =  \int\limits^L_0  \, d\theta

\theta =  \int\limits^L_0  \, 32tx.dx / πGd⁴

\theta = 32t / πGd⁴  \int\limits^L_0  \, xdx

\theta = 32t / πGd⁴ [ L²/2]

\theta = 16tL² / πGd⁴  

Therefore,  the angle of twist between the ends of the bar is 16tL² / πGd⁴  

7 0
3 years ago
PLEASE HELP<br> Due in a few minutes!!!!!<br> Have not finished yet :
Zigmanuir [339]
The North Star, or Polaris, is the brightest star in the constellation Ursa Minor, the little bear (also known as the Little Dipper). As viewed by observers in the Northern Hemisphere, Polaris occupies a special place
5 0
3 years ago
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