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Anastaziya [24]
3 years ago
9

What happens when a roller coaster car moves down from the top of a hill?

Physics
2 answers:
Andrew [12]3 years ago
5 0

Answer:

Gravitational potential energy is converted into kinetic energy

Explanation:

At the top of the hill, the roller coaster car has gravitational potential energy, which is given by:

U=mgh

where m is the mass of the car, g is the gravitational acceleration, h is the height of the hill.

As the car descends the hill, it gains speed (v), so it also gains kinetic energy, given by:

K=\frac{1}{2}mv^2

where v is the speed of the car.

The mechanical energy of the car is the sum of its potential energy and kinetic energy:

E=U+K

In absence of friction, the law of conservation of energy states that the mechanical energy of the car is constant. Therefore, since as the car moves down the hill the potential energy decreases (because the height, h, decreases), it means that the kinetic energy must increase (and this is wy the speed increases). So we can say that gravitational potential energy is converted into kinetic energy.

VLD [36.1K]3 years ago
3 0
When the roller coaster moves down from the top of the hill, all of its stored potential energy is converted into kinetic energy to move it and when it goes back up the hill it turns kinetic into potential.

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A diffraction grating with 230 lines per mm is used in an experiment to study the visible spectrum of a gas discharge tube. At w
mr_godi [17]

Answer:

θ₁ = 5.4°

θ₂ = 10.86°

Explanation:

The angle ca be found by using grating equation:

mλ = d Sinθ

where,

m = order of diffraction

λ = wavelength = 405.3 nm = 4.053 x 10⁻⁷ m

d = grating element = 1/230 lines/mm = 0.0043 mm/line = 4.3 x 10⁻⁶ m/line

θ = angle = ?

FOR m = 1:

(1)(4.053 x 10⁻⁷ m) = (4.3 x 10⁻⁶ m/line) Sin θ₁

Sin θ₁ = 0.09425

θ₁ = Sin⁻¹(0.09425)

<u>θ₁ = 5.4°</u>

<u></u>

FOR m = 2:

(2)(4.053 x 10⁻⁷ m) = (4.3 x 10⁻⁶ m/line) Sin θ₁

Sin θ₂ = 0.1885

θ₂ = Sin⁻¹(0.1885)

<u>θ₂ = 10.86°</u>

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3 years ago
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Public television station KQED in San Francisco broadcasts a sinusoidal radio signal at a power of 777 kW. Assume that the wave
ehidna [41]

Answer:

A) P = 3.3 × 10^(-11) Pa

B) Amplitude of electric field = 1.931 N/C

Amplitude of magnetic field = 6.44 × 10^(-9) T

C) μ_av = 1.65 × 10^(-11) J/m³

D) 50% each for the electric and magnetic field

Explanation:

A) First of all let's calculate intensity.

I = P_av/A

We are given;

P_av = 777 KW = 777,000 W

Distance = 5 km = 5000 m

Thus;

I = 777000/(2π × 5000²)

I = 0.00495 W/m²

Now, the average pressure would be given by the formula;

P = 2I/C

Where C is speed of light = 3 × 10^(8) m/s

P = (2 × 0.00495)/(3 × 10^(8))

P = 3.3 × 10^(-11) Pa

B) Formula for the amplitude of the electric field is gotten from;

E_max = √[2I/(εo•c)].

Where εo is the Permittivity of free space with a constant value of 8.85 × 10^(−12) c²/N.mm²

I and c remain as before.

Thus;

E_max = √[(2 × 0.00495)/(8.85 × 10^(−12) × 3 × 10^(8))]

E_max = √3.72881355932

E_max = 1.931 N/C

Formula for amplitude of magnetic field is gotten from;

B_max = E_max/c

B_max = 1.931/(3 × 10^(8))

B_max = 6.44 × 10^(-9) T

C) Formula for average density is;

μ_av = εo(E_rms)²

Now, E_rms = E_max/√2

Thus;

E_rms = 1.931/√2

μ_av = 8.85 × 10^(−12) × (1.931/√2)²

μ_av = 1.65 × 10^(-11) J/m³

D) The energy density for the electric and magnetic field is the same. So both of them will have 50% of the energy density.

4 0
3 years ago
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Leno4ka [110]

Answer:

by reducing friction.....

6 0
3 years ago
Is eureka a better conductor than copper
cricket20 [7]
Nope Copper is a better conductor
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