Physics High School

A children's roller coaster has a horizontal, circular loop of radius 4.00 m. Cars enter the loop with a speed of 11.5 m/s. How long does it take for a car to complete the circular loop?
0.488 s
0.655 s
3.05 s
0.347 s
2.19 s

Answers

Answer 1

The time required for a car to complete the circular loop in the children's roller coaster is approximately 2.19 seconds.

The time it takes for the car to complete the circular loop using the given value of 11.5 m/s as the initial velocity.

The formula to calculate the time is:

T = (2 π r) / v

Plugging in the values, we have:

T = (2 π × 4.00 m) / 11.5 m/s

T = (2 × 3.14  × 4.00 m) / 11.5 m/s

T ≈ 2.19 s

Therefore, the correct answer is approximately 2.19 seconds.

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Related Questions

Question 8 (F): There is a spherical conductor (radius a) with a total (free) charge Q on it. It is centered on the origin, and surrounded by a linear, isotropic, homogeneous dielectric (Xe) that fills the space a

Answers

The question involves a spherical conductor with a charge Q and a radius a, surrounded by a linear, isotropic, homogeneous dielectric (Xe).

Explanation: In this scenario, the spherical conductor acts as a source of electric field due to the charge Q. The dielectric material, in this case xenon (Xe), influences the electric field by altering its strength. The dielectric is linear, isotropic, and homogeneous, meaning it behaves uniformly in all directions and has constant properties throughout its volume.

When a dielectric is introduced, it affects the electric field by reducing the overall strength of the field within the material. This effect is quantified by the relative permittivity or dielectric constant (ε_r) of the material, which characterizes how much the electric field is weakened compared to a vacuum. The dielectric constant of xenon (Xe) determines the extent to which it weakens the electric field. The presence of the dielectric also alters the capacitance of the conductor, which relates the charge on the conductor to the potential difference across it. Overall, the introduction of the linear, isotropic, homogeneous dielectric (Xe) influences the electric field and capacitance of the spherical conductor with charge Q, leading to a modified electrostatic behavior in the surrounding space.

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use the formula to calculate the relativistic length of a 100 m long spaceship travelling at 3000 m s-1.

Answers

The relativistic length of a 100 m long spaceship traveling at 3000 m/s is approximately 99.9995 m.

The relativistic length contraction formula is given by: L=L0√(1-v^2/c^2)Where L is the contracted length.L0 is the original length. v is the velocity of the object. c is the speed of light. The formula to calculate the relativistic length of a 100 m long spaceship traveling at 3000 m/s is: L=L0√(1-v^2/c^2)Given, L0 = 100 mV = 3000 m/sc = 3 × 10^8 m/sSubstituting the values in the formula:L = 100 × √(1-(3000)^2/(3 × 10^8)^2)L = 100 × √(1 - 0.00001)L = 100 × √0.99999L = 100 × 0.999995L ≈ 99.9995 m.

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Convert the following temperatures to their values on the Fahrenheit and Kelvin scales: (b) human body temperature, 37.0°C.

Answers

The human body temperature is 98.6 °F and 310.15 K when converted to Fahrenheit and Kelvin scales respectively

The human body temperature is 37.0°C. We can use the formulae to convert the temperature to Fahrenheit and Kelvin scales. The formulae are given below:Fahrenheit scale: F = (9/5)*C + 32

Kelvin scale: K = C + 273.15where C is the temperature in Celsius scale.On the Fahrenheit scale:F = (9/5)*37 + 32= 98.6 °FTherefore, the human body temperature is 98.6 °F.On the Kelvin scale:K = 37 + 273.15= 310.15 K.

Therefore, the human body temperature is 310.15 K. In summary, the human body temperature is 98.6 °F and 310.15 K when converted to Fahrenheit and Kelvin scales respectively.

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3. a capacitor is connected across an oscillating emf. the peak current through the capacitor is 2.0 a. what is the peak current if: a. the capacitance c is doubled? b. the peak emf e0 is doubled? c. the frequency v is doubled?

Answers

Doubling the capacitance would halve the peak current, but the changes in peak emf and frequency would not directly impact the peak current without additional information about the circuit configuration.

To determine the effects on the peak current in a capacitor when certain parameters are changed, we can analyze each scenario separately:

a. If the capacitance (C) is doubled:

  The peak current (I) through a capacitor in an oscillating circuit is given by the equation:

  I = C * dV/dt

  Where dV/dt represents the rate of change of voltage across the capacitor.

  Doubling the capacitance while keeping the rate of change of voltage constant would result in a halving of the peak current. Therefore, the peak current would become 1.0 A.

b. If the peak emf (E0) is doubled:

  The peak current (I) in an oscillating circuit is also influenced by the peak emf. The relationship between peak current and peak emf depends on the circuit parameters and is determined by Ohm's Law and the impedance of the circuit.

  Without specific information about the circuit configuration, it is difficult to determine the exact relationship between the peak current and peak emf. Therefore, we cannot determine the new value of the peak current without additional information.

c. If the frequency (v) is doubled:

  Doubling the frequency in an oscillating circuit would not directly affect the peak current through the capacitor. The peak current is primarily determined by the capacitance, voltage, and circuit impedance. Therefore, doubling the frequency would not change the peak current.

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