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Which affects racing car stability? Force F produces extension e in wire. If length & radius doubled, extension =?

Physics

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Part 46

Which affects racing car stability?

  • Stability in a racing car is enhanced by a low center of gravity (CG) and a wide base. A low CG reduces the tendency to roll over, and a wide base provides a larger support area, making the car less likely to tip.

The final answer is B. Low CG & wide base\boxed{\text{B. Low CG \& wide base}}.

Part 47

Product of force and perpendicular distance from pivot is

  • The product of a force and its perpendicular distance from a pivot point is the definition of moment (also known as torque). It measures the turning effect of a force.

The final answer is B. Moment\boxed{\text{B. Moment}}.

Part 48· 3 steps

Force F produces extension e in wire. If length & radius doubled, extension =

  1. Recall the formula for extension based on Young's Modulus.

    Young's Modulus (YY) is given by:

    Y=StressStrain=F/Ae/LY = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{e/L}

    where FF is the force, AA is the cross-sectional area, ee is the extension, and LL is the original length.
    From this, the extension ee can be expressed as:

    e=FLAYe = \frac{FL}{AY}

    Since the cross-sectional area A=πr2A = \pi r^2, we have:

    e=FLπr2Ye = \frac{FL}{\pi r^2 Y}
  2. Define the initial and new conditions.

    Let the initial extension be e1e_1, initial length L1L_1, and initial radius r1r_1.

    e1=FL1πr12Ye_1 = \frac{FL_1}{\pi r_1^2 Y}

    The new length is L2=2L1L_2 = 2L_1.
    The new radius is r2=2r1r_2 = 2r_1.
    The force FF and Young's Modulus YY remain constant.

  3. Calculate the new extension e2e_2.

    Substitute the new length and radius into the formula:

    e2=F(2L1)π(2r1)2Ye_2 = \frac{F(2L_1)}{\pi (2r_1)^2 Y} e2=2FL1π(4r12)Ye_2 = \frac{2FL_1}{\pi (4r_1^2) Y} e2=24(FL1πr12Y)e_2 = \frac{2}{4} \left( \frac{FL_1}{\pi r_1^2 Y} \right) e2=12e1e_2 = \frac{1}{2} e_1

    The new extension is half of the original extension.

    The final answer is C. e/2\boxed{\text{C. e/2}}.

Part 49· 4 steps

Block of mass M slides down frictionless plane inclined 30°. Acceleration (g = 10 m/s²) =

  1. Identify the forces acting on the block.

    On a frictionless inclined plane, the only force component causing acceleration down the plane is the component of gravity parallel to the plane.

  2. Apply Newton's Second Law.

    The force parallel to the plane is Fparallel=Mgsin⁡θF_{parallel} = Mg \sin \theta.
    According to Newton's Second Law, Fparallel=MaF_{parallel} = Ma, where aa is the acceleration.

    Ma=Mgsin⁡θMa = Mg \sin \theta
  3. Solve for acceleration aa.

    Cancel out the mass MM from both sides:

    a=gsin⁡θa = g \sin \theta
  4. Substitute the given values.

    Given g=10 m/s2g = 10 \text{ m/s}^2 and θ=30∘\theta = 30^\circ.

    a=10 m/s2×sin⁡30∘a = 10 \text{ m/s}^2 \times \sin 30^\circ

    We know that sin⁡30∘=0.5\sin 30^\circ = 0.5.

    a=10 m/s2×0.5a = 10 \text{ m/s}^2 \times 0.5 a=5 m/s2a = 5 \text{ m/s}^2

    The acceleration of the block is 5 m/s25 \text{ m/s}^2.

    The final answer is C. 5 m/s2\boxed{\text{C. 5 m/s}^2}.

Part 50

Point beyond which a material does not return to original shape is

  • The elastic limit is the maximum stress a material can withstand and still return to its original shape once the stress is removed. If the material is stressed beyond this point, it undergoes permanent (plastic) deformation and will not fully return to its original shape.

The final answer is B. Elastic limit\boxed{\text{B. Elastic limit}}.

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