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MTEL Physics (69) study materials
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MTEL Physics (69) practice questions

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Question 1Mechanics and Properties of Matter

A ball is released from rest near Earth's surface and falls freely a distance of 45 m. Using g = 10 m/s² and ignoring air resistance, how long does the ball take to fall?

  1. 4.5 s
  2. 9.0 s
  3. 3.0 s Correct answer.
  4. 2.0 s

Explanation

For free fall from rest, the distance fallen is h = ½gt². Solving for time gives t = √(2h/g) = √(2 × 45 / 10) = √9 = 3.0 s. The result assumes constant acceleration g and negligible air resistance.

Question 2Mechanics and Properties of Matter

A 5.0 kg block slides down a 30° incline with a coefficient of kinetic friction of 0.20. Using g = 10 m/s², what is the magnitude of the block's acceleration along the incline?

  1. 3.3 m/s² Correct answer.
  2. 5.0 m/s²
  3. 1.7 m/s²
  4. 6.7 m/s²

Explanation

The component of gravity along the incline is g sinθ, and kinetic friction opposes motion with magnitude μg cosθ. Thus a = g(sinθ − μ cosθ) = 10(0.50 − 0.20 × 0.866) ≈ 3.3 m/s². The mass cancels and does not affect the acceleration.

Question 3Mechanics and Properties of Matter

A car travels around a flat circular curve of radius 50 m at a constant speed of 20 m/s. What is the magnitude of the car's centripetal acceleration?

  1. 8.0 m/s² Correct answer.
  2. 10. m/s²
  3. 0.40 m/s²
  4. 4.0 m/s²

Explanation

Centripetal acceleration for uniform circular motion is a_c = v²/r. Substituting v = 20 m/s and r = 50 m gives a_c = (20²)/50 = 400/50 = 8.0 m/s². The acceleration points toward the center of the curve even though the speed is constant.

Question 4Mechanics and Properties of Matter

A spring with spring constant 200 N/m is compressed 0.10 m and releases a 0.50 kg block on a frictionless horizontal surface. What is the block's speed as it leaves the spring?

  1. 4.0 m/s
  2. 1.0 m/s
  3. 2.0 m/s Correct answer.
  4. 8.0 m/s

Explanation

Elastic potential energy stored in the spring is ½kx², and it converts entirely into kinetic energy ½mv² on a frictionless surface. Canceling ½ gives v = x√(k/m) = 0.10 × √(200/0.50) = 0.10 × 20 = 2.0 m/s. The block's mass matters because the same energy is shared with a larger or smaller inertia.

Question 5Mechanics and Properties of Matter

A 2.0 kg cart moving at 3.0 m/s to the right collides with a stationary 4.0 kg cart, and the two carts stick together. What is their common speed immediately after the collision?

  1. 2.0 m/s
  2. 1.5 m/s
  3. 1.0 m/s Correct answer.
  4. 0.50 m/s

Explanation

Momentum is conserved in the collision: (2.0 kg)(3.0 m/s) + 0 = (2.0 kg + 4.0 kg)v. Solving 6.0 = 6.0v gives v = 1.0 m/s. Kinetic energy is not conserved because the carts stick together, so an inelastic analysis is required.

Question 6Mechanics and Properties of Matter

A uniform meter stick is pivoted at the 30 cm mark. A 2.0 N weight hangs from the stick at the 10 cm mark. What is the magnitude of the torque produced by this weight about the pivot?

  1. 2.0 N·m
  2. 0.40 N·m Correct answer.
  3. 0.60 N·m
  4. 0.20 N·m

Explanation

Torque magnitude is τ = rF sinθ, where r is the distance from the pivot to the point of force application. Here r = 0.30 m − 0.10 m = 0.20 m, and the weight acts perpendicular to the stick, so τ = (0.20 m)(2.0 N) = 0.40 N·m. The pivot location, not the stick's full length, sets the lever arm.

Question 7Mechanics and Properties of Matter

A solid block floats in water with 60% of its volume below the surface. What is the density of the block? (Density of water = 1.0 g/cm³.)

