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kvasek [131]
3 years ago
12

Where is potential energy the greatest on a rollercoaster?

Physics
1 answer:
lidiya [134]3 years ago
4 0
On a roller coaster, the greatest potential energy is at the highest point of the roller coaster
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Two particles oscillate in simple harmonic motion along a common straight-line segment of length 1.0 m. Each particle has a peri
igor_vitrenko [27]

Answer:

a) the particles are <em>0.217 m </em>apart

b) <em>the particles are moving in the same direction</em>.

Explanation:

a) The amplitude of the oscillations is A/2 and the period of each particle is

T = 1.5 s however, they differ by a phase of π/6 rad. Let the phase of the first particle be zero so that the phase of the second particle is π/6. So we can write the coordinates of each of the particles as,

x₁ = A/2 cos(ωt)

x₂ = A/2 cos(ωt + π/6)

we can write the angular frequency ω, as

ω = 2π / T

so,

x₁ = A/2 cos(2π / T)

x₂ = A/2 cos(2π / T + π/6)

Thus, the coordinates of the particles at t = 0.45 s are,

x₁ = A/2 cos((2π × 0.45) / 1.5)) = -0.155 A

x₂ = A/2 cos((2π × 0.45) / 1.5) + π/6) = -0.372 A

Their separation at that time is, therefore,

Δx = x₁ - x₂

    = -0.155 A + 0.372 A

    = 0.217 A

since A = 1 m

Thus,

<em>Δx  = 0.217 m</em>

<em></em>

<em></em>

b) In order to find their directions, we must take the derivatives at t = 0.45 s.

Therefore,

v₁ = dx₁ / dt

   = (-πA / T) sin(2πt / T)

   = -(π(1) / 1.5) sin(2π(0.45) / 1.5)

   = -1.99

and,

v₂ = dx₂ / dt

   = (-πA / T) sin((2πt / T) + π/6)

   = -(π(1) / 1.5) sin((2π(0.45) / 1.5) + π/6)

   = -1.40

Since both v₁ and v₂ are negative, this shows that <em>the particles are moving in the same direction</em>.

6 0
3 years ago
The sky is blue because: Select one: a. the index of refraction for air is slightly larger for blue than for red b. atomic hydro
lapo4ka [179]

Answer:

a. the index of refraction for air is slightly larger for blue than for red

5 0
2 years ago
The smallest detail visible with ground-based solar telescopes is about 1 arc second. How large a region (in km) does this repre
mestny [16]

Answer:

x = 727.5 km

Explanation:

With the conditions given using trigonometry, we can find the tangent

       tan θ = CO / CA

With CO the opposite leg and CE is the adjacent leg which is the distance from the Tierral to Sun

   

        D =150 10⁶ km (1000m / 1 km)

        D = 150 10⁹ m.

We must take the given angle to radians.

       1º  = 3600 arc s  

       π rad = 180º

       θ = 1 arc s (1º / 3600 s arc) (pi rad / 180º) =

       θ = 4.85 10⁻⁶ rad

That angle is extremely small, so we can approximate the tangent to the angle

     

       θ = x / D

       x = θ D

       x = 4.85 10-6  150 109

       x = 727.5 103 m

       x = 727.5 km

4 0
3 years ago
An object is dropped from rest from a 70.6 m tower. Air resistance is negligible. After 0.32 seconds, what is magnitude and dire
dem82 [27]

Answer:

<em>1,378.9ms²</em>

Explanation:

Given the following

Distance S = 70.6m

Time t = 0.32secs

Initial velocity = 0m/s

Required

Acceleration

Using the equation of motion

S = ut+1/2at²

Substitute

70.6 = 0+1/2a(0.32)²

70.6 = 0.0512a

a = 70.6/0.0512

a = 1,378.9

<em>Hence the acceleration is 1,378.9ms²</em>

7 0
2 years ago
A car initially traveling 7 m/s speeds up uniformly at a rate of 3 m/s2 until it reaches a velocity of 22 m/s. How much time did
dezoksy [38]

Answer:

t = 5 s

Explanation:

Data:

  • Initial Velocity (Vo) = 7 m/s
  • Acceleration (a) = 3 m/s²
  • Final Velocity (Vf) = 22 m/s
  • Time (t) = ?

Use formula:

  • \boxed{t=\frac{Vf - Vo}{a}}

Replace:

  • \boxed{t=\frac{22\frac{m}{s} -7\frac{m}{s}}{3\frac{m}{s^{2}}}}

Solve the subtraction of the numerator:

  • \boxed{t=\frac{15\frac{m}{s}}{3\frac{m}{s^{2}}}}

It divides:

  • \boxed{t=5\ s}

How much time did it take the car to reach this final velocity?

It took a time of <u>5 seconds.</u>

8 0
2 years ago
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