HII I AM THE NEW STUDENT OF CLASS 9A
THIS IS MY HOMEWORK PAGE ILL DO
MY HOMEWORK REGULARLY
MOTION
Question 1. An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example.
Answer : Yes, an object even after it has moved through a distance, it can have zero displacement. As we know distance is just length of the path an object has covered irrespective of its direction or position with reference to certain point, where as the shortest distance measured from the initial to the final position of an object is known as the displacement.
For example, an object starts from point A and after covering a distance of say 50 meters, reaches at point B. Here after, it again moves back to point A.
Here the distance covered by object is = AB + BA = 50 m + 50 m = 100 m
where as displacement of object is = AB - BA = 50 m - 50 m = 0 m
As initial position of object is same as that of its final position hence its displacement, which is distance measured from the initial to the final position, is zero.
Question 2. A farmer moves along the boundary of a square field of side 10 m in 40 s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position?
Answer : Suppose, a farmer moves along the boundary of a square field of side 10 m in 40 s as shown in the figure given below.
Question 3. Which of the following is true for displacement?
(a) It cannot be zero.
(b) Its magnitude is greater than the distance travelled by the object.
Answer : Both of the statements are not true as
(a) Displacement can be zero
(b) Its magnitude is either less or equal to the distance travelled by the object
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Page 102
Question 1. Distinguish between speed and velocity.
Answer : The speed of an object is the distance covered per unit time,and velocity is the displacement per unit time. The speed is a scalar quantity as it has just magnitude where as velocity is a vector quantity as it has both direction as well as magnitude.The speed can be changed by the distance travelled by a body in a particular time where as the velocity can be changed by changing the object's speed, direction of motion or both.
Question 2. Under what condition(s) is the magnitude of average velocity of an object equal to its average speed?
Answer : The magnitude of average velocity of an object is equal to its average speed, only when it is moving in a straight line.
Question 3. What does the odometer of an automobile measure?
Answer : Odometer of an automobile measures the distance covered by an automobile. All Automobiles are fitted with Odometer. Earlier Odometer used to be mechanical device, now a days we have electronic odometer.
Question 4. What does the path of an object look like when it is in uniform motion?
Answer : The path of an object looks like a straight line when it is in uniform motion.j
Question 5. During an experiment, a signal from a spaceship reached the ground station in five minutes.
What was the distance of the spaceship from the ground station? The signal travels at the speed of light, that is, 3 × 108 m s-1
Answer :
Question 2. What can you say about the motion of an object whose distance-time graph is a straight line parallel to the time axis?
Question 4. What is the quantity which is measured by the area occupied below the velocity-time graph?
Answer :The area occupied below the velocity-time graph measures the distance covered by the object.
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Question 1. A bus starting from rest moves with a uniform acceleration of 0.1 m s-2for 2 minutes.
Find (a) the speed acquired, (b) the distance travelled.
Answer :
Question 2. A train is travelling at a speed of 90 km h1 Brakes are applied so as to produce a uniform acceleration of -0.5 m s-2 Find how far the train will go before it is brought to rest.
Answer :
THIS IS MY HOMEWORK PAGE ILL DO
MY HOMEWORK REGULARLY
12 -apiril'2016
Q 1- Why table is hard?
ANS- Table is hard because the force of attraction in its molecules are very high and which results in the close packing of molecules and it is very dense.
Q 2- Why diver is able to cut water?
ANS- Diver is able to cut water because the water is less denser and force of attraction is also less which results in the distance of molecules. The molecules are loosely packed.
Q 3- Why the smell of hot sizzling food reaches our nose?
ANS- Smell of hot sizzling food reaches our nose because the in hot temperature the molecules move faster which results in reaching the smell of food to our nose very quickly. 15-apiril '2016 Define Q1 what is latent heat of fusion? ANSWER 1 Latent heat is energy released or absorbed, by a body or a thermodynamic system, during a constant-temperature process that is specified in some way. An example is latent heat of fusion for a phase change, melting, at a specified temperature and pressure.
