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ch4aika [34]
3 years ago
14

Why are the loops on a roller coaster tear-drop shaped instead of circular? It would be simpler to design & build tracks wit

h a circular shape, so why do roller coaster manufacturers use a tear-drop shape instead? Be specific, and include a discussion of the forces acting on the rider, their centripetal acceleration, etc. as needed to make your point.
Physics
1 answer:
VLD [36.1K]3 years ago
5 0
The centripetal force acting on the rider is much greater if the roller coaster has a circular loop, rather than oval. This is because the change in direction is much sharper throughout the loop, causing the rider to experience a much more intense G-Force throughout the loop. 
A teardrop loop features a more gradual change of direction: the cart spends time moving upward, briefly changes direction, and spends the rest of the time moving downward and flattening to a horizontal path. This means the riders experience the majority of the force on the way down as the car levels out, rather than an intense G-force throughout the ride. 

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The electric field in a certain region of Earth’s atmo- sphere is directed vertically down. At an altitude of 300 m the field ha
I am Lyosha [343]

Answer:

Net charge contained in the cubeq= 3.536×10^-6C

Explanation:

Formular for total flux in a cube is given as:

Total flux= E300Acos(180) + E200Acos(0)

Where A is crossectional area

Total flux= A(E200-E300)

Total flux= q/Eo

q= Eo×total flux

q=(8.84×10^-12)×(100)^2×(100-60)

q= 3.536×10^-6C

6 0
4 years ago
Read 2 more answers
An Argon laser (λ = 5.0×102nm) shines down a silica glass fiber-optic cable with index of refraction 1.46. What is the speed of
Yanka [14]

Answer:

The speed of the laser light in the cable, c_f=2.1\times 10^8\ m/s

Explanation:

It is given that,

Wavelength of Argon laser, \lambda=5\times 10^2\ nm=5\times 10^{-7}\ m

Refractive index, n = 1.46

Let c_f is the speed of the laser light in the cable. The speed of light in a medium is given by :

c_f=\dfrac{c}{n}

c_f=\dfrac{3\times 10^8\ m/s}{1.46}

c_f=2.05\times 10^8\ m/s

or

c_f=2.1\times 10^8\ m/s

So, the speed of the laser light is 2.1\times 10^8\ m/s. Hence, this is the required solution.

4 0
4 years ago
Read 2 more answers
8. A 3.0 x 10' g nerf projectile is fired from a 1.5 kg nerf gun. If the velocity of the projectile is
tatuchka [14]

Answer:

v_g=0.2\ m.s^{-1}

Explanation:

Given:

  • mass of nerf projectile, m_p=30\ g=0.03\ kg
  • velocity of nerf projectile, v_p=10\ m.s^{-1}
  • mass of the gun, m_g=1.5\ kg

Now in this case the collision can be assumed to be perfectly elastic.

<u>For an elastic collision:</u>

m_p.v_p=m_g.v_g

where:

v_g= recoil velocity of the gun

0.03\times 10=1.5\times v_g

v_g=0.2\ m.s^{-1}

4 0
3 years ago
what are the horizontal and vertical components of a vector that is 25units long with an angle of 130 degrees​
AlekseyPX

Answer:

The horizontal component of the vector ≈ -16.06

The vertical component of the vector ≈ 19.15

Explanation:

The magnitude of the vector, \left | R \right | = 25 units

The direction of the vector, θ = 130°

Therefore, we have;

The horizontal component of the vector, Rₓ = \left | R \right | × cos(θ)

∴ Rₓ = 25 × cos(130°) ≈ -16.06

<em>The horizontal component of the vector, Rₓ ≈ -16.06</em>

The vertical component of the vector, R_y = \left | R \right | × sin(θ)

∴  R_y = 25 × sin(130°) ≈ 19.15

<em>The vertical component of the vector, R</em>_y<em> ≈ 19.15</em>

(The vector, R = Rₓ + R_y

\underset{R}{\rightarrow} = Rₓ·i + R_y·j

∴ \underset{R}{\rightarrow} ≈ -16.07·i + 19.15j)

5 0
3 years ago
When the material in the mantle cools off near the surface then sinks
MatroZZZ [7]
I’m guessing convection currents since you mention mantle and core, reminds me of heat
5 0
3 years ago
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