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lisov135
13 days ago
13

A new roller coaster contains a loop-the-loop in which the car and rider are completely upside down. If the radius of the loop i

s 13.2 m, with what minimum speed must the car traverse the loop so that the rider does not fall out while upside down at the top? Assume the rider is not strapped to the car. Group of answer choices 10.1 m/s 11.4 m/s 14.9 m/s 12.5 m/s
Physics
1 answer:
Sav [1.1K]13 days ago
8 0

Respuesta:

11.4 m/s

Explicación:

La fórmula para la aceleración centrípeta es:

a=\frac{v^2}{R}

donde, a es la aceleración, v la velocidad alrededor de la circunferencia y R el radio del círculo.

En este problema,

a = g = aceleración debida a la gravedad en la cima = 9.81\ m/s^2

v = ?

R = 13.2 m

Por lo tanto,

9.81=\frac{v^2}{13.2}

v^2=9.81\times {13.2}

v = 11.4 m/s

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Ostrovityanka [942]

Answer:

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Explanation:

The power output from a car engine is equivalent to that of a bicycle since both perform the same amount of work over time. Both raul and katrina shared a frozen meal, heating each portion in different microwaves. Katrina's portion was warm in one minute, whereas raul's portion required two minutes. Therefore, the power utilized by raul's microwave aligns with that of katrina's, given that it took longer to achieve the same result.

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5 days ago
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Charge is placed on two conducting spheres that are very far apart and connected by a long thin wire. The radius of the smaller
Softa [913]

Answer:

σ₁ = 3.167 * 10^{-6} C/m²

σ₂ = 7.6 * 10 ^{-6}  C/m²

Explanation:

Provided Information:

i) Smaller sphere's radius ( r ) = 5 cm.

ii) Larger sphere's radius ( R ) = 12 cm.

iii) Electric field at the larger sphere's surface  ( E₁ ) = 358 kV/m, which is equivalent to 358 * 1000 v/m

E_{1} = (\frac{1}{4\pi\epsilon }) (\frac{Q_{1} }{R^{2} } )

358000 = 9 * 10^{9 } *\frac{Q_{1} }{0.12^{2} }

Charge (Q₁) = 572.8 * 10^{-9} C

Since the electric field inside a conductor is zero, the electric potential ( V ) remains constant.

V = constant

∴\frac{Q_{1} }{R} = \frac{Q_{2} }{r}

Q_{2} = \frac{r}{R} *Q_{1}

Q_{2} = \frac{5}{12} *572.8*10^{-9}   = 238.666 *10^{-9} C

Surface charge density ( σ₁ ) for the larger sphere.

Calculated Area ( A₁ )  = 4 * π * R²  = 4 * 3.14 * 0.12 = 0.180864 m².

σ₁  = \frac{Q_{1} }{A_{1} } = \frac{572.8 *10^{-9} }{0.180864} = 3.167 * 10^{-6}  C/m².

Surface charge density ( σ₂ ) for the smaller sphere.

Calculated Area ( A₂ )  = 4 * π * r²  = 4 * 3.14 * 0.05²  = 0.0314 m².

σ₂ =\frac{Q_{2} }{A_{2} } = \frac{238.66 *10^{-9} }{0.0314} = 7.6 * 10 ^{-6} C/m²

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4 days ago
tas watches as his uncle changes a flat tire on a car. his uncle raises the car using a machine called a jack. each time his unc
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Machines are essential for people to amplify their strength; without them, lifting a car would be impossible.

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Utilizing a greater lever allows for extensive movement with minimal force, resulting in the opposite side moving shorter distances with an increased force.

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An object is attached to a hanging unstretched ideal and massless spring and slowly lowered to its equilibrium position, a dista
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Answer:

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Explanation:

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distance = 6.4 cm

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