  1. 1.0 g/cm³
  2. 0.60 g/cm³ Correct answer.
  3. 0.40 g/cm³
  4. 0.80 g/cm³

Explanation

For a floating object, the buoyant force equals the object's weight: ρ_water V_submerged g = ρ_block V_total g. Thus ρ_block = ρ_water × (V_submerged/V_total) = 1.0 × 0.60 = 0.60 g/cm³. The 40% above water is consistent with this density ratio.

Question 8Mechanics and Properties of Matter

A simple pendulum has a period of 2.0 s near Earth's surface. Using g = 9.8 m/s², what is the approximate length of the pendulum?

  1. 1.0 m Correct answer.
  2. 0.25 m
  3. 2.0 m
  4. 0.50 m

Explanation

The period of a simple pendulum is T = 2π√(L/g). Rearranging gives L = g(T/2π)² = 9.8 × (2.0/6.283)² ≈ 9.8 × 0.101 ≈ 0.99 m. Thus the length is approximately 1.0 m.

Question 9Mechanics and Properties of Matter

A satellite moves in a circular orbit of radius r around Earth with orbital speed v. If it is moved to a circular orbit of radius 2r, what is its new orbital speed?

  1. 2v
  2. v/√2 Correct answer.
  3. v/4
  4. v/2

Explanation

For a circular orbit, gravitational force supplies the centripetal force: GMm/r² = mv²/r, so v = √(GM/r). Speed therefore scales as 1/√r, and doubling r gives v_new = v/√2 ≈ 0.707v. The satellite's mass cancels out of the orbital-speed expression.

Question 10Mechanics and Properties of Matter

A uniform beam of weight 200 N and length 4.0 m is supported at both ends. A 300 N load hangs 1.0 m from the left end. What is the upward force exerted by the right support?

  1. 125 N
  2. 325 N
  3. 175 N Correct answer.
  4. 500 N

Explanation

Take torques about the left end so the left support force has zero lever arm. The 300 N load at 1.0 m and the 200 N beam weight at its center (2.0 m) create clockwise torques totaling 300 × 1.0 + 200 × 2.0 = 700 N·m. The right support at 4.0 m must provide an equal counterclockwise torque, so F_right = 700/4.0 = 175 N.

Question 11Mechanics and Properties of Matter

A projectile is launched at an angle above the horizontal and follows an ideal parabolic path with negligible air resistance. At the highest point of its trajectory, which statement about the projectile is correct?

  1. The vertical component of its velocity is zero. Correct answer.
  2. Its acceleration is zero.
  3. The horizontal component of its velocity is zero.
  4. Both velocity components are zero.

Explanation

In ideal projectile motion, the horizontal velocity component remains constant, while the vertical velocity component decreases under gravity and becomes zero at the highest point. The acceleration is g downward throughout the flight, not zero.

Question 12Mechanics and Properties of Matter

Within the elastic limit, a metal wire is stretched by a tensile force. The ratio of tensile stress to tensile strain for the wire is called which property?

  1. Poisson's ratio
  2. Bulk modulus
  3. Shear modulus
  4. Young's modulus Correct answer.

Explanation

Young's modulus is defined as tensile stress divided by tensile strain for a material in the linear elastic region. It characterizes stiffness under tension or compression, while the other moduli describe different types of deformation.

Question 13Mechanics and Properties of Matter

A diver is 3.0 m below the surface of a freshwater lake. Using a water density of 1000 kg/m³ and g = 10 m/s², what is the gauge pressure at the diver's location?

  1. 3.0 × 10^3 Pa
  2. 300 Pa
  3. 3.0 × 10^4 Pa Correct answer.
  4. 1.3 × 10^5 Pa

Explanation

Gauge pressure depends only on the fluid depth: P_gauge = ρgh = (1000 kg/m³)(10 m/s²)(3.0 m) = 3.0 × 10^4 Pa. Atmospheric pressure is not included in a gauge reading.

Question 14Mechanics and Properties of Matter

A 60 kg student stands on a scale in an elevator that is accelerating upward at 2.0 m/s². Using g = 10 m/s², what is the scale reading in newtons?

  1. 480 N
  2. 600 N
  3. 720 N Correct answer.
  4. 840 N

Explanation

The scale reads the normal force on the student. Newton's second law gives N − mg = ma, so N = m(g + a) = 60 kg(10 + 2.0) m/s² = 720 N. Upward acceleration increases the apparent weight.