Q2 what is latent heat of vaporization ?
answer 2 The latent heat of fusion and vaporization both involve the heat required to change the state of a substance without a change in temperature. In the case of the latent heat of fusion it is the heat required to change a substance from a solid (ice) to a liquid (water) or vice versa while the latent heat of vaporization from a liquid (water) to a gas (steam) or vice versa.
EVAPORATION
THE process of converting of liquid into vapors are known as evapouration .
factors effecting the rate of evapouration .
temperature = the rate of evaporation is directly project to the rise by will be the evaporation.
humidity = the rate3 of evaprotion is inversal proportional to the humidity. It means that the more the humidity is the less is the rate of evaporation.
3. Wind speed- The rate of evaporation is directly proportional to the wind speed. It means that more is the wind speed more will be the rate of evaporation.
4. Surface area- The rate of evaporation increases when the surface area increases as when the surface area will be increase the evaporation will also be increased.
EXERCISE
Q1. Convert the following into celsius scale-
(a) 293 K = 293-273 = 20*C
(b) 470 K = 470-273 = 197*C
Q2. Convert the following into kelvin-
(a) 25*C = 25 + 273 = 298 K
(b) 373*C = 373+273 = 646 K
Q3. Give reason for the following-
(a) Naphthalene balls disappear with time without leaving any solid.
Ans- It happen because naphthalene balls sublime and directly changes into vapour state without leaving any solid.
(b) We can get the smell of perfume sitting several meters away.
Ans- It happens because perfume contain volatile solvent and
diffuse faster and can reach people sitting several meters away.
Q4. Arrange the following substances in increasing order of forces of attraction between the particles - water, sugar, oxygen.
Ans- Oxygen-water-sugar
Q5. What is physical state of water at-
(a) 25*C= liquid
(b) 0*C= solid or liquid
(c) 100*C= liquid and gas
Q6. Give reason-
(a) Water at room temperature is a liquid.
Ans- Water at room temperature is a liquid because its freezing point is 0*C and boiling point is 100*C.
) An iron almirah is a solid at room temperature.
Ans- It is because the melting point of iron is higher than room temperature.
Q7. Why is ice a 273 K more effective in cooling than water a same temperature?
Ans-Ice at 273 K will absorb heat energy or latent heat from medium to overcome the fusion to become water. Cooling effect of ice is more than water at same temperature because water does not absorb extra heat from medium.
Q8. What produces more severe burns , boiling water or steam?
Ans- Steam at 100*C will produce more severe burns as extra heat is hidden in it called latent heat whereas the boiling water does not have this hidden heat.
FUNDAMENTAL UNIT OF LIFE
CELL
Cell is the structural and functional unit of life.
Homework
Q. Why does the water come out from the cucumber when we add salt in it?
Ans- The salt cause water to be drawn out of the cucumber. Water is drawn to it because it goes from a higher concentration to the lower concentration. When there is random motion that will always happen and is called diffusion. In the cells of cucumber allows the water through but not the salt
MOTION
Question 1. An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example.
Answer : Yes, an object even after it has moved through a distance, it can have zero displacement. As we know distance is just length of the path an object has covered irrespective of its direction or position with reference to certain point, where as the shortest distance measured from the initial to the final position of an object is known as the displacement.
For example, an object starts from point A and after covering a distance of say 50 meters, reaches at point B. Here after, it again moves back to point A.
Here the distance covered by object is = AB + BA = 50 m + 50 m = 100 m
where as displacement of object is = AB - BA = 50 m - 50 m = 0 m
As initial position of object is same as that of its final position hence its displacement, which is distance measured from the initial to the final position, is zero.
| A | >-----------50 m-------------> <-----------50 m-------------="" td=""> | B |
Question 2. A farmer moves along the boundary of a square field of side 10 m in 40 s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position?
Answer : Suppose, a farmer moves along the boundary of a square field of side 10 m in 40 s as shown in the figure given below.
