Q. If a river's flow rate is 500 cubic meters per second, how many liters of water flow through the river in one hour? (2000)
A.
1800000
B.
3600000
C.
1200000
D.
2400000
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Solution
Flow rate in liters = 500 m³/s × 3600 s = 1800000 liters.
Correct Answer:
B
— 3600000
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Q. If a river's flow rate is 500 cubic meters per second, how much water flows in one hour? (2021)
A.
1,800,000 m³
B.
1,500,000 m³
C.
2,000,000 m³
D.
2,500,000 m³
Show solution
Solution
Water flow in one hour = Flow rate x Time = 500 m³/s x 3600 s = 1,800,000 m³.
Correct Answer:
A
— 1,800,000 m³
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Q. If a river's sediment load is 200 kg/m³ and the river has a flow rate of 10 m³/s, what is the total sediment transported in one hour? (2022)
A.
720,000 kg
B.
600,000 kg
C.
1,200,000 kg
D.
1,440,000 kg
Show solution
Solution
Sediment transported = sediment load × flow rate × time = 200 kg/m³ × 10 m³/s × 3600 s = 720,000 kg.
Correct Answer:
D
— 1,440,000 kg
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Q. If a river's sediment load is 300 kg/m³ and it flows at a rate of 10 m³/s, what is the total sediment load carried by the river in one hour? (2020)
A.
108000 kg
B.
1080000 kg
C.
180000 kg
D.
360000 kg
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Solution
Sediment load = 300 kg/m³ × 10 m³/s × 3600 s = 10800000 kg.
Correct Answer:
B
— 1080000 kg
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Q. If a river's sediment load is 5 kg/m³ and the river has a flow rate of 100 m³/s, what is the total sediment transported in one hour? (2019)
A.
18,000 kg
B.
20,000 kg
C.
15,000 kg
D.
12,000 kg
Show solution
Solution
Total sediment = Sediment load × Flow rate × Time = 5 kg/m³ × 100 m³/s × 3600 s = 18,000 kg.
Correct Answer:
B
— 20,000 kg
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Q. If a river's sediment transport capacity is 15 tons per cubic meter of water, how much sediment can it carry if the discharge is 100 m³/s? (2000)
A.
1500 tons
B.
1000 tons
C.
2000 tons
D.
3000 tons
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Solution
Sediment carried = Discharge × Sediment transport capacity = 100 m³/s × 15 tons/m³ = 1500 tons.
Correct Answer:
A
— 1500 tons
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Q. If a river's width increases from 50 meters to 100 meters while maintaining the same depth of 3 meters, what is the change in cross-sectional area? (2022)
A.
150 m²
B.
200 m²
C.
300 m²
D.
400 m²
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Solution
Initial area = 50 m × 3 m = 150 m²; Final area = 100 m × 3 m = 300 m²; Change = 300 m² - 150 m² = 150 m².
Correct Answer:
C
— 300 m²
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Q. If a river's width is doubled and the depth remains the same, how does this affect the cross-sectional area? (2021)
A.
It remains the same
B.
It doubles
C.
It triples
D.
It quadruples
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Solution
Cross-sectional area = Width x Depth. If width is doubled, the area also doubles.
Correct Answer:
B
— It doubles
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Q. If a riverbank erodes at a rate of 2 meters per year, how much will it erode in 5 years? (2023)
A.
5 m
B.
10 m
C.
15 m
D.
20 m
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Solution
Erosion = rate × time = 2 m/year × 5 years = 10 m.
Correct Answer:
B
— 10 m
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Q. If a rolling object has a mass m and radius R, what is the expression for its rotational kinetic energy?
A.
(1/2)Iω^2
B.
(1/2)mv^2
C.
(1/2)mv^2/R^2
D.
(1/2)mv^2 + (1/2)Iω^2
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Solution
The rotational kinetic energy is given by (1/2)Iω^2, where I is the moment of inertia.
Correct Answer:
A
— (1/2)Iω^2
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Q. If a rolling object has a mass m and radius r, what is the expression for its total kinetic energy?
