67486 + Interview Questions in Secondary School in Physics

1.	If a cell with potential difference is between 2 resistors whether they are in series or parallel
Answer» RESISTORS in PARALLEL When resistors are CONNECTED in parallel, the supply current is equal to the sum of the currents through each RESISTOR. ... When resistors are connected in parallel, they have the same potential difference across them.

1.

If a cell with potential difference is between 2 resistors whether they are in series or parallel

Answer»

When resistors are CONNECTED in parallel, the supply current is equal to the sum of the currents through each RESISTOR. ... When resistors are connected in parallel, they have the same potential difference across them.

Discussion

2.	A hiker begins a trip by waking 25.0 km South-East from her base camp. On the second day she walks 40.0 km in direction o 60 North to East, at which point she discovers a forest ranger's tower? (i) Determine the component of the hiker's displacements in the first and second days. (ii) Determine the component of the hiker's total displacement for the trip. (iii) Find the magnitude and direction of the displacement from base camp
Answer» Answer: Here are the answers to all the parts Explanation: (I) Ax = 17.7km, Ay = −17.7km, Bx = 20.0km, and By = 34.6km. (ii)RX = 37.7km and Ry = 16.9km. (iii) \|R~ \| = 41.3km, 24.4◦ north of east from the base camp. Here displacement is the TOTAL distance COVERED with a change in the direction. The HIKER covered a total distance of 34.6Km.

2.

A hiker begins a trip by waking 25.0 km South-East from her base camp. On the second day she walks 40.0 km in direction o 60 North to East, at which point she discovers a forest ranger's tower? (i) Determine the component of the hiker's displacements in the first and second days. (ii) Determine the component of the hiker's total displacement for the trip. (iii) Find the magnitude and direction of the displacement from base camp

Answer»

Answer:

Here are the answers to all the parts

Explanation:

(I) Ax = 17.7km, Ay = −17.7km, Bx = 20.0km, and By = 34.6km.

(ii)RX = 37.7km and Ry = 16.9km.

(iii) |R~ | = 41.3km, 24.4◦ north of east from the base camp.

Here displacement is the TOTAL distance COVERED with a change in the direction. The HIKER covered a total distance of 34.6Km.

Discussion

3.	Fill in the blanks.1. The capacity of a container is the ...... of theliquid it can hold.2. The unit of capacity is the same as .......3. A substance of relative density greater than 1is ...... than water.4. The mass of a body is the amount of ...... inthe body.5. The greater the mass of a body, the ..... is itsweight.
Answer» ANSWER: mass calories LIQUID weight greater

3.

Fill in the blanks.1. The capacity of a container is the ...... of theliquid it can hold.2. The unit of capacity is the same as .......3. A substance of relative density greater than 1is ...... than water.4. The mass of a body is the amount of ...... inthe body.5. The greater the mass of a body, the ..... is itsweight.

Answer»

ANSWER:

mass

calories

LIQUID

weight

greater

Discussion

4.	A glass tube is inverted with the open ends of the straight limbs of diameters o.5mm and 1.00mm redpectively, b er low the surface of water in a beaker.thr air pressure in thevupper psrt increased until one meniscus level with eater outside.gind the heifht of the water inbthe other limb
Answer» ANSWER: the ANS is 20000 mm Explanation:

4.

A glass tube is inverted with the open ends of the straight limbs of diameters o.5mm and 1.00mm redpectively, b er low the surface of water in a beaker.thr air pressure in thevupper psrt increased until one meniscus level with eater outside.gind the heifht of the water inbthe other limb

Answer»

ANSWER:

the ANS is 20000 mm

Explanation:

Discussion

5.	A uniform metre rule is balanced horizontally on a wedge placed under the 40cm mark by a weight of 0.5N hanging from the 20cm mark. a) What is the weight of the metre rule? b) the wedge is moved under the 60cm mark. What weight must now be placed at the end of the rule to keep it horizontal?
Answer» Answer: I ASSUME you meant to SAY the length of the stick is 100 cm. Set up equilibrium of TORQUES about the pivot POINT. Apply the weight of the stick through its midpoint at 50 cm. M = mass of the stick MG[(50 - 42.5) cm] = (40.0 grams)g[(42.5 - 20) cm] M = 40.0(42.5 - 20) / (50 - 42.5) grams = 120 grams

5.

