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100+ Free Junkao Science Knowledge Practice Questions
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Sample Junkao Science Knowledge Practice Questions
Try these sample questions to test your Junkao Science Knowledge exam readiness. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.
1An artillery projectile is accelerated uniformly from rest along a gun barrel of length 6.0 m and exits the muzzle with a velocity of 1,200 m/s. Assuming the acceleration is constant throughout the barrel, what is the magnitude of the projectile's acceleration?
A.1.2 × 10^5 m/s²
B.2.4 × 10^5 m/s²
C.1.0 × 10^5 m/s²
D.7.2 × 10^4 m/s²
Explanation: Using the kinematic relation for uniform acceleration with zero initial velocity, v² = 2as, we can solve for acceleration: a = v² / (2s) = (1,200 m/s)² / (2 × 6.0 m) = 1,440,000 / 12.0 = 1.2 × 10^5 m/s². This demonstrates the tremendous mechanical acceleration experienced by munitions during internal ballistics.
2A flare is launched vertically upward from ground level with an initial speed of 30 m/s. Neglecting air resistance and taking the acceleration due to gravity as g = 10 m/s², what is the maximum height reached by the flare above the launch point?
A.30 m
B.45 m
C.60 m
D.90 m
Explanation: At maximum height, the vertical velocity is zero (v = 0). Using the formula v² = v₀² - 2gh, we set 0 = (30 m/s)² - 2(10 m/s²)h, which gives h = 900 / 20 = 45 m. Alternatively, conservation of energy gives (1/2)mv₀² = mgh, yielding the same 45 m.
3A military transport truck traveling along a straight road at an initial speed of 20 m/s brakes with a constant deceleration of 4.0 m/s² until it comes to a complete stop. What is the total stopping distance of the truck?
A.40 m
B.50 m
C.80 m
D.100 m
Explanation: The stopping time is t = v₀ / a = (20 m/s) / (4.0 m/s²) = 5.0 s. The stopping distance is given by s = v₀² / (2a) = (20 m/s)² / (2 × 4.0 m/s²) = 400 / 8.0 = 50 m. This calculation is essential for safe military convoy following distances.
4A military supply crate with a mass of 50 kg rests on a flat, horizontal concrete surface where the coefficient of kinetic friction is μ = 0.20. A soldier applies a continuous horizontal pulling force of 200 N to the crate. Taking g = 10 m/s², what is the acceleration of the crate?
A.1.0 m/s²
B.2.0 m/s²
C.3.0 m/s²
D.4.0 m/s²
Explanation: The normal force on the crate is N = mg = 50 kg × 10 m/s² = 500 N. The kinetic friction force is f_k = μN = 0.20 × 500 N = 100 N. The net horizontal force is F_net = F_applied - f_k = 200 N - 100 N = 100 N. By Newton's second law, acceleration is a = F_net / m = 100 N / 50 kg = 2.0 m/s².
5A field equipment container of mass m slides down an inclined ramp angled at θ = 30° to the horizontal. If the coefficient of kinetic friction between the ramp and the container is μ = 0.10, and taking g = 10 m/s² (with sin 30° = 0.50, cos 30° ≈ 0.866), what is the acceleration of the container down the ramp?
A.4.13 m/s²
B.5.00 m/s²
C.3.27 m/s²
D.5.87 m/s²
Explanation: The component of gravitational force down the incline is F_g = mg sin θ, and the normal force is N = mg cos θ. The kinetic friction force opposing motion is f_k = μN = μmg cos θ. The net force down the slope is F_net = mg sin θ - μmg cos θ, so acceleration is a = g(sin θ - μ cos θ) = 10(0.50 - 0.10 × 0.866) = 10(0.50 - 0.0866) = 10(0.4134) ≈ 4.13 m/s².
6According to Newton's third law of motion, when a soldier stands motionless on horizontal ground, which of the following pairs of forces constitutes an action-reaction pair?
A.The gravitational pull of the Earth on the soldier and the upward normal support force of the ground on the soldier
B.The downward contact force exerted by the soldier's boots on the ground and the upward normal support force exerted by the ground on the soldier's boots
C.The gravitational pull of the Earth on the soldier and the friction between the soldier's boots and the ground
D.The upward support force of the ground on the soldier and the atmospheric pressure pressing down on the soldier
Explanation: Newton's third law action-reaction pairs act on two different bodies, are of the exact same physical nature (both contact normal forces), and are equal in magnitude and opposite in direction. The force exerted by the soldier on the ground and the force exerted by the ground on the soldier form an action-reaction pair. The Earth's gravitational pull on the soldier is balanced by the ground's normal force on the soldier (two forces on the same body, representing equilibrium, not an action-reaction pair).
