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Peru · Grado 11 · Practice Sheet 35

Sheet code: PERU-G11-S35 · 20 questions

Name: ______________
Date: ______________
  1. 1.

    S/ 2,500 is invested at 8% compound interest per annum for 4 years. Find the final amount (to 2 d.p.).

    Formula:Compound Interest
    A=P(1+r100)tA = P\left(1 + \dfrac{r}{100}\right)^{t}
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  2. 2.

    Find the gravitational potential energy of a 50 kg object raised 19 m. (g = 9.8 m/s²)

    Formula:Gravitational Potential Energy
    PE=mghPE = mgh
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  3. 3.

    In a triangle, a = 18, b = 10 and the included angle C = 100°. Find side c (to 2 d.p.).

    Formula:Cosine Rule
    c=a2+b22abcosCc = \sqrt{a^2 + b^2 - 2ab\cos C}
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  4. 4.

    A body starts at 20 m/s and accelerates at 2.5 m/s² for 8 s. Find its final velocity.

    Formula:Kinematics (v = u + at)
    v=u+atv = u + at
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  5. 5.

    S/ 1,500 is invested at 8% compound interest per annum for 4 years. Find the final amount (to 2 d.p.).

    Formula:Compound Interest
    A=P(1+r100)tA = P\left(1 + \dfrac{r}{100}\right)^{t}
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  6. 6.

    A current of 5 A flows through a 19 Ω resistor. Find the voltage across it.

    Formula:Ohm's Law
    V=IRV = I R
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  7. 7.

    Two masses 6,300 kg and 570 kg are 15 m apart. Find the gravitational force between them. (G = 6.674×10⁻¹¹)

    Formula:Newton’s Law of Gravitation
    F=Gm1m2r2F = G\dfrac{m_1 m_2}{r^2}
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  8. 8.

    Find the force needed to accelerate a 39 kg mass at 0.8 m/s².

    Formula:Newton's Second Law
    F=maF = ma
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  9. 9.

    Two masses 5,600 kg and 330 kg are 20 m apart. Find the gravitational force between them. (G = 6.674×10⁻¹¹)

    Formula:Newton’s Law of Gravitation
    F=Gm1m2r2F = G\dfrac{m_1 m_2}{r^2}
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  10. 10.

    A body starts at 14 m/s and accelerates at 2.5 m/s² for 8 s. Find its final velocity.

    Formula:Kinematics (v = u + at)
    v=u+atv = u + at
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  11. 11.

    An object with initial velocity 14 m/s accelerates at 3.5 m/s² for 3 s. Find the distance travelled.

    Formula:Kinematics (s = ut + ½at²)
    s=ut+12at2s = ut + \tfrac{1}{2}at^2
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  12. 12.

    A cylinder has radius 2 cm and height 26 cm. Find its volume. (π ≈ 3.14159)

    Formula:Volume of a Cylinder
    V=πr2hV = \pi r^2 h
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  13. 13.

    A triangle has sides a = 6 cm and b = 13 cm with an included angle C = 45°. Find its area (to 2 d.p.).

    Formula:Area of a Triangle (Sine Rule)
    A=12absinCA = \tfrac{1}{2} ab\sin C
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  14. 14.

    Find the population standard deviation of: 4, 2, 12, 2, 16 (to 2 d.p.).

    Formula:Standard Deviation (Population)
    σ=(xxˉ)2n\sigma = \sqrt{\dfrac{\sum (x - \bar{x})^2}{n}}
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  15. 15.

    Two masses 4,700 kg and 510 kg are 3 m apart. Find the gravitational force between them. (G = 6.674×10⁻¹¹)

    Formula:Newton’s Law of Gravitation
    F=Gm1m2r2F = G\dfrac{m_1 m_2}{r^2}
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  16. 16.

    In a triangle, a = 8, b = 11 and the included angle C = 100°. Find side c (to 2 d.p.).

    Formula:Cosine Rule
    c=a2+b22abcosCc = \sqrt{a^2 + b^2 - 2ab\cos C}
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  17. 17.

    Find the force needed to accelerate a 14 kg mass at 1 m/s².

    Formula:Newton's Second Law
    F=maF = ma
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  18. 18.

    Find the population standard deviation of: 18, 16, 7, 20, 3 (to 2 d.p.).

    Formula:Standard Deviation (Population)
    σ=(xxˉ)2n\sigma = \sqrt{\dfrac{\sum (x - \bar{x})^2}{n}}
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  19. 19.

    A cone has base radius 3 cm and height 10 cm. Find its volume. (π ≈ 3.14159)

    Formula:Volume of a Cone
    V=13πr2hV = \tfrac{1}{3} \pi r^2 h
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  20. 20.

    A current of 4.1 A flows through a 58 Ω resistor. Find the voltage across it.

    Formula:Ohm's Law
    V=IRV = I R
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