Question 15Mechanics and Properties of Matter

A 4.0 kg crate is initially at rest on a frictionless horizontal floor. A constant horizontal force of 20 N is applied over a distance of 3.0 m. What is the crate's final speed?

  1. 7.7 m/s
  2. 5.5 m/s Correct answer.
  3. 15 m/s
  4. 3.9 m/s

Explanation

The applied force does work W = Fd = 20 N × 3.0 m = 60 J. By the work–energy theorem, this equals the change in kinetic energy, so ½(4.0 kg)v² = 60 J and v ≈ 5.5 m/s.

Question 16Mechanics and Properties of Matter

A 0.20 kg ball approaches a bat horizontally at 15 m/s and rebounds horizontally at 10 m/s in the opposite direction. If the ball is in contact with the bat for 0.010 s, what is the magnitude of the average force exerted by the bat on the ball?

  1. 1000 N
  2. 100 N
  3. 300 N
  4. 500 N Correct answer.

Explanation

Taking the initial direction as positive, the velocity change is −10 m/s − 15 m/s = −25 m/s. The impulse magnitude is m|Δv| = 0.20 kg × 25 m/s = 5.0 N·s, so the average force is 5.0 N·s / 0.010 s = 500 N.

Question 17Mechanics and Properties of Matter

A uniform solid disk of mass 2.0 kg and radius 0.50 m rotates about an axis through its center. A net torque of 1.5 N·m is applied to the disk. What is the magnitude of its angular acceleration?

  1. 3.0 rad/s²
  2. 6.0 rad/s² Correct answer.
  3. 12 rad/s²
  4. 1.5 rad/s²

Explanation

For a uniform solid disk about its central axis, I = ½MR² = ½(2.0 kg)(0.50 m)² = 0.25 kg·m². Newton's second law for rotation gives α = τ/I = 1.5 N·m / 0.25 kg·m² = 6.0 rad/s².

Question 18Mechanics and Properties of Matter

A 0.50 kg ball is attached to a string of length 1.0 m and swings in a vertical circle. At the highest point of the path, the ball's speed is 5.0 m/s. Using g = 10 m/s², what is the magnitude of the tension in the string at that point?

  1. 12.5 N
  2. 5.0 N
  3. 2.5 N
  4. 7.5 N Correct answer.

Explanation

At the highest point, both tension and weight point toward the center of the circle. Thus T + mg = mv²/r, and T = (0.50 kg)(5.0 m/s)²/(1.0 m) − (0.50 kg)(10 m/s²) = 12.5 N − 5.0 N = 7.5 N.

Question 19Mechanics and Properties of Matter

A block of volume 0.30 m³ and density 600 kg/m³ is held completely submerged in water by a vertical rope. Using water density 1000 kg/m³ and g = 10 m/s², what is the tension in the rope?

  1. 3000 N
  2. 1800 N
  3. 600 N
  4. 1200 N Correct answer.

Explanation

The buoyant force is ρ_waterVg = (1000)(0.30)(10) = 3000 N upward, and the block's weight is ρ_blockVg = (600)(0.30)(10) = 1800 N downward. For equilibrium while held submerged, the rope tension is 3000 N − 1800 N = 1200 N.

Question 20Mechanics and Properties of Matter

At a distance of one Earth radius above Earth's surface, what is the gravitational field strength compared with its value at Earth's surface?

  1. one-half
  2. one-fourth Correct answer.
  3. one-eighth
  4. the same

Explanation

Gravitational field strength follows the inverse-square law, g = GM/r². At one Earth radius above the surface, the distance from Earth's center is 2R rather than R, so the field is (1/2)² = 1/4 as strong.

Question 21Mechanics and Properties of Matter

A spacecraft is far from any star and moves with its engines off at a constant velocity of 200 m/s. What is the net force acting on the spacecraft?

  1. 200 N in the direction of motion
  2. a force equal to the spacecraft's weight
  3. 0 N Correct answer.
  4. a force that decreases as the spacecraft moves

Explanation

With the engines off and gravity negligible, no forward force is needed to keep the spacecraft moving. Constant velocity means zero acceleration, so Newton's first law requires the net force to be zero.