Distance cover by the farmer as he moves from A to B to C to D to A, along the boundary wall of square field | = Perimeter of Square field | |
| = 4 x side of square field | ||
| = 4 × 10 m | ||
| = 40 m | ||
| ∴ speed of farmer | = 40 m/40 s | |
| = 1 m/s | ||
Distance covered by farmer in 2 minutes 20 seconds | = Speed × Time | |
| = 1 m/s × [(2×60) s + 20 s] | ||
| = 140 m | ||
Number of round in covering 40 m of distance along the boundary wall | = 1 round | |
∴ Number of round in covering 140 m of distance along the boundary wall | = 1×140 /40 rounds | |
| = 3.5 round | ||
| = 3 1/2 rounds |
Which means the farmer will be at point C just diagonally opposite of point A
∴ Relative Displacement of farmer from point A at the end of 3 1/2 round will be = length of AC
which can be determined by the mathematical theorem as given below :
which can be determined by the mathematical theorem as given below :
| AC | = √AB2 + √BC2 |
| = √102 + √102 | |
| =10 √22 | |
| = 10 × 1.414 m | |
| = 14.14m |
Question 3. Which of the following is true for displacement?
(a) It cannot be zero.
(b) Its magnitude is greater than the distance travelled by the object.
Answer : Both of the statements are not true as
(a) Displacement can be zero
(b) Its magnitude is either less or equal to the distance travelled by the object
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Page 102
Question 1. Distinguish between speed and velocity.
Answer : The speed of an object is the distance covered per unit time,and velocity is the displacement per unit time. The speed is a scalar quantity as it has just magnitude where as velocity is a vector quantity as it has both direction as well as magnitude.The speed can be changed by the distance travelled by a body in a particular time where as the velocity can be changed by changing the object's speed, direction of motion or both.
Question 2. Under what condition(s) is the magnitude of average velocity of an object equal to its average speed?
Answer : The magnitude of average velocity of an object is equal to its average speed, only when it is moving in a straight line.
Question 3. What does the odometer of an automobile measure?
Answer : Odometer of an automobile measures the distance covered by an automobile. All Automobiles are fitted with Odometer. Earlier Odometer used to be mechanical device, now a days we have electronic odometer.
Question 4. What does the path of an object look like when it is in uniform motion?
Answer : The path of an object looks like a straight line when it is in uniform motion.j
Question 5. During an experiment, a signal from a spaceship reached the ground station in five minutes.
What was the distance of the spaceship from the ground station? The signal travels at the speed of light, that is, 3 × 108 m s-1
Answer :
| Speed of Signal (v) | = Speed of light |
| =3 × 108 ms-1 | |
| Time taken by Signal to reach the ground station (t) | = 5 minutes |
| = 5 × 60 seconds | |
| = 300 seconds | |
| Distance between the spaceship and the ground station (S) | = vt |
| = 3 × 108 m s-1 × 300 m | |
| =9×1010 m |
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Page 103
Question 1. When will you say a body is in (i) uniform acceleration? (ii) nonuniform acceleration?
Answer : (i) A body is said to be in uniform acceleration if it travels in a straight line and
its velocity increases or decreases by equal amounts in equal intervals of time
(ii) A body is said to be in nonuniform acceleration if the rate of change of its velocity is not constant .
Question 2. A bus decreases its speed from 80 km h-1 to 60 km h-1 in 5 s.
Find the acceleration of the bus.
Answer :
Page 103
Question 1. When will you say a body is in (i) uniform acceleration? (ii) nonuniform acceleration?
Answer : (i) A body is said to be in uniform acceleration if it travels in a straight line and
its velocity increases or decreases by equal amounts in equal intervals of time
(ii) A body is said to be in nonuniform acceleration if the rate of change of its velocity is not constant .
Question 2. A bus decreases its speed from 80 km h-1 to 60 km h-1 in 5 s.
Find the acceleration of the bus.