A.
(1/2)mv^2
B.
(1/2)mv^2 + (1/2)Iω^2
C.
(1/2)mv^2 + (1/2)mr^2ω^2
D.
(1/2)mv^2 + (1/2)(2/5)mr^2(ω^2)
Show solution
Solution
The total kinetic energy of a rolling object is the sum of translational and rotational kinetic energy, which can be expressed as (1/2)mv^2 + (1/2)(2/5)mr^2(ω^2).
Correct Answer:
D
— (1/2)mv^2 + (1/2)(2/5)mr^2(ω^2)
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Q. If a rolling object has a radius of R and rolls with a speed v, what is its kinetic energy?
A.
(1/2)mv^2
B.
(1/2)mv^2 + (1/2)Iω^2
C.
(1/2)mv^2 + (1/2)(1/2)mR^2(v/R)^2
D.
None of the above
Show solution
Solution
The total kinetic energy of a rolling object is the sum of translational and rotational kinetic energy, which simplifies to (1/2)mv^2 + (1/4)mv^2 = (3/4)mv^2.
Correct Answer:
C
— (1/2)mv^2 + (1/2)(1/2)mR^2(v/R)^2
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Q. If a rolling object has a radius R and rolls with an angular velocity ω, what is its linear velocity?
A.
Rω
B.
2Rω
C.
R/2ω
D.
3Rω
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Solution
The linear velocity v of a rolling object is given by v = Rω.
Correct Answer:
A
— Rω
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Q. If a rolling object has a translational speed of v and a rotational speed of ω, what is the relationship between them for rolling without slipping?
A.
v = ωR
B.
v = 2ωR
C.
v = ω/R
D.
v = R/ω
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Solution
For rolling without slipping, the relationship is v = ωR, where v is the translational speed and ω is the angular speed.
Correct Answer:
A
— v = ωR
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Q. If a rotating body has an angular momentum of L and its moment of inertia is I, what is the angular velocity ω of the body?
A.
L/I
B.
I/L
C.
L^2/I
D.
I^2/L
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Solution
Angular momentum L = Iω, thus ω = L/I.
Correct Answer:
A
— L/I
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Q. If a rotating object has a moment of inertia of 4 kg·m² and is spinning with an angular velocity of 3 rad/s, what is its angular momentum?
A.
12 kg·m²/s
B.
4 kg·m²/s
C.
1 kg·m²/s
D.
7 kg·m²/s
Show solution
Solution
Angular momentum L = Iω = 4 kg·m² * 3 rad/s = 12 kg·m²/s.
Correct Answer:
A
— 12 kg·m²/s
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Q. If a rotating object has a moment of inertia of 5 kg·m² and is rotating with an angular velocity of 3 rad/s, what is its angular momentum?
A.
15 kg·m²/s
B.
5 kg·m²/s
C.
8 kg·m²/s
D.
10 kg·m²/s
Show solution
Solution
Angular momentum L is given by L = Iω. Thus, L = 5 kg·m² * 3 rad/s = 15 kg·m²/s.
Correct Answer:
A
— 15 kg·m²/s
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Q. If a rotating object has a moment of inertia of 5 kg·m² and is spinning with an angular velocity of 3 rad/s, what is its angular momentum?
A.
15 kg·m²/s
B.
5 kg·m²/s
C.
8 kg·m²/s
D.
10 kg·m²/s
Show solution
Solution
Angular momentum L = Iω = 5 kg·m² * 3 rad/s = 15 kg·m²/s.
Correct Answer:
A
— 15 kg·m²/s
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Q. If a rotating object has a moment of inertia of I and is rotating with an angular velocity ω, what is its rotational kinetic energy?
A.
1/2 Iω
B.
1/2 Iω^2
C.
Iω^2
D.
Iω
Show solution
Solution
The rotational kinetic energy is given by KE = 1/2 Iω^2.
Correct Answer:
B
— 1/2 Iω^2
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Q. If a ruler has a least count of 0.1 cm, what is the maximum error in measurement? (2023)
A.