Question 22Mechanics and Properties of Matter

A solid cube has sides of length 0.10 m and a mass of 0.80 kg. What is its density?

  1. 8.0 kg/m³
  2. 8000 kg/m³
  3. 800 kg/m³ Correct answer.
  4. 80 kg/m³

Explanation

The cube's volume is (0.10 m)³ = 1.0 × 10⁻³ m³. Density is mass divided by volume, so 0.80 kg ÷ 1.0 × 10⁻³ m³ = 800 kg/m³.

Question 23Mechanics and Properties of Matter

A book rests on a table. Earth pulls downward on the book with a gravitational force. According to Newton's third law, which force is the reaction partner to this gravitational pull?

  1. the normal force exerted by the table on the book
  2. the gravitational pull exerted by the book on Earth Correct answer.
  3. the normal force exerted by the book on the table
  4. the weight of the book

Explanation

Newton's third-law pairs act on different objects, are equal in magnitude, opposite in direction, and are the same type of force. The reaction to Earth pulling the book is therefore the book pulling Earth, not a normal force.

Question 24Mechanics and Properties of Matter

A 2.0 kg box is pulled from rest across a rough horizontal floor by a 12 N horizontal force. A kinetic friction force of 4.0 N acts on the box. After the box has moved 5.0 m, what is its speed?

  1. 6.3 m/s Correct answer.
  2. 4.5 m/s
  3. 10 m/s
  4. 8.9 m/s

Explanation

The net horizontal force is 12 N − 4.0 N = 8.0 N, so the net work is 8.0 N × 5.0 m = 40 J. By the work-energy theorem, ½(2.0 kg)v² = 40 J, giving v = √40 ≈ 6.3 m/s.

Question 25Mechanics and Properties of Matter

A 3.0 kg mass and a 6.0 kg mass are connected by a light rod of length 1.2 m. How far from the 3.0 kg mass is the center of mass of the system?

  1. 0.40 m
  2. 0.80 m Correct answer.
  3. 0.60 m
  4. 1.0 m

Explanation

Taking the 3.0 kg mass at x = 0, the center of mass is (3.0 kg × 0 + 6.0 kg × 1.2 m) ÷ 9.0 kg = 0.80 m. The heavier mass pulls the center of mass closer to its end, two-thirds of the way along the rod.

Question 26Mechanics and Properties of Matter

Water flows through a horizontal pipe that has a radius of 2.0 cm at one section and 1.0 cm at a narrower section. If the speed in the 2.0 cm section is 1.5 m/s, what is the speed in the 1.0 cm section?

  1. 0.38 m/s
  2. 6.0 m/s Correct answer.
  3. 3.0 m/s
  4. 12 m/s

Explanation

The continuity equation for an incompressible fluid is A₁v₁ = A₂v₂. Halving the radius reduces the cross-sectional area to one-fourth, so the speed must become four times larger: 4 × 1.5 m/s = 6.0 m/s.

Question 27Mechanics and Properties of Matter

A hydraulic lift has a small piston with radius 5.0 cm and a large piston with radius 25 cm. If a force of 100 N is applied to the small piston, what force is exerted by the large piston?

  1. 2500 N Correct answer.
  2. 250 N
  3. 500 N
  4. 20 N

Explanation

Pascal's principle says the pressure is transmitted undiminished, so F₁/A₁ = F₂/A₂. The area ratio is (25 cm / 5.0 cm)² = 25, so the output force is 25 × 100 N = 2500 N.

Question 28Mechanics and Properties of Matter

A 0.50 kg mass attached to a horizontal spring oscillates with a period of 0.80 s. What is the spring constant of the spring?

  1. 31 N/m Correct answer.
  2. 39 N/m
  3. 15 N/m
  4. 7.7 N/m

Explanation

For a mass-spring oscillator, T = 2π√(m/k). Rearranging gives k = 4π²m/T² = 4π²(0.50 kg)/(0.80 s)² ≈ 31 N/m.

Question 29Mechanics and Properties of Matter

A spinning ice skater pulls her arms inward, reducing her moment of inertia to one-half of its original value. No external torque acts on her. By what factor does her rotational kinetic energy change?

