Physics — Semester A
Free Practice · 10 Questions · 20 min
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Question 1 of 10
TEKS 1A-4CEasy Word

A student measures the mass of a sample as 0.00284 kg. Which expression correctly represents this mass in scientific notation with proper SI units?

A2.84 × 10⁻² g
B2.84 × 10³ kg
C28.4 × 10⁻⁴ kg
D2.84 × 10⁻³ kg
Explanation
📌 Move decimal 3 places right: 0.00284 → 2.84 × 10⁻³. Coefficient must satisfy 1 ≤ a < 10, and SI mass unit is kg. Distractor B has wrong exponent sign; C violates scientific notation convention; D uses wrong unit (2.84 × 10⁻² g = 0.0000284 kg, two orders off).
Question 2 of 10
TEKS 5F-5HEasy Word

According to Newton’s law of universal gravitation, if the distance between two objects is doubled while their masses remain the same, the gravitational force between them becomes:

Two objects — distance changesBeforem₁m₂dAfter (distance doubled)m₁m₂2d
Ahalved
Bfour times the original
Cone-fourth of the original
Ddoubled
Explanation
📌 F = GmM/d². Force is inversely proportional to d². Double d → d² becomes 4d² → F becomes F/4. Distractor A treats it as direct proportion; B applies 1/d instead of 1/d² (forgot the square); D squares wrong direction (direct instead of inverse).
Question 3 of 10
TEKS 5A-5EMedium Calc Word Diagram

The position-vs-time graph shows an object's motion. Between t = 2 s and t = 6 s, which best describes the object's motion?

Position vs. time0246810t (s)0102030x (m)(2, 10)(6, 10)focus interval
AThe object is moving backward.
BThe object is accelerating.
CThe object is moving at constant velocity.
DThe object is at rest.
Explanation
📌 A horizontal line on a position-vs-time graph means position is not changing — the object is stationary (at rest). Distractor A confuses horizontal line with acceleration; B confuses horizontal with constant motion; D confuses horizontal (no motion) with negative slope (backward motion).
Question 4 of 10
TEKS 5F-5HMedium Calc Word Diagram

A 4.0-kg block slides down a frictionless ramp inclined at 30° to the horizontal. What is the magnitude of the block’s acceleration along the ramp? Use g = 9.8 m/s².

Block on frictionless incline (θ = 30°)30°m = 4 kgW = mgN
A4.9 m/s²
B8.5 m/s²
C2.0 m/s²
D9.8 m/s²
Explanation
📌 On a frictionless incline, acceleration along the ramp = g·sin θ = 9.8 × sin(30°) = 9.8 × 0.5 = 4.9 m/s². Mass cancels — same rate for any object at same angle. Distractor A unnecessarily divided by mass; C used cos(30°) instead of sin(30°) — the sin/cos swap; D used full g without decomposition.
Question 5 of 10
TEKS 5F-5HMedium Calc Word Diagram

A 5.0-kg block on a horizontal surface is pushed with a horizontal applied force of 30 N. A frictional force of 12 N opposes motion. What is the block’s acceleration?

Horizontal push with friction (free-body view)m = 5.0 kgF = 30 Nf = 12 NNW = mg
A3.6 m/s²
B2.4 m/s²
C6.0 m/s²
D8.4 m/s²
Explanation
📌 Net force = 30 − 12 = 18 N right. a = F_net/m = 18/5.0 = 3.6 m/s². Distractor A ignored applied force; C ignored friction; D added forces instead of subtracting (sign error — friction opposes motion).
Question 6 of 10
TEKS 5A-5EEasy Word

A cyclist travels 240 m in 30 s at constant velocity. What is her average velocity?

A8.0 m
B8.0 m/s
C7,200 m/s
D0.125 m/s
Explanation
📌 Average velocity v = d/t = 240 m / 30 s = 8.0 m/s. Distractor A multiplied (÷/× swap); C took reciprocal; D forgot velocity units (m instead of m/s).
Question 7 of 10
TEKS 5A-5EHard Calc Word Diagram

A projectile is launched from ground level with an initial speed of 20 m/s at 30° above horizontal. What is the maximum height reached? Use g = 9.8 m/s² and ignore air resistance.

Projectile launch: v0 = 20 m/s, θ = 30°v0 = 20 m/sθ = 30°apex (max height)
A15.3 m
B5.10 m
C20.4 m
D2.55 m
Explanation
📌 Vertical component v_iy = v sin θ = 20 × sin(30°) = 10 m/s. At max height v_y = 0. Use v_fy² − v_iy² = −2g·h → 0 − 100 = −19.6h → h = 5.10 m. Distractor A used v_iy/2 (halved velocity); C used cos(30°) instead of sin(30°) — the classic sin/cos swap; D ignored angle entirely, used full v² / 2g.
Question 8 of 10
TEKS 7D-7EHard Calc Word Diagram

A 2.0-kg cart moving at 4.0 m/s collides with a stationary 3.0-kg cart. The two carts stick together after the collision. What is the velocity of the combined mass immediately after the collision?

Two carts collide and stick together on a horizontal trackBeforem1 = 2 kgv1 = 4 m/sm2 = 3 kg(at rest)After (stuck together)m1 + m2v′ = ?
A4.0 m/s
B2.0 m/s
C6.7 m/s
D1.6 m/s
Explanation
📌 Conservation of momentum (perfectly inelastic): m₁v₁ + m₂v₂ = (m₁+m₂)v_f. (2.0)(4.0) + 0 = (5.0)v_f → v_f = 1.6 m/s. Distractor B averaged velocities without mass weighting; C ignored mass conservation; D used wrong combined mass. Note: kinetic energy is NOT conserved here — 15 J is lost to heat, sound, and deformation.
Question 9 of 10
TEKS 1A-4CEasy Word

The SI unit of electric current is:

AWatt
BAmpere
CCoulomb
DVolt
Explanation
📌 The ampere (A) is the SI unit of current. 1 A = 1 C/s. The other units name different quantities: the coulomb is charge, the volt is potential, and the watt is power.
Question 10 of 10
TEKS 7A-7CMedium Calc Word

A person pushes a box with a horizontal force of 50 N over a distance of 8.0 m. If the person applies the force parallel to the direction of motion, how much work is done on the box?

Box pushed horizontally: 50 N over 8.0 mboxF = 50 Nd = 8.0 m
A3,200 J
B400 J
C6.25 J
D42 J
Explanation
📌 W = F·d·cos θ, where θ = 0° for parallel force (cos 0 = 1). W = 50 × 8.0 × 1 = 400 J. Distractor A divided (÷ instead of ×); B subtracted; D multiplied one extra time.

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