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Furkat [3]
3 years ago
7

A roller coaster car of mass m= 300 kg is released from rest at the top of a 60 m high hill (position A), and rolls with a negli

gible friction down the hill, through a circular loop of radius 20 m (Positions B,C,and D) and along a horizontal track (to position E). Describe the scientific principles involved in this scenario.
Physics
1 answer:
Andrews [41]3 years ago
4 0

Answer: The principle of conservation of energy, angular speed and centripetal force

Explanation:

At point A, the car experienced maximum of potential energy

As it moves down the hill, the potential energy decreases while the kinetic energy increases.

The maximum kinetic energy of the car is needed for the attainment of enough centripetal force to help the car move through the loop without falling .

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Viktor [21]

Answer:

Things float when they are positively buoyant, or less dense than the fluid in which they are sitting. This does not mean that an object has to be lighter than the fluid, as in the case of a boat; objects just need to have a greater ratio of empty space to mass than the fluid.

Explanation:

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3 years ago
The pressure exerted by a gas is 2.0 atm while it has a volume of 350 mL. What would be the volume of this sample of gas at stan
34kurt

Answer:

700 mL or 0.0007 m³

Explanation:

P₁ = Initial pressure = 2 atm

V₁ = Initial volume = 350 mL

P₂ = Final pressure = 1 atm

V₂ = Final volume

Here the temperature remains constant. So, Boyle's law can be applied here.

P₁V₁ = P₂V₂

\frac{P_1V_1}{P_2}=V_2\\\Rightarrow V_2=\frac{2\times 350}{1}\\\Rightarrow V_2=700\ mL

So, volume of this sample of gas at standard atmospheric pressure would be 700 mL or 0.0007 m³

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What is the momentum of an object with 5.22 kg and 1.7 m/s
BlackZzzverrR [31]

Answer:

8.874

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You need to times 5.22 kg and 1.7 m/s to get 8.874.

8 0
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The melting point of a substance occurs at the same temperatures as it’s blank point
azamat

I think it's a.

Explanation:

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Read 2 more answers
Three crates with various contents are pulled by a force Fpull=3615 N across a horizontal, frictionless roller‑conveyor system.
SIZIF [17.4K]

The question is incomplete. Here is the complete question.

Three crtaes with various contents are pulled by a force Fpull=3615N across a horizontal, frictionless roller-conveyor system.The group pf boxes accelerates at 1.516m/s2 to the right. Between each adjacent pair of boxes is a force meter that measures the magnitude of the tension in the connecting rope. Between the box of mass m1 and the box of mass m2, the force meter reads F12=1387N. Between the box of mass m2 and box of mass m3, the force meter reads F23=2304N. Assume that the ropes and force meters are massless.

(a) What is the total mass of the three boxes?

(b) What is the mass of each box?

Answer: (a) Total mass = 2384.5kg;

               (b) m1 = 915kg;

                   m2 = 605kg;

                   m3 = 864.5kg;

Explanation: The image of the boxes is described in the picture below.

(a) The system is moving at a constant acceleration and with a force Fpull. Using Newton's 2nd Law:

F_{pull}=m_{T}.a

m_{T}=\frac{F_{pull}}{a}

m_{T}=\frac{3615}{1.516}

m_{T}=2384.5

Total mass of the system of boxes is 2384.5kg.

(b) For each mass, analyse each box and make them each a free-body diagram.

<u>For </u>m_{1}<u>:</u>

The only force acting On the m_{1} box is force of tension between 1 and 2 and as all the system is moving at a same acceleration.

m_{1} = \frac{F_{12}}{a}

m_{1} = \frac{1387}{1.516}

m_{1} = 915kg

<u>For </u>m_{2}<u>:</u>

There are two forces acting on m_{2}: tension caused by box 1 and tension caused by box 3. Positive referential is to the right (because it's the movement's direction), so force caused by 1 is opposing force caused by 3:

m_{2} = \frac{F_{23}-F_{12}}{a}

m_{2} = \frac{2304-1387}{1.516}

m_{2} = 605kg

<u>For </u>m_{3}<u>:</u>

m_{3} = m_{T} - (m_{1}+m_{2})

m_{3} = 2384.5-1520.0

m_{3} = 864.5kg

8 0
3 years ago
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