  1. It quadruples.
  2. It doubles. Correct answer.
  3. It is unchanged.
  4. It halves.

Explanation

With no external torque, angular momentum L = Iω is conserved. Halving I doubles ω, and since rotational kinetic energy is L²/(2I), halving I doubles the kinetic energy; the extra energy comes from work done by the skater pulling her arms inward.

Question 30Mechanics and Properties of Matter

A metal wire of length L and cross-sectional radius r stretches by ΔL when a load F hangs from it. A second wire made of the same metal has length 2L and radius 2r. What load will produce the same stretch ΔL?

  1. 4F
  2. 2F Correct answer.
  3. F/2
  4. F

Explanation

The stretch is ΔL = FL/(AY). For the same material and the same stretch, F must be proportional to A/L. Doubling the radius quadruples the area, while doubling the length doubles L, so the required load is 2F.

Question 31Mechanics and Properties of Matter

A student walks 3.0 m east and then 4.0 m north. What is the magnitude of the student's displacement from the starting point?

  1. 1.0 m
  2. 4.0 m
  3. 5.0 m Correct answer.
  4. 7.0 m

Explanation

Displacement is the straight-line vector from start to finish. The 3.0 m east and 4.0 m north components are perpendicular, so the magnitude is √(3.0² + 4.0²) = 5.0 m. The 7.0 m value is the distance traveled, not the displacement.

Question 32Mechanics and Properties of Matter

An astronaut has a mass of 80 kg on Earth. On the Moon, where the gravitational field strength is about one-sixth that at Earth's surface, which quantity remains unchanged?

  1. The astronaut's acceleration due to gravity
  2. The astronaut's weight
  3. The astronaut's mass Correct answer.
  4. The gravitational force on the astronaut

Explanation

Mass is the amount of matter in an object and does not depend on the local gravitational field. Weight, gravitational force, and the acceleration due to gravity all depend on g, so they are smaller on the Moon.

Question 33Mechanics and Properties of Matter

A 10 kg crate rests on a horizontal floor. The coefficient of static friction between the crate and the floor is 0.40, and g = 10 m/s². What is the maximum horizontal force that can be applied before the crate begins to slide?

  1. 100 N
  2. 40 N Correct answer.
  3. 25 N
  4. 4.0 N

Explanation

The maximum static friction is f_s,max = μ_s N. On a horizontal floor the normal force equals mg = (10 kg)(10 m/s²) = 100 N, so f_s,max = 0.40 × 100 N = 40 N. A larger applied force would cause sliding.

Question 34Mechanics and Properties of Matter

Two blocks of masses 3.0 kg and 5.0 kg hang from opposite ends of a light string passing over a frictionless pulley. Using g = 10 m/s², what is the magnitude of the acceleration of the blocks?

  1. 5.0 m/s²
  2. 2.5 m/s² Correct answer.
  3. 1.25 m/s²
  4. 10 m/s²

Explanation

For an Atwood machine, the net driving force is the difference in weights, (5.0 − 3.0)(10) = 20 N, and the total mass being accelerated is 8.0 kg. Newton's second law gives a = 20 N / 8.0 kg = 2.5 m/s².

Question 35Mechanics and Properties of Matter

A sled is pulled 10 m along a horizontal surface by a rope that makes a 30° angle above the horizontal. The tension in the rope is 50 N. How much work does the tension do on the sled?

  1. 250 J
  2. 0 J
  3. 500 J
  4. 430 J Correct answer.

Explanation

Work is W = F d cos θ, where θ is the angle between the force and the displacement. Here W = (50 N)(10 m)(cos 30°) ≈ 433 J, which rounds to 430 J.

Question 36Mechanics and Properties of Matter

A pendulum bob of mass 0.50 kg is released from rest at a height of 0.80 m above its lowest point. Neglecting air resistance, what is the speed of the bob at the lowest point? Use g = 10 m/s².

  1. 16 m/s
  2. 8.0 m/s
  3. 2.0 m/s
  4. 4.0 m/s Correct answer.

Explanation

Mechanical energy is conserved, so the potential energy at release equals the kinetic energy at the bottom: mgh = ½mv². The mass cancels, giving v = √(2gh) = √(2 × 10 × 0.80) = 4.0 m/s.

