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agasfer [191]
2 years ago
6

At what minimum speed must a roller coaster be traveling when upside down at the top of a circle so that the passengers do not f

all out? Assume a radius of curvature of 7.5
Physics
1 answer:
worty [1.4K]2 years ago
8 0

Answer:

v_{min} \approx 17.153\,\frac{m}{s}

Explanation:

The roller coaster begins with maximum kinetic energy and no gravitational potential energy. The gravitational potential energy reaches its maximum when roller coaster is upside down at the top of the circle. The physical model for the roller coaster is constructed by means of the Principle of Energy Conservation:

\frac{1}{2}\cdot m \cdot v_{min}^{2} = m\cdot g \cdot h

The minimum velocity is:

v_{min} = \sqrt{2\cdot g \cdot h}

Let assume that radio of curvature is measured in meters. Hence:

v_{min} = \sqrt{2\cdot (9.807\,\frac{m}{s^{2}} )\cdot(15\,m)}

v_{min} \approx 17.153\,\frac{m}{s}

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A 7.5 cm object is 14 cm from a concave lens, which has a focal length of –7 cm. Its image is 5 cm in front of the lens. What is
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Given conditions:
height of object = 7.5cmdistance of object from mirror  = 14 cmfocus length = -7 cmimage distance =  ?
Using mirror formula: 
1/(focus length) = 1/(object distance) + 1/(image distance)
or, -1/7 = 1/14 + 1/(image distance)
or, image distance = -4.66cm (the image formed is a virtual image)


Also, magnification of image is:
image height /height of object = - image distance /object distance
or, image height = - image distance / object distance * height of object
or, image height = -(-4.66) / 14 * 7.5 = 2.49 = 3(nearest whole number)
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2 years ago
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Answer:

A. Particles in air move in circles as the wave moves forward.

B. Particles in air move forward but not backward as the wave moves

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C. Particles in air move up and down as the wave moves forward.

✔ D. Particles in air move forward and backward as the wave moves

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Explanation:

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a girl pushes a cart to the left with a 100-N force. A boy pushes it to the right with a 50-N force. what is the net force exert
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Answer:

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Explanation:

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A child on a sled slides (starting from rest) down an icy slope that makes an angle of 15◦ with the horizontal. After sliding 20
statuscvo [17]

Answer:

A) v₁ = 10.1 m/s t₁= 4.0 s

B) x₂= 17.2 m

C) v₂=7.1 m/s

D) x₂=7.5 m

Explanation:

A)

  • Assuming no friction, total mechanical energy must keep constant, so the following is always true:

       \Delta K + \Delta U = (K_{f} - K_{o}) +( U_{f} - U_{o}) = 0  (1)

  • Choosing the ground level as our zero reference level, Uf =0.
  • Since the child starts from rest, K₀ = 0.
  • From (1), ΔU becomes:
  • \Delta U = 0- m*g*h = -m*g*h (2)  
  • In the same way, ΔK becomes:
  • \Delta K = \frac{1}{2}*m*v_{1}^{2}  (3)      
  • Replacing (2) and (3) in (1), and simplifying, we get:

       \frac{1}{2}*v_{1}^{2}  = g*h  (4)

  • In order to find v₁, we need first to find h, the height of the slide.
  • From the definition of sine of an angle, taking the slide as a right triangle, we can find the height h, knowing the distance that the child slides down the slope, x₁, as follows:

       h = x_{1} * sin \theta_{1} = 20.0 m * sin 15 = 5.2 m (5)

       Replacing (5) in (4) and solving for v₁, we get:

      v_{1} = \sqrt{2*g*h} = \sqrt{2*9.8m/s2*5.2m} = 10.1 m/s  (6)

  • As this speed is achieved when all the energy is kinetic, i.e. at the bottom of the first slide, this is the answer we were looking for.
  • Now, in order to finish A) we need to find the time that the child used to reach to that point, since she started to slide at the its top.
  • We can do this in more than one way, but a very simple one is using kinematic equations.
  • If we assume that the acceleration is constant (which is true due the child is only accelerated by gravity), we can use the following equation:

       v_{1}^{2} - v_{o}^{2} = 2*a* x_{1}  (7)

  • Since v₀ = 0 (the child starts from rest) we can solve for a:

       a = \frac{v_{1}^{2}}{2*x_{1} } = \frac{(10.1m/s)^{2}}{2* 20.0m} = 2.6 m/s2  (8)

  • Since v₀ = 0, applying the definition of acceleration, if we choose t₀=0, we can find t as follows:

       t_{1} =\frac{v_{1} }{a} =\frac{10.1m/s}{2.6m/s2} = 4.0 s  (9)

B)

  • Since we know the initial speed for this part, the acceleration, and the time, we can use the kinematic equation for displacement, as follows:

       x_{2} = v_{1} * t_{2} + \frac{1}{2} *a_{2}*t_{2}^{2}  (10)

  • Replacing the values of v₁ = 10.1 m/s, t₂= 2.0s and a₂=-1.5m/s2 in (10):

       x_{2} = 10.1m/s * 2.0s + \frac{1}{2} *(-1.5m/s2)*(2.0s)^{2}  = 17.2 m (11)

C)  

  • From (6) and (8), applying the definition for acceleration, we can find the speed of the child whem she started up the second slope, as follows:

       v_{2} = v_{1} + a_{2} *t_{2} = 10.1m/s - 1.5m/s2*2.0s = 7.1 m/s (12)

D)

  • Assuming no friction, all the kinetic energy when she started to go up the second slope, becomes gravitational potential energy when she reaches to the maximum height (her speed becomes zero at that point), so we can write the following equation:

       \frac{1}{2}*v_{2}^{2}  = g*h_{2}   (13)

  • Replacing from (12) in (13), we can solve for h₂:

       h_{2} =\frac{v_{2} ^{2}}{2*g} = \frac{(7.1m/s) ^{2}}{2*9.8m/s2} = 2.57 m  (14)

  • Since we know that the slide makes an angle of 20º with the horizontal, we can find the distance traveled up the slope applying the definition of sine of an angle, as follows:

       x_{3} = \frac{h_{2} }{sin 20} = \frac{2.57m}{0.342} = 7.5 m (15)

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when there is an increase in thermal energy, the kinetic energy if the matter increases as well.

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