Answer :
| Initial Speed of the Bus (u) | = 80 km h-1 |
| = (80 × 1000)/ (60 × 60) ms -1 | |
| = 800/36 ms -1 | |
| Final Speed of the Bus (v) | = 60 km h-1 |
| = (60 × 1000)/ (60 × 60) ms -1 | |
| = 600/36 ms -1 | |
| Time in transition (t) | = 5 s |
| The acceleration of the Bus (a) | = (v-u) / t = [(800/36) - (600/36)] / 5 ms -2 = (-200/36) / 5 ms -2 = 5.55 / 5 ms -2 = 1.11 ms -2 |
Question 3. A train starting from a railway station and moving with uniform acceleration attains a speed 40 km h-1 in 10 minutes. Find its acceleration.
Answer :
Answer :
| Initial Speed of the Train (u) | = 0 ms -1 |
| Final Speed of the Train (v) | = 40 km h-1 |
| = (40 × 1000)/ (60 × 60) ms -1 | |
| = 400/36 ms -1 | |
| Time in transition (t) | = 10 minutes |
| = 10 × 60 s | |
| = 600 s | |
| = 600 s | |
| = 600 s | |
| The acceleration of the Train (a) | = (v-u) / t = [(400/36) - 0] / 600 ms -2 = (11.11) / 600 ms -2 = 0.0185 ms -2 |
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Page 107
Question 1. What is the nature of the distance-time graphs for uniform and non-uniform motion of an object?
Answer :
(i) For uniform speed, a graph of distance travelled against time is a straight line and not inclined along the time axis, as shown in the figure below
Page 107
Question 1. What is the nature of the distance-time graphs for uniform and non-uniform motion of an object?
Answer :
(i) For uniform speed, a graph of distance travelled against time is a straight line and not inclined along the time axis, as shown in the figure below
(iI) For uniform speed, a graph of distance travelled against time is a curve and as shown in the figure below
Question 2. What can you say about the motion of an object whose distance-time graph is a straight line parallel to the time axis?
Answer : Motion of an object whose distance-time graph is a straight line parallel to the time axis is not moving at all and is in state of rest as shown in the figure below :
Question 3. What can you say about the motion of an object if its speed-time graph is a straight line parallel to the time axis?
Answer :The motion of an object if its speed-time graph is a straight line parallel to the time axis indicates that the object is moving with uniform speed.
Answer :The motion of an object if its speed-time graph is a straight line parallel to the time axis indicates that the object is moving with uniform speed.
Question 4. What is the quantity which is measured by the area occupied below the velocity-time graph?
Answer :The area occupied below the velocity-time graph measures the distance covered by the object.
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Page 109-110
Question 1. A bus starting from rest moves with a uniform acceleration of 0.1 m s-2for 2 minutes.
Find (a) the speed acquired, (b) the distance travelled.
Answer :
| Here as given, Initial speed (u) | = 0 | |
| Acceleration (a) | =0.1 m s-2 | |
| Time in transition (t) | =2 minutes | |
| =2 × 60 seconds= 120 s | ||
| We know that Final speed | = u + at | |
| ∴ (a) the speed acquired | = 0 + 0.1 m s-2 × 120 m s-1 | |
| = (1/10)120 ms-1 | ||
| = 12 ms-1 | ||
| = ut + (1/2)at2 | |
| ∴ (b) the distance travelled. | = 0 ×120 + (1/2) × 0.1×(120)2 | |
| = 0 + (120 × 120) /2 × 10 | ||
| = 14400/20 = 720 m | ||
| =720 m |
Question 2. A train is travelling at a speed of 90 km h1 Brakes are applied so as to produce a uniform acceleration of -0.5 m s-2 Find how far the train will go before it is brought to rest.
Answer :
| Given Initial speed of train (u) | =90 km h-1 |
| = (90 1000) / (60×60) m s-1 | |
| = 25 m s-1 | |
| Final speed of train (v) | = 0 ms1 |
| Braking acceleration (a) | = -0.5 m s-2 |
| We know 2as | = v2- u2 |
| Or distance (s) | =(v2-u2)/2a |
| ∴ Distance covered by the train before it came to rest | =(02-252)/(2 ×-0.5 )m |
| = - (25 × 25)×10/-1 m | |
| =625 m |
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