myclass24
myclass24your class. your pace.
Q&A BANK

every question.
answered.

1,129 answers across CBSE, ICSE, and State boards — Class 1 to 12, every subject.

CBSEICSEState
SUBJECT
CLASS

Filtered results

1,129 TOTAL
PhysicsClass 9CBSE

A ball is dropped from a height. Describe its motion in terms of velocity and acceleration, neglecting air resistance.

When a ball is dropped from a height, neglecting air resistance:Acceleration: The acceleration of the ball is constant and equal to the acceleration due to gravity (g), which is approximately 9.8 m/s². Its direction is always vertically downwards.Velocity: The ball starts with an initial velocity of zero (dropped from rest). As it falls, its velocity continuously increases in the downward direction due to the constant downward acceleration. This means its speed increases steadily.

PhysicsClass 9CBSE

Distinguish between instantaneous speed and average speed. When are they equal?

Instantaneous speed is the speed of an object at a particular instant in time. It is what a speedometer in a car measures. It can vary from moment to moment.Average speed is the total distance covered by an object divided by the total time taken for the entire journey.Instantaneous speed and average speed are equal when an object is moving with uniform speed (constant speed) throughout its motion. In this case, the speed at any instant is the same as the overall average speed.

PhysicsClass 9CBSE

A car travels from city A to city B at a speed of 40 km/h and returns from city B to city A at a speed of 60 km/h. What is the average velocity of the car for the entire journey?

For the entire journey from city A to city B and back to city A, the initial position is city A and the final position is also city A. Therefore, the total displacement of the car is zero.Average Velocity = Total Displacement / Total TimeSince Total Displacement = 0,Average Velocity = 0 / Total Time = 0 km/h.

PhysicsClass 9CBSE

Look at the given velocity-time graph. Describe the motion of the object in each segment (OA, AB, BC).

Segment OA: The velocity-time graph is a straight line passing through the origin with a positive slope. This indicates that the object is moving with uniform positive acceleration, starting from rest.Segment AB: The velocity-time graph is a straight line parallel to the time axis. This indicates that the object is moving with a constant velocity (uniform velocity), meaning its acceleration is zero.Segment BC: The velocity-time graph is a straight line with a negative slope, ending at zero velocity. This indicates that the object is moving with uniform negative acceleration (deceleration or retardation) and eventually comes to rest.

PhysicsClass 9CBSE

A car is traveling at 25 m/s. The driver applies the brakes, and the car decelerates uniformly at 5 m/s². How long does it take for the car to stop, and what distance does it cover during this time?

Initial velocity (u) = 25 m/sFinal velocity (v) = 0 m/s (comes to stop)Acceleration (a) = -5 m/s² (deceleration)Part 1: Time to stop (t)Using v = u + at:0 = 25 + (-5) * t0 = 25 - 5t5t = 25t = 5 seconds.Part 2: Distance covered (s)Using v² = u² + 2as:0² = 25² + 2 * (-5) * s0 = 625 - 10s10s = 625s = 62.5 meters.Alternatively, using s = ut + (1/2)at²:s = (25 * 5) + (1/2) * (-5) * (5)²s = 125 + (1/2) * (-5) * 25s = 125 - 62.5s = 62.5 meters.

PhysicsClass 9CBSE

Does an object in uniform circular motion have acceleration? If yes, in which direction does it act?

Yes, an object in uniform circular motion does have acceleration. This is because, even though its speed is constant, its velocity is continuously changing due to the continuous change in direction. This acceleration is called centripetal acceleration, and it always acts towards the center of the circular path.

PhysicsClass 9CBSE

In uniform circular motion, is the speed constant? Is the velocity constant? Explain your answer.

In uniform circular motion, the speed is constant because the magnitude of the velocity remains the same. However, the velocity is not constant because its direction continuously changes as the object moves along the circular path. Since velocity is a vector quantity (magnitude and direction), a change in direction means a change in velocity.

PhysicsClass 9CBSE

A boy is sitting on a merry-go-round which is moving. With respect to whom is the boy at rest, and with respect to whom is he in motion?

