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Evgen [1.6K]
3 years ago
11

Which would best help a student determine the net force acting on a rollercoaster car as it moves from one point on its track to

another?
Physics
1 answer:
Lelu [443]3 years ago
6 0

Answer:

you can show them a vid

Explanation:

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A projectile of mass 2.0 kg is fired in the air at an angle of 40.0 ° to the horizon at a speed of 50.0 m/s. At the highest poin
tekilochka [14]

Answer:

a) The fragment speeds of 0.3 kg is 33.3 m / s on the y axis

                                         0.7 kg is 109.4 ms on the x axis

b)  Y = 109.3 m

Explanation:

This is a moment and projectile launch exercise.

a) Let's start by finding the initial velocity of the projectile

       sin 40 = voy / v₀

       v_{oy} = v₀ sin 40

       v_{oy} = 50.0 sin40

       v_{oy} = 32.14 m / s

       cos 40 = v₀ₓ / V₀

       v₀ₓ = v₀ cos 40

       v₀ₓ = 50.0 cos 40

       v₀ₓ = 38.3 m / s

Let us define the system as the projectile formed t all fragments, for this system the moment is conserved in each axis

Let's write the amounts

Initial mass of the projectile M = 2.0 kg

Fragment mass 1 m₁ = 1.0 kg and its velocity is vₓ = 0 and v_{y} = -10.0 m / s

Fragment mass 2 m₂ = 0.7 kg moves in the x direction

Fragment mass 3 m₃ = 0.3 kg moves up (y axis)

Moment before the break

X axis

     p₀ₓ = m v₀ₓ

Y Axis y

    p_{oy} = 0

After the break

X axis

   p_{fx} = m₂ v₂

Axis y

     p_{fy} = m₁ v₁ + m₃ v₃

Let's write the conservation of the moment and calculate

Y Axis  

     0 = m₁ v₁ + m₃ v₃

Let's clear the speed of fragment 3

     v₃ = - m₁ v₁ / m₃

     v₃ = - (-10) 1 / 0.3

     v₃ = 33.3 m / s

X axis

     M v₀ₓ = m₂ v₂

     v₂ = v₀ₓ M / m₂

     v₂ = 38.3  2 / 0.7

     v₂ = 109.4 m / s

The fragment speeds of 0.3 kg is 33.3 m / s on the y axis

                                         0.7 kg is 109.4 ms on the x axis

b) The speed of the fragment is 33.3 m / s and has a starting height of where the fragmentation occurred, let's calculate with kinematics

       v_{fy}² = v_{oy}² - 2 gy

       0 =  v_{oy}²-2gy

       y =  v_{oy}² / 2g

       y = 32.14² / 2 9.8

       y = 52.7 m

This is the height where the break occurs, which is the initial height for body movement of 0.3 kg

      v_{f}² =  v_{y}² - 2 g y₂

      0 =  v_{y}² - 2 g y₂

     y₂ =  v_{y}² / 2g

     y₂ = 33.3²/2 9.8

     y₂ = 56.58 m

Total body height is

      Y = y + y₂

      Y = 52.7 + 56.58

     Y = 109.3 m

8 0
3 years ago
Some fuel cells are powered by hydrogen. Scientists are looking into the decomposition of water (H2O) to make hydrogen fuel with
mr_godi [17]

A challenge scientists face with this process is the use of ultrathin iron oxide, to pull protons off water and produce hydrogen gas, which itself is a poor electrical conductor.

5 0
3 years ago
Read 2 more answers
Buoyancy increases with the increase in the density of a) Submerged body b) Fluid​
AlekseyPX

Answer:

Submerged body

Explanation:

  • If buoyancy is greater than weight then object will float.
  • If buoyancy is less then weight then object will sink.
  • If buoyancy=weight then objects remains stable
4 0
3 years ago
A car initially at rest undergoes uniform acceleration for 6.32 seconds and covers a distance of 120 meters. What is the approxi
Alja [10]
The approximate acceleration of the car would be <span>3.00

</span>How? We have to use the formula to find the velocity.

v = d/t = 120/6.32

v = 19

Then plug the factor into the acceleration formula to find the acceleration:

a = v/t = 19/6.32

a = 3.00

So, the approximate acceleration of the car would be 3.00
4 0
4 years ago
While carrying a heavy box up the stairs,you set the box on a step and rest. Then you pick up the box and carry it to the top of
slega [8]

Answer: When the box is on a step and in a resting position then in this situation, the forces acting on the box is said to be balanced forces. While carrying the box to the top of the stair then the forces acting on it will be unbalanced forces.

Explanation:

When the box is on a step and in a resting position then the forces acting on the box are acting in an equal in magnitude and opposite in direction. Therefore, these forces are balanced forces.

While carrying the box to the top of the stair then the forces acting on it will be unbalanced forces because the box is changing its state. There is an unbalanced force which opposes the motion of the box.

6 0
4 years ago
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