AP Calculus Quiz On Formulas!

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AP Calculus Quiz On Formulas! - Quiz

Advanced Placement Calculus is a set of two distinct Advanced Placement calculus courses and exams offered by College Board. AP Calculus AB covers limits, derivatives, and integrals. AP Calculus BC covers all AP Calculus AB topics plus additional topics (including more integration techniques such as integration by parts, Taylor series, parametric equations, polar coordinate functions, and curve interpolations).
This is a preparatory test ahead of the Calculus test.


Questions and Answers
  • 1. 

    Transposing I = V/R for resistance R gives

    • A.

      I-V

    • B.

      V-I

    • C.

      I/V

    • D.

      VI

    Correct Answer
    B. V-I
    Explanation
    The given equation, I = V/R, represents Ohm's Law, which states that the current (I) flowing through a conductor is equal to the voltage (V) across the conductor divided by the resistance (R) of the conductor. To transpose this equation for resistance (R), we can rearrange it as R = V/I. Therefore, the correct answer is V-I.

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  • 2. 

    When two resistors R1 and R2 are connected in parallel the formula 1/Rr = 1/R1 + 1/R2 the formula is used to determine the total resistance Rr. If R1 = 470 ohms, and R2 = 2.7k ohms, RT (in 3 significant figures) =

    • A.

      2.068 ohms

    • B.

      400 ohms

    • C.

      473 ohms

    • D.

      3170 ohms

    Correct Answer
    B. 400 ohms
    Explanation
    The formula 1/Rr = 1/R1 + 1/R2 is used to calculate the total resistance Rr when two resistors are connected in parallel. In this case, R1 is given as 470 ohms and R2 is given as 2.7k ohms. To calculate the total resistance, we substitute these values into the formula and solve for Rr. By doing the calculations, we find that Rr is equal to 400 ohms.

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  • 3. 

    The formula for a focal length f of a convex lens is 1/f = 1/u + 1/v, where f = 4 and u = 6, v is

    • A.

      -2

    • B.

      1/12

    • C.

      12

    • D.

      -1/2

    Correct Answer
    C. 12
    Explanation
    The formula for a focal length of a convex lens states that 1/f = 1/u + 1/v, where f is the focal length, u is the object distance, and v is the image distance. In this question, the given values are f = 4 and u = 6. By substituting these values into the formula, we can solve for v. Rearranging the formula, we get 1/v = 1/f - 1/u. Plugging in the given values, we have 1/v = 1/4 - 1/6. Simplifying this expression gives us 1/v = 3/12 - 2/12 = 1/12. Therefore, the value of v is 1/12.

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  • 4. 

    Volume = mass/density, the density in (kg/m³) when mass is 2.532kg and volume is 162cm³ is:

    • A.

      0.01563kg/m3

    • B.

      410.2kg/m3

    • C.

      15630kg/m3

    • D.

      64.0kg/m3

    Correct Answer
    C. 15630kg/m3
    Explanation
    The given formula for density is density = mass/volume. To find the density, we need to divide the given mass of 2.532kg by the volume of 162cm³. However, since the volume is given in cm³, we need to convert it to m³ by dividing by 100³. The result is 0.00162m³. Dividing the mass by the converted volume gives us a density of approximately 15630kg/m³.

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  • 5. 

    PV = mRT is the characteristic gas equation. When P = 100 × 10³, V = 4.0, R = 288, and T = 300, what is the value of m?

    • A.

      4.630

    • B.

      313600

    • C.

      0.216

    • D.

      100592

    Correct Answer
    A. 4.630
    Explanation
    The given equation is PV = mRT, where P is pressure, V is volume, R is the gas constant, T is temperature, and m is the molar mass. In this question, we are given P = 100 × 10^3, V = 4.0, R = 288, and T = 300. To find the value of m, we rearrange the equation as m = PV / RT. Substituting the given values, we get m = (100 × 10^3 × 4.0) / (288 × 300) = 4.630. Therefore, the value of m is 4.630.

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  • 6. 

    The volume V2 of a material when temperature is increased is by V2 = V1 [1 + y (t2 – t1)]. The value of t2 when V2 = 61.5cm³, V1 = 60cm³, y = 54 × 10^-6 and t1 = 250 is

    • A.

      213

    • B.

      463

    • C.

      713

    • D.

      28028

    Correct Answer
    C. 713
  • 7. 

    The current i amperes flowing in a capacitor at time t seconds is given by i = 10(1 − e^−t/CR), where resistance R is 25 × 10³ ohms and capacitance C is 16 × 10^−6 farads. When current i reaches 7 amperes, the time t is:

    • A.

      -0.48s

    • B.

      0.14s

    • C.

      0.21s

    • D.

      0.48s

    Correct Answer
    D. 0.48s
    Explanation
    The given equation represents the current in a capacitor as a function of time. To find the time at which the current reaches 7 amperes, we can substitute i = 7 into the equation and solve for t.

    7 = 10(1 - e^(-t/CR))

    Dividing both sides by 10:

    0.7 = 1 - e^(-t/CR)

    Rearranging the equation:

    e^(-t/CR) = 1 - 0.7

    Taking the natural logarithm of both sides:

    -t/CR = ln(0.3)

    Solving for t:

    t = -CR * ln(0.3)

    Substituting the given values for R and C:

    t = - (25 * 10^3) * (16 * 10^(-6)) * ln(0.3)

    Calculating the value:

    t ≈ 0.48s

    Therefore, the correct answer is 0.48s.

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  • 8. 

    A formula used for calculating the resistance of a cable is R = ρl/a . A cable’s resistance R = 0.50 ohms, its length l is 5000 m and its cross-sectional area a is 4 × 10-⁴ m². The resistivity ρ of the material is:

    • A.

      6.25 x 107 ohms m

    • B.

      4 x 10-8 ohms m

    • C.

      2.5 x 107 ohms m

    • D.

      3.2 x 10-7 ohms m

    Correct Answer
    B. 4 x 10-8 ohms m
    Explanation
    The formula for resistance of a cable is R = ρl/a, where R is the resistance, ρ is the resistivity of the material, l is the length of the cable, and a is the cross-sectional area of the cable. In this question, the resistance R is given as 0.50 ohms, the length l is 5000 m, and the cross-sectional area a is 4 x 10-⁴ m². Plugging these values into the formula, we can solve for ρ. The correct answer is 4 x 10-8 ohms m, which is the only option that gives the correct value for ρ when substituted into the formula.

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  • 9. 

    Current I in an electrical circuit is given by I = E − e/ R + r. Transposing for R gives:

    • A.

      (E - e - Ir)/I

    • B.

      (E - e)/(I + r)

    • C.

      (E - e)(I + r)

    • D.

      (E - e)/Ir

    Correct Answer
    A. (E - e - Ir)/I
    Explanation
    The correct answer is (E - e - Ir)/I. This is the correct transposition of the given equation. In order to solve for R, we need to isolate it on one side of the equation. By subtracting e and Ir from both sides and then dividing by I, we obtain the correct expression for R.

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  • 10. 

    Transposing t = 2π √l/g for g gives:

    • A.

      (t − 2π)² / l

    • B.

      (2π/ t) l²

    • C.

      (√t/2π) / l

    • D.

      4π²l / t²

    Correct Answer
    D. 4π²l / t²
    Explanation
    The given expression is t = 2π√l/g. To transpose the equation for g, we need to isolate g on one side of the equation. By rearranging the equation, we can get g = 4π²l/t². This can be done by squaring both sides of the equation, then multiplying both sides by l, and finally dividing both sides by t². Therefore, the correct answer is 4π²l / t².

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