0.05 cm
B.
0.1 cm
C.
0.2 cm
D.
0.01 cm
Show solution
Solution
The maximum error is typically taken as the least count of the measuring instrument, which is 0.1 cm.
Correct Answer:
B
— 0.1 cm
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Q. If a ruler has a least count of 0.1 cm, what is the smallest measurement it can accurately provide? (2023)
A.
0.1 cm
B.
0.05 cm
C.
0.01 cm
D.
1 cm
Show solution
Solution
The least count of a measuring instrument is the smallest value that can be measured accurately, which in this case is 0.1 cm.
Correct Answer:
A
— 0.1 cm
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Q. If a runner completes a 10 km race in 40 minutes, what is his average speed in km/h? (2021)
A.
12 km/h
B.
13 km/h
C.
14 km/h
D.
15 km/h
Show solution
Solution
Time = 40 minutes = 40/60 hours = 2/3 hours. Speed = Distance / Time = 10 km / (2/3) hours = 10 * (3/2) = 15 km/h.
Correct Answer:
C
— 14 km/h
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Q. If a salary is increased by 25% and the new salary is $50,000, what was the original salary?
A.
$40,000
B.
$45,000
C.
$35,000
D.
$30,000
Show solution
Solution
Let the original salary be x. Then, x + 0.25x = 50000 => 1.25x = 50000 => x = 50000 / 1.25 = $40,000.
Correct Answer:
A
— $40,000
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Q. If a salary of $50,000 is increased by 10% and then decreased by 10%, what is the final salary?
A.
$50,000
B.
$49,500
C.
$51,000
D.
$52,000
Show solution
Solution
New salary after increase = 50000 * 1.10 = $55,000. After decrease = 55000 * 0.90 = $49,500.
Correct Answer:
B
— $49,500
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Q. If a satellite is in a geostationary orbit, what is its orbital period?
A.
24 hours
B.
12 hours
C.
6 hours
D.
1 hour
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Solution
A geostationary satellite has an orbital period equal to the Earth's rotation period, which is 24 hours.
Correct Answer:
A
— 24 hours
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Q. If a satellite is in a stable orbit, what can be said about the net force acting on it?
A.
It is zero
B.
It is equal to the gravitational force
C.
It is equal to the centripetal force
D.
It is equal to the sum of gravitational and centripetal forces
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Solution
In a stable orbit, the net force acting on the satellite is zero because the gravitational force provides the necessary centripetal force.
Correct Answer:
A
— It is zero
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Q. If a satellite is launched into a higher orbit, how does its potential energy change compared to its initial orbit?
A.
It decreases
B.
It remains the same
C.
It increases
D.
It becomes zero
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Solution
The potential energy of a satellite increases when it is moved to a higher orbit due to the increase in distance from the Earth's center.
Correct Answer:
C
— It increases
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Q. If a satellite is moved to a higher orbit, what happens to its orbital period?
A.
It decreases.
B.
It increases.
C.
It remains the same.
D.
It becomes zero.
Show solution
Solution
The orbital period of a satellite increases when it is moved to a higher orbit, according to Kepler's third law.
Correct Answer:
B
— It increases.
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Q. If a satellite is moving in a circular orbit, what is the relationship between its centripetal acceleration and gravitational acceleration?
A.
Centripetal = Gravitational
B.
Centripetal > Gravitational
C.
Centripetal < Gravitational
D.
No relationship
Show solution
Solution
For a satellite in a stable circular orbit, the centripetal acceleration is equal to the gravitational acceleration.
Correct Answer:
A
— Centripetal = Gravitational
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Q. If a satellite is moving in a circular orbit, what type of energy does it possess?
A.
Only kinetic energy
B.
Only potential energy
C.
Both kinetic and potential energy
D.
Neither kinetic nor potential energy
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Solution
A satellite in a circular orbit possesses both kinetic energy due to its motion and potential energy due to its position in the gravitational field.
Correct Answer:
C
— Both kinetic and potential energy
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