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MariettaO [177]
3 years ago
14

If the line graph had a broken line between the points

Physics
1 answer:
viktelen [127]3 years ago
6 0
It seems like your sentence was not completed, but I think you were wondering what it is called when there is a broken line passing through two points. It would be a graphed inequality, and if the inequality has a greater than symbol ( > ), the right side of the graph would be shaded in, and if the inequality has a less than symbol ( < ), the left side of the graph would be shaded in. Inequalities with greater than or equal to and less than or equal to symbols have solid lines going through two points with no sides shaded.

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The sun can continue to exist in its present stable state for about another
Wewaii [24]

Answer:

5.5 billion years

Explanation:

The answer

3 0
3 years ago
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A farmer hitches her tractor to a sled loaded with firewood and pulls it a distance
Delvig [45]

(a) The work done by the force applied by the tractor is 79,968.47 J.

(b) The work done by the frictional force on the tractor is 55,977.93 J.

(c) The total work done by  all the forces is 23,990.54 J.

<h3>Work done by the applied force</h3>

The work done by the force applied by the tractor is calculated as follows;

W = Fd cosθ

W = (5000 x 20) x cos(36.9)

W = 79,968.47 J

<h3>Work done by frictional force</h3>

W = Ffd cosθ

W = (3500 x 20) x cos(36.9)

W = 55,977.93 J

<h3>Net work done by all the forces on the tractor</h3>

W(net) = work done by applied force  -  work done by friction force

W(net) = 79,968.47 J -  55,977.93 J

W(net) = 23,990.54 J

Learn more about work done here: brainly.com/question/25573309

#SPJ1

4 0
1 year ago
Assume that the driver begins to brake the car when the distance to the wall is d=107m, and take the car's mass as m-1400kg, its
Evgen [1.6K]

Answer:

Explanation:

a ) Let let the frictional force needed be F

Work done by frictional force = kinetic energy of car

F x 107 = 1/2 x 1400 x 35²

F = 8014 N

b )

maximum possible static friction

= μ mg

where μ is coefficient of static friction

= .5 x 1400 x 9.8

= 6860 N

c )

work done by friction for μ = .4

= .4 x 1400 x 9.8 x 107

= 587216 J

Initial Kinetic energy

= .5 x 1400 x 35 x 35

= 857500 J

Kinetic energy at the at of collision

= 857500 - 587216

= 270284 J

So , if v be the velocity at the time of collision

1/2 mv² = 270284

v = 19.65 m /s

d ) centripetal force required

= mv₀² / d which will be provided by frictional force

= (1400 x 35 x 35) / 107

= 16028 N

Maximum frictional force possible

= μmg

= .5 x 1400 x 9.8

= 6860 N

So this is not possible.

4 0
3 years ago
Give two examples of common force fields and name the sources of these fields.
algol13
Electric force from electomagnetic force and force of gravity from gravitational force
4 0
3 years ago
As an object rolls downhill, some of the energy is
Charra [1.4K]
When the object is at the top of the hill it has the most potential energy. If it is sitting still, it has no kinetic energy. As the object begins to roll down the hill, it loses potential energy, but gains kinetic energy. The potential energy of the position of the object at the top of the hill is getting converted into kinetic energy. Hope this helped. :)


7 0
3 years ago
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