Question 37Mechanics and Properties of Matter

A 2.0 kg cart moving at 6.0 m/s collides with a stationary 4.0 kg cart. The carts stick together. What is their common speed immediately after the collision?

  1. 3.0 m/s
  2. 1.0 m/s
  3. 2.0 m/s Correct answer.
  4. 6.0 m/s

Explanation

In a perfectly inelastic collision, momentum is conserved but kinetic energy is not. The initial momentum is (2.0 kg)(6.0 m/s) = 12 kg·m/s, and after sticking the total mass is 6.0 kg, so v = 12/6.0 = 2.0 m/s.

Question 38Mechanics and Properties of Matter

A spring is compressed 0.20 m and stores 4.0 J of elastic potential energy. What is the spring constant of the spring?

  1. 20 N/m
  2. 40 N/m
  3. 200 N/m Correct answer.
  4. 100 N/m

Explanation

Elastic potential energy is U = ½kx², where x is the compression. Rearranging gives k = 2U/x² = 2(4.0 J)/(0.20 m)² = 8.0/0.040 = 200 N/m.

Question 39Mechanics and Properties of Matter

A uniform solid sphere is released from rest and rolls without slipping down an incline that makes an angle of 30° with the horizontal. Using g = 10 m/s², what is the linear acceleration of the sphere's center of mass?

  1. 10 m/s²
  2. 3.6 m/s² Correct answer.
  3. 7.1 m/s²
  4. 5.0 m/s²

Explanation

For rolling without slipping, the acceleration down an incline is a = g sin θ / (1 + I/(mr²)). For a solid sphere, I = (2/5)mr², so the denominator is 1 + 2/5 = 7/5. Thus a = (5/7)g sin 30° = (5/7)(10)(0.5) ≈ 3.6 m/s².

Question 40Mechanics and Properties of Matter

A car travels around a circular track of radius 50 m at a speed of 10 m/s while speeding up with a tangential acceleration of 2.0 m/s². What is the magnitude of the car's total acceleration at that instant?

  1. 10 m/s²
  2. 2.0 m/s²
  3. 2.8 m/s² Correct answer.
  4. 4.0 m/s²

Explanation

The radial (centripetal) acceleration is a_r = v²/r = (10 m/s)²/50 m = 2.0 m/s². The tangential acceleration is also 2.0 m/s², and these perpendicular components add vectorially: a_total = √(2.0² + 2.0²) ≈ 2.8 m/s².

Question 41Mechanics and Properties of Matter

A car moves around a circular track at a constant speed of 15 m/s. Which statement describes the car's motion at any instant?

  1. Its velocity is constant because its speed is constant.
  2. Its acceleration points opposite the direction of motion.
  3. Its acceleration points toward the center of the circle. Correct answer.
  4. Its acceleration is zero because its speed is constant.

Explanation

For uniform circular motion, the speed is constant but the velocity vector changes direction continuously. That changing velocity means the car has a centripetal acceleration directed toward the center of the circle. The net force on the car also points toward the center, not along the direction of motion.

Question 42Mechanics and Properties of Matter

A bus stops suddenly, and a standing passenger continues moving forward relative to the bus. Which concept best explains this observation?

  1. The conservation of mechanical energy
  2. The normal force from the bus floor
  3. The passenger's inertia Correct answer.
  4. The gravitational force between the passenger and Earth

Explanation

Newton's first law states that an object continues in its state of motion unless a net external force acts on it. When the bus brakes, the passenger's body tends to keep its forward velocity, a property called inertia. The lurch is not caused by a new forward force on the passenger.

Question 43Mechanics and Properties of Matter

A 0.20 kg book is lifted from the floor to a shelf 1.5 m above the floor. Using g = 9.8 m/s², what is the increase in the book's gravitational potential energy?

  1. 2.9 J Correct answer.
  2. 1.5 J
  3. 29 J
  4. 0.30 J

Explanation

The increase in gravitational potential energy near Earth's surface is ΔU = mgΔh. Substituting gives (0.20 kg)(9.8 m/s²)(1.5 m) = 2.94 J, which rounds to 2.9 J.