The boy is at rest with respect to the merry-go-round itself (e.g., the seat he is sitting on). He is in motion with respect to an observer standing on the ground, or with respect to the trees and buildings around the merry-go-round.

PhysicsClass 9CBSE

A bus moving at 36 km/h is brought to rest by applying brakes. If the brakes produce a uniform retardation of 0.5 m/s², how far does the bus travel before coming to rest?

Initial velocity (u) = 36 km/hConvert u to m/s: 36 km/h * (1000 m / 1 km) * (1 h / 3600 s) = 10 m/sFinal velocity (v) = 0 m/s (comes to rest)Acceleration (a) = -0.5 m/s² (retardation is negative acceleration)Using the third equation of motion: v² = u² + 2as0² = 10² + 2 * (-0.5) * s0 = 100 - ss = 100 meters.

PhysicsClass 9CBSE

A car accelerates uniformly from 10 m/s to 20 m/s in 5 seconds. What is the distance covered by the car during this time?

Initial velocity (u) = 10 m/sFinal velocity (v) = 20 m/sTime (t) = 5 sFirst, calculate acceleration (a) using v = u + at:20 = 10 + a * 510 = 5aa = 2 m/s²Now, calculate displacement (s) using s = ut + (1/2)at²:s = (10 * 5) + (1/2) * 2 * (5)²s = 50 + (1/2) * 2 * 25s = 50 + 25s = 75 meters.Alternatively, using v² = u² + 2as:20² = 10² + 2 * 2 * s400 = 100 + 4s300 = 4ss = 300 / 4 = 75 meters.

PhysicsClass 9CBSE

A train starting from rest attains a velocity of 72 km/h in 5 minutes. Assuming the acceleration is uniform, calculate the acceleration in m/s².

Initial velocity (u) = 0 m/s (since it starts from rest)Final velocity (v) = 72 km/hConvert v to m/s: 72 km/h * (1000 m / 1 km) * (1 h / 3600 s) = 20 m/sTime (t) = 5 minutesConvert t to seconds: 5 minutes * 60 seconds/minute = 300 secondsUsing the first equation of motion: v = u + at20 = 0 + a * 30020 = 300aa = 20 / 300 = 1 / 15 m/s² ≈ 0.0667 m/s².

PhysicsClass 9CBSE

A car travels a distance of 100 km in 2 hours and then another 150 km in 3 hours. Calculate the average speed of the car for the entire journey.

Total Distance = 100 km + 150 km = 250 kmTotal Time = 2 hours + 3 hours = 5 hoursAverage Speed = Total Distance / Total Time = 250 km / 5 hours = 50 km/h.

PhysicsClass 9CBSE

Derive the third equation of motion (v² = u² + 2as) graphically.

PROMPT: A velocity-time graph showing uniform acceleration. The y-axis represents velocity (v) and the x-axis represents time (t). A straight line starts from initial velocity 'u' on the y-axis at t=0 and goes up to final velocity 'v' at time 't'. A horizontal dashed line extends from 'u' to the point (t,u) and a vertical dashed line extends from 'v' to the point (t,v). A point 'A' is at (0,u), 'B' is at (t,v), 'C' is at (t,u), and 'D' is at (0,0). The area under the graph OABD is calculated as the area of a trapezium.Consider an object moving with uniform acceleration 'a'. Let its initial velocity be 'u' at time t=0, and its final velocity be 'v' at time 't'.From the velocity-time graph, the displacement (s) is given by the area under the v-t graph, which is the area of trapezium OABD.Area of trapezium = (1/2) × (Sum of parallel sides) × heights = (1/2) × (OA + BD) × ODFrom the graph, OA = u, BD = v, and OD = t.So, s = (1/2) × (u + v) × t --- (Equation 1)From the first equation of motion, we know that v = u + at, which can be rearranged to find t:t = (v - u) / a --- (Equation 2)Substitute the value of 't' from Equation 2 into Equation 1:s = (1/2) × (u + v) × [(v - u) / a]s = (1/2a) × (v + u)(v - u)Using the algebraic identity (x + y)(x - y) = x² - y²:s = (1/2a) × (v² - u²)2as = v² - u²Therefore, v² = u² + 2as.

PhysicsClass 9CBSE

Derive the second equation of motion (s = ut + (1/2)at²) graphically.

