justify your answer about which car if either completes one trip around the track in less tame quuantitatively with appropriate equations

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Answer 1

To determine which car completes one trip around the track in less time, we can analyze their respective velocities and the track distance.

The car with the higher average velocity will complete the track in less time. Let's denote the velocity of Car A as VA and the velocity of Car B as VB. The track distance is given as d.

We can use the equation:

Time = Distance / Velocity

For Car A:

Time_A = d / VA

For Car B:

Time_B = d / VB

To compare the times quantitatively, we need more information about the velocities of the cars.

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Related Questions

(ii) a skateboarder, with an initial speed of 2.0 ms, rolls virtually friction free down a straight incline of length 18 m in 3.3 s. at what angle u is the incline oriented above the horizontal?

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A skateboarder, with an initial speed of 2.0 ms, rolls virtually friction free down a straight incline of length 18 m in 3.3 s.The incline is oriented approximately 11.87 degrees above the horizontal.

To determine the angle (θ) at which the incline is oriented above the horizontal, we need to use the equations of motion. In this case, we'll focus on the motion in the vertical direction.

The skateboarder experiences constant acceleration due to gravity (g) along the incline. The initial vertical velocity (Viy) is 0 m/s because the skateboarder starts from rest in the vertical direction. The displacement (s) is the vertical distance traveled along the incline.

We can use the following equation to relate the variables:

s = Viy × t + (1/2) ×g ×t^2

Since Viy = 0, the equation simplifies to:

s = (1/2) × g × t^2

Rearranging the equation, we have:

g = (2s) / t^2

Now we can substitute the given values:

s = 18 m

t = 3.3 s

Plugging these values into the equation, we find:

g = (2 × 18) / (3.3^2) ≈ 1.943 m/s^2

The acceleration due to gravity along the incline is approximately 1.943 m/s^2.

To find the angle (θ), we can use the relationship between the angle and the acceleration due to gravity:

g = g ×sin(θ)

Rearranging the equation, we have:

θ = arcsin(g / g)

Substituting the value of g, we find:

θ = arcsin(1.943 / 9.8)

the angle θ is approximately 11.87 degrees.

Therefore, the incline is oriented approximately 11.87 degrees above the horizontal.

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

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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.

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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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An operational amplifier has to be designed for an on-chip audio band pass IGMF filter. Explain using appropriate mathematical derivations what the impact of reducing the input impedance (Zin), and reducing the open loop gain (A) of the opamp will have for the general opamps performance. What effect would any changes to (Zin) or (A) have on the design of an IGMF band pass filter?

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Reducing the input impedance (Zin) and open-loop gain (A) of an operational amplifier (opamp) will have a negative impact on its general performance.

Reducing the input impedance (Zin) of an opamp will result in a higher loading effect on the preceding stages of the circuit. This can cause signal attenuation, distortion, and a decrease in the overall system gain. Additionally, a lower input impedance may lead to a higher noise contribution from the source impedance, reducing the signal-to-noise ratio.

Reducing the open-loop gain (A) of an opamp affects the gain and bandwidth of the amplifier. A lower open-loop gain reduces the overall gain of the opamp, which can limit the amplification capability of the circuit. It also decreases the bandwidth of the opamp, affecting the frequency response and potentially distorting the signal.

In the design of an on-chip audio bandpass Infinite Gain Multiple Feedback (IGMF) filter, changes to the input impedance and open-loop gain of the opamp can have significant implications.

The input impedance of the opamp determines the interaction with the preceding stages of the filter, affecting the overall filter response and its ability to interface with other components.

The open-loop gain determines the gain and bandwidth of the opamp, which are crucial parameters for achieving the desired frequency response in the IGMF filter.

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Consider a radioactive sample. Determine the ratio of the number of nuclei decaying during the first half of its halflife to the number of nuclei decaying during the second half of its half-life.

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The ratio is 2. To determine the ratio of the number of nuclei decaying during the first half of the half-life to the number of nuclei decaying during the second half of the half-life, we need to understand the concept of half-life.



The half-life of a radioactive substance is the time it takes for half of the radioactive nuclei in a sample to decay. Let's say the half-life of the radioactive substance in question is represented by "t".

During the first half-life (t/2), half of the nuclei in the sample will decay. So, if we start with "N" nuclei, after the first half-life, we will have "N/2" nuclei remaining.

During the second half-life (t/2), another half of the remaining nuclei will decay. So, starting with "N/2" nuclei, after the second half-life, we will have "N/2" divided by 2, which is "N/4" nuclei remaining.

Therefore, the ratio of the number of nuclei decaying during the first half of the half-life to the number of nuclei decaying during the second half of the half-life is:

(N/2) / (N/4)

Simplifying this expression, we get:

(N/2) * (4/N)

This simplifies to:

2

So, the ratio is 2.

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