Question 44Mechanics and Properties of Matter

A diver is 5.0 m below the surface of fresh water. Using density 1.0×10³ kg/m³ and g = 9.8 m/s², what is the gauge pressure due to the water at this depth?

  1. 1.5×10⁵ Pa
  2. 4.9×10⁴ Pa Correct answer.
  3. 4.9×10² Pa
  4. 9.8×10³ Pa

Explanation

Gauge pressure due to a liquid column is P = ρgh. Thus P = (1.0×10³ kg/m³)(9.8 m/s²)(5.0 m) = 4.9×10⁴ Pa. Gauge pressure excludes atmospheric pressure.

Question 45Mechanics and Properties of Matter

A mechanic applies a force of 80 N perpendicular to the handle of a wrench at a distance of 0.25 m from the bolt. What is the magnitude of the torque about the bolt?

  1. 32 N·m
  2. 320 N·m
  3. 20 N·m Correct answer.
  4. 80 N·m

Explanation

Torque has magnitude τ = rF sinθ. Because the force is perpendicular to the handle, sinθ = 1, so τ = (0.25 m)(80 N) = 20 N·m.

Question 46Mechanics and Properties of Matter

A ball is launched horizontally from a 20 m high cliff with an initial horizontal speed of 12 m/s. Neglecting air resistance and using g = 9.8 m/s², how long does the ball take to reach the ground?

  1. 4.1 s
  2. 2.0 s Correct answer.
  3. 2.9 s
  4. 1.4 s

Explanation

The vertical motion determines the fall time and is independent of the horizontal launch speed. With zero initial vertical velocity, h = ½gt², so t = sqrt(2h/g) = sqrt(2(20 m)/(9.8 m/s²)) ≈ 2.0 s.

Question 47Mechanics and Properties of Matter

A 0.15 kg baseball strikes a wall horizontally with a speed of 20 m/s and rebounds horizontally with a speed of 15 m/s. What is the magnitude of the impulse delivered to the ball by the wall?

  1. 5.3 kg·m/s Correct answer.
  2. 3.0 kg·m/s
  3. 0.75 kg·m/s
  4. 7.5 kg·m/s

Explanation

Impulse equals the change in momentum, and momentum reverses direction on rebound. Taking the initial direction as positive, Δp = m(v_f - v_i) = (0.15 kg)(-15 m/s - 20 m/s) = -5.25 kg·m/s, so the magnitude is about 5.3 kg·m/s.

Question 48Mechanics and Properties of Matter

A simple pendulum has a length of 2.0 m. Using g = 9.8 m/s² and the small-angle approximation, what is its period?

  1. 1.4 s
  2. 0.45 s
  3. 2.8 s Correct answer.
  4. 2.0 s

Explanation

For a simple pendulum with small angular displacement, T = 2πsqrt(L/g). Substituting L = 2.0 m and g = 9.8 m/s² gives T ≈ 2.8 s.

Question 49Mechanics and Properties of Matter

A uniform horizontal beam of length 4.0 m weighs 200 N and is hinged at its left end. The right end is supported by a cable that makes a 30° angle with the beam. What is the tension in the cable?

  1. 100 N
  2. 200 N Correct answer.
  3. 800 N
  4. 400 N

Explanation

Take torques about the hinge so the hinge force has no torque. The beam's 200 N weight acts at its center, 2.0 m from the hinge, giving 400 N·m clockwise. The cable's vertical component is T sin30°, acting 4.0 m from the hinge, so 4.0(T)(0.5) = 400 N·m and T = 200 N.

Question 50Mechanics and Properties of Matter

A solid block floats in water with 60% of its volume below the surface. The same block floats in an unknown oil with 80% of its volume below the surface. What is the density of the oil? (Density of water = 1.0×10³ kg/m³.)

  1. 1.3×10³ kg/m³
  2. 8.0×10² kg/m³
  3. 7.5×10² kg/m³ Correct answer.
  4. 6.0×10² kg/m³

Explanation

For a floating object, the submerged fraction equals the ratio of the object's density to the fluid's density. From water, ρ_block = 0.60(1.0×10³) = 6.0×10² kg/m³. In the oil, 0.80 = ρ_block/ρ_oil, so ρ_oil = 6.0×10²/0.80 = 7.5×10² kg/m³.