Physics — Semester A
Free Practice · 10 Questions · 20 min
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Question 1 of 10
TEKS 5F-5HEasy Word Diagram

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
Afour times the original
Bone-fourth of the original
Chalved
Ddoubled
Explanation
📌 F = GmM/d². Force is inversely proportional to d². Double d → d² becomes 4d² → F becomes F/4. Common wrong answers come from treats it as direct proportion, applies 1/d instead of 1/d² (forgot the square), and squares wrong direction (direct instead of inverse).
Question 2 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
B28.4 × 10⁻⁴ kg
C2.84 × 10³ 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. Common wrong answers come from has wrong exponent sign, violates scientific notation convention, and uses wrong unit (2.84 × 10⁻² g = 0.0000284 kg, two orders off).
Question 3 of 10
TEKS 7D-7EEasy Word

A 3.0-kg object moves at a velocity of 5.0 m/s. What is its momentum?

A45 kg·m/s
B15 kg·m/s
C8.0 kg·m/s
D0.6 kg·m/s
Explanation
📌 p = mv = 3.0 × 5.0 = 15 kg·m/s. Common wrong answers come from dividing (÷ instead of ×), adding, and using wrong formula.
Question 4 of 10
TEKS 7A-7CEasy Calc Word Diagram

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
A6.25 J
B3,200 J
C400 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. Common wrong answers come from dividing (÷ instead of ×), subtracting, and multiplying one extra time.
Question 5 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/s
B7,200 m/s
C8.0 m
D0.125 m/s
Explanation
📌 Average velocity v = d/t = 240 m / 30 s = 8.0 m/s. Common wrong answers come from multiplying (÷/× swap), taking reciprocal, and forgetting velocity units (m instead of m/s).
Question 6 of 10
TEKS 5A-5EEasy Calc Word

Which quantity is a scalar?

Aspeed
Bacceleration
Cvelocity
Ddisplacement
Explanation
📌 Speed = magnitude of velocity, no direction → scalar. Velocity, displacement, and acceleration are vectors (have direction).
Question 7 of 10
TEKS 1A-4CEasy Word

What is the SI unit of energy?

APascal (Pa)
BNewton (N)
CWatt (W)
DJoule (J)
Explanation
📌 The joule (J) is the SI unit of energy, defined as 1 N·m. Common wrong answers come from force, power (J/s), and pressure (N/m²).
Question 8 of 10
TEKS 1A-4CEasy Word

The SI unit of electric current is:

AAmpere
BWatt
CVolt
DCoulomb
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 9 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
B2.0 m/s²
C8.5 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. Common wrong answers come from unnecessarily divided by mass, using cos(30°) instead of sin(30°) — the sin/cos swap, and using full g without decomposition.
Question 10 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
A8.4 m/s²
B3.6 m/s²
C6.0 m/s²
D2.4 m/s²
Explanation
📌 Net force = 30 − 12 = 18 N right. a = F_net/m = 18/5.0 = 3.6 m/s². Common wrong answers come from ignoring applied force, ignoring friction, and adding forces instead of subtracting (sign error — friction opposes motion).

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