7An electric winch on a military recovery vehicle lifts a pallet of equipment of mass 500 kg vertically at a constant speed of 2.0 m/s. Taking g = 10 m/s² and assuming no energy losses, what is the output power delivered by the winch motor?
A.5.0 kW
B.10 kW
C.20 kW
D.2.5 kW
Explanation: Since the pallet is lifted at a constant velocity, the tension force exerted by the winch cable equals the gravitational weight: F = mg = 500 kg × 10 m/s² = 5,000 N. The mechanical power is P = F × v = 5,000 N × 2.0 m/s = 10,000 W = 10 kW.
8An aerial test payload of mass 10 kg is dropped from rest from a helicopter hovering at a height of 80 m above a level target area. Neglecting air resistance and taking g = 10 m/s², what is the kinetic energy of the payload immediately before it impacts the ground?
A.800 J
B.4.0 kJ
C.8.0 kJ
D.16 kJ
Explanation: By the law of conservation of mechanical energy, in the absence of non-conservative forces like air resistance, the entire initial gravitational potential energy is converted into kinetic energy: E_k = E_p = mgh = 10 kg × 10 m/s² × 80 m = 8,000 J = 8.0 kJ. Alternatively, v = √(2gh) = √(1600) = 40 m/s, so E_k = (1/2)mv² = (1/2)(10)(1600) = 8,000 J.
9A recoil buffer spring in an automatic weapon mechanism has a spring constant of k = 2,000 N/m. When the bolt carrier compresses the spring by x = 0.10 m from its uncompressed equilibrium length, how much elastic potential energy is stored in the spring?
A.10 J
B.20 J
C.100 J
D.200 J
Explanation: The elastic potential energy stored in an ideal Hookean spring is given by E_p = (1/2)kx². Substituting the given values: E_p = (1/2) × (2,000 N/m) × (0.10 m)² = 1,000 × 0.010 = 10 J.
10A service rifle of mass M = 4.0 kg is freely suspended and fires a bullet of mass m = 8.0 g (0.0080 kg) horizontally with a muzzle velocity of v = 800 m/s. Assuming the system is initially at rest, what is the initial horizontal recoil velocity of the rifle?
A.0.80 m/s
B.1.6 m/s
C.3.2 m/s
D.6.4 m/s
Explanation: By the law of conservation of linear momentum, total initial momentum is zero: P_initial = 0 = m_bullet × v_bullet + M_rifle × V_recoil. Therefore, M_rifle × |V_recoil| = m_bullet × v_bullet = 0.0080 kg × 800 m/s = 6.4 kg·m/s. Solving for recoil speed gives |V_recoil| = 6.4 / 4.0 = 1.6 m/s.
About the Junkao Science Knowledge Exam
Science Knowledge Comprehensive is one of the named cultural subjects in the current soldier military-academy examination. The official paper is in standard written Chinese. This independent English-language MCQ bank supports curriculum review; it is not an official translation, current-paper replica, or substitute for Chinese-language and written problem-solving practice.
Assessment
Written comprehensive subject covering physics, chemistry, and history; current detailed blueprint and authorized materials are provided through the military examination system.
Time Limit
150 minutes in the published 2023 schedule; verify the current annual notice.
Passing Score
Competitive composite selection; no standalone permanent public pass mark.
Exam Fee
No public individual fee schedule located for this internal military selection. (Central Military Commission training and military academy admissions authorities)
Junkao Science Knowledge Exam Content Outline
not-published
Physics
Mechanics, thermodynamics, electricity and magnetism, optics, and related applications.
not-published
Chemistry
Inorganic, organic, analytical, electrochemical, materials, and safety concepts.
not-published
History
Chinese and world history in the authorized educational scope.
How to Pass the Junkao Science Knowledge Exam
What You Need to Know
- Passing score: Competitive composite selection; no standalone permanent public pass mark.
- Assessment: Written comprehensive subject covering physics, chemistry, and history; current detailed blueprint and authorized materials are provided through the military examination system.
- Time limit: 150 minutes in the published 2023 schedule; verify the current annual notice.
- Exam fee: No public individual fee schedule located for this internal military selection.
Keys to Passing
- Work through all 100 available questions
- Review every answer and explanation
- Track weak areas and revisit them
- Use our AI tutor for tough concepts
Junkao Science Knowledge Study Tips from Top Performers
1Show every calculation and check units.
2Review experiment and chemical-safety reasoning, not only recall.
3Use current authorized Chinese materials for live-exam scope.
Frequently Asked Questions
Which disciplines are covered?
Official sources identify physics, chemistry, and history within Science Knowledge Comprehensive.
Are public percentage weights current?
No stable current public blueprint with fixed weights was located; this page does not invent them.
What language is official?
The official paper is in Chinese; this bank is an English-language study adaptation.