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Flura [38]
2 years ago
14

A train travels from the bottom of a hill to the top of a hill, its moves slower by the time it reaches the top is an example of

____. A. positive acceleration B. velocity C. negative acceleration D. speed
Physics
2 answers:
Neko [114]2 years ago
5 0
The answer is letter C. <span>A train travels from the bottom of a hill to the top of a hill, its moves slower by the time it reaches the top is an example of negative acceleration.

In physics, acceleration can be described as an objects change of velocity. When an object gains velocity, it is positive acceleration, and negative acceleration for the opposite.

</span>

Thank you for posting your question. I hope you found what you were after. Please feel free to ask me more.

<span> </span>

hjlf2 years ago
5 0
<span>Negative Acceleration is when the acceleration of an object slows down, like throwing something in the air. It will get slower going up, and then it will stop, that is the negative acceleration, and then the object will come back down, and will speed up, which is positive acceleration.
</span>
A train travels from the bottom of a hill to the top of a hill, its moves slower by the time it reaches the top is an example of <span>negative acceleration.
Best Choice : </span><span> C. negative acceleration</span>

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The Assignment: A fixed quantity of an ideal gas (R 0.28 kJ/kgK; Cv-0.71kJ/kgK) is expanded from an initial condition of 35 bar,
Nikolay [14]

Answer:

Index of expansion: 4.93

Δu = -340.8 kJ/kg

q = 232.2 kJ/kg

Explanation:

The index of expansion is the relationship of pressures:

pi/pf

The ideal gas equation:

p1*v1/T1 = p2*v2/T2

p2 = p1*v1*T2/(T2*v2)

500 C = 773 K

20 C = 293 K

p2 = 35*0.1*773/(293*1.3) = 7.1 bar

The index of expansion then is 35/7.1 = 4.93

The variation of specific internal energy is:

Δu = Cv * Δt

Δu = 0.71 * (20 - 500) = -340.8 kJ/kg

The first law of thermodynamics

q = l + Δu

The work will be the expansion work

l = p2*v2 - p1*v1

35 bar = 3500000 Pa

7.1 bar = 710000 Pa

q = p2*v2 - p1*v1 + Δu

q = 710000*1.3 - 3500000*0.1 - 340800 = 232200 J/kg = 232.2 kJ/kg

7 0
3 years ago
Plot StartRoot 1.5 EndRoot and StartRoot 1.9 EndRoot on the number line to find which inequalities are true. Check all that appl
Vera_Pavlovna [14]

Answer:

A, B, C, E

Explanation:

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4 0
2 years ago
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SCALCET8 3.9.018.MI. A spotlight on the ground shines on a wall 12 m away. If a man 2 m tall walks from the spotlight toward the
Firlakuza [10]

Answer:

The length of his shadow is decreasing at a rate of 1.13 m/s

Explanation:

The ray of light hitting the ground forms a right angled triangle of height H, which is the height of the building and width, D which is the distance of the tip of the shadow from the building.

Also, the height of the man, h which is parallel to H forms a right-angled triangle of width, L which is the length of the shadow.

By similar triangles,

H/D = h/L

L = hD/H

Also, when the man is 4 m from the building, the length of his shadow is L = D - 4

So, D - 4 = hD/H

H(D - 4) = hD

H = hD/(D - 4)

Since h = 2 m and D = 12 m,

H = 2 m × 12 m/(12 m - 4 m)

H = 24 m²/8 m

H = 3 m

Since L = hD/H

and h and H are constant, differentiating L with respect to time, we have

dL/dt = d(hD/H)/dt

dL/dt = h(dD/dt)/H

Now dD/dt = velocity(speed) of man = -1.7 m/s ( negative since he is moving towards the building in the negative x - direction)

Since h = 2 m and H = 3 m,

dL/dt = h(dD/dt)/H

dL/dt = 2 m(-1.7 m/s)/3 m

dL/dt = -3.4/3 m/s

dL/dt = -1.13 m/s

So, the length of his shadow is decreasing at a rate of 1.13 m/s

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2 years ago
All of the following are types of symbiotic relationships except for
nevsk [136]
The answer is D. the way I remember it is they all end with -alism.

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Answer:

The micturition reflex can be voluntarily controlled by the relaxation of the external urethral sphincter.

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