PROMPT: A velocity-time graph showing uniform acceleration. The y-axis represents velocity (v) and the x-axis represents time (t). A straight line starts from initial velocity 'u' on the y-axis at t=0 and goes up to final velocity 'v' at time 't'. A horizontal dashed line extends from 'u' to the point (t,u) and a vertical dashed line extends from 'v' to the point (t,v). A point 'A' is at (0,u), 'B' is at (t,v), 'C' is at (t,u), and 'D' is at (0,0). The area under the graph OABD represents displacement. This area is divided into a rectangle OACD and a triangle ABC.Consider an object moving with uniform acceleration 'a'. Let its initial velocity be 'u' at time t=0, and its final velocity be 'v' at time 't'.From the velocity-time graph, the displacement (s) is given by the area under the v-t graph.Area of trapezium OABD = Area of rectangle OACD + Area of triangle ABCArea of rectangle OACD = length × breadth = OA × OD = u × tArea of triangle ABC = (1/2) × base × height = (1/2) × AC × BCFrom the graph, AC = OD = t.And BC = BD - CD = v - u.We know from the first equation of motion that v - u = at.So, Area of triangle ABC = (1/2) × t × (at) = (1/2)at².Therefore, Total Displacement (s) = ut + (1/2)at².

PhysicsClass 9CBSE

Derive the first equation of motion (v = u + at) graphically.

PROMPT: A velocity-time graph showing uniform acceleration. The y-axis represents velocity (v) and the x-axis represents time (t). A straight line starts from initial velocity 'u' on the y-axis at t=0 and goes up to final velocity 'v' at time 't'. A horizontal dashed line extends from 'u' to the point (t,u) and a vertical dashed line extends from 'v' to the point (t,v). A point 'A' is at (0,u), 'B' is at (t,v), 'C' is at (t,u), and 'D' is at (0,0). The slope of the line AB represents acceleration.Consider an object moving with uniform acceleration 'a'. Let its initial velocity be 'u' at time t=0, and its final velocity be 'v' at time 't'.From the velocity-time graph, acceleration is the slope of the velocity-time graph.Slope = (Change in velocity) / (Change in time)a = (v - u) / (t - 0)a = (v - u) / tat = v - uTherefore, v = u + at.

PhysicsClass 9CBSE

State the third equation of motion and explain the meaning of each term in it.

The third equation of motion is: v² = u² + 2asWhere:v = final velocity of the objectu = initial velocity of the objecta = acceleration of the objects = displacement of the object

PhysicsClass 9CBSE

State the second equation of motion and explain the meaning of each term in it.

The second equation of motion is: s = ut + (1/2)at²Where:s = displacement of the objectu = initial velocity of the objecta = acceleration of the objectt = time taken for the displacement 's'

PhysicsClass 9CBSE

State the first equation of motion and explain the meaning of each term in it.

The first equation of motion is: v = u + atWhere:v = final velocity of the objectu = initial velocity of the objecta = acceleration of the objectt = time taken for the velocity to change from u to v

BiologyClass 9CBSE

A boy runs along a circular track of radius 7 meters. If he completes one full round in 22 seconds, calculate his average speed and the magnitude of his average velocity for this round.

For one full round:Distance covered = Circumference = 2πr = 2 * (22/7) * 7 m = 44 meters.Time taken = 22 seconds.Average Speed = Distance / Time = 44 m / 22 s = 2 m/s.For one full round, the boy returns to his starting position. Therefore, his displacement is zero.Average Velocity = Displacement / Time = 0 m / 22 s = 0 m/s.The magnitude of his average velocity is 0 m/s.

PhysicsClass 9CBSE

Explain what is meant by positive acceleration and negative acceleration (retardation). Give one example for each.

Positive acceleration occurs when the velocity of an object increases in the direction of motion. For example, a car speeding up from a traffic light. Negative acceleration, also known as retardation or deceleration, occurs when the velocity of an object decreases, or its speed reduces, in the direction of motion. For example, a car applying brakes to slow down.

STILL STUCK?

Get a tutor for just your question.

One-on-one help, verified tutors. Matched within 24 hours.

Get Started →