A uniform metre rule is balanced horizontally on a wedge placed under the 40cm mark by a weight of 0.5N hanging from the 20cm mark. a) What is the weight of the metre rule? b) the wedge is moved under the 60cm mark. What weight must now be placed at the end of the rule to keep it horizontal?

Answer»

Answer:

I ASSUME you meant to SAY the length of the stick is 100 cm. Set up equilibrium of TORQUES about the pivot POINT. Apply the weight of the stick through its midpoint at 50 cm.

M = mass of the stick

MG[(50 - 42.5) cm] = (40.0 grams)g[(42.5 - 20) cm]

M = 40.0(42.5 - 20) / (50 - 42.5) grams = 120 grams

Discussion

6.	Using your own ideas on your own understanding try to frame a defination on 'motion'
Answer» ANSWER: MOTION is a MOVEMENT or a WAY of MOVING

6.

Using your own ideas on your own understanding try to frame a defination on 'motion'

Answer»

ANSWER:

MOTION is a MOVEMENT or a WAY of MOVING

Discussion

7.	An athletic can complete one round of a truck of radius R in 40sec. His displacement after 2 minutes 20 seconds will be : (a) 2R (b) 7R (c) 2πR (d) 7πR
Answer» Explanation: althletic need 40s to complete one round therefore althletic need 2 min to complete 5 rounds displacement is zero in 5 rounds as starting and ending point is same in next 20 SEC he travel half round displacement=2r hope you LIKE answer

7.

An athletic can complete one round of a truck of radius R in 40sec. His displacement after 2 minutes 20 seconds will be : (a) 2R (b) 7R (c) 2πR (d) 7πR

Answer»

Explanation:

althletic need 40s to complete one round

therefore althletic need 2 min to complete 5 rounds

displacement is zero in 5 rounds as starting and ending point is same

in next 20 SEC he travel half round

displacement=2r

hope you LIKE answer

Discussion

8.	What is fromulae of force
Answer» ANSWER: EXPLANATION: F = ma

8.

What is fromulae of force

Answer»

ANSWER:

EXPLANATION:

F = ma

Discussion

9.	How to find reading of voltmeter when it is connected with a resistance
Answer» EXPLANATION: Now the resistances 200 ohms, 100 ohms and the combined resistance of 200 ohms are all CONNECTED in series. So the total resistance is 200+100+200 = 500 ohms. Now, the current through the CIRCUIT is CALCULATED as I=V/R = 100/500 = 0.2 A. The voltmeter of 600 ohms is connected across the 300 ohms resistance

9.

How to find reading of voltmeter when it is connected with a resistance

Answer»

EXPLANATION:

Now the resistances 200 ohms, 100 ohms and the combined resistance of 200 ohms are all CONNECTED in series. So the total resistance is 200+100+200 = 500 ohms. Now, the current through the CIRCUIT is CALCULATED as I=V/R = 100/500 = 0.2 A. The voltmeter of 600 ohms is connected across the 300 ohms resistance

Discussion

10.	Please give right answer how you how much you can know
Answer» ANSWER: SYMBIOSIS is an ASSOCIATION between TWO species where both of them are benefited ex; leech

10.

Please give right answer how you how much you can know

Answer»

ANSWER:

SYMBIOSIS is an ASSOCIATION between TWO species where both of them are benefited

ex; leech

Discussion

11.	Odd one out with reason please....
Answer» ANSWER: Sit Resultant of force Streamlined shape Explanation:

11.

Odd one out with reason please....

Answer»

ANSWER:

Sit

Resultant of force

Streamlined shape

Explanation:

Discussion

12.	Two masses of 1g and 4g are moving with equal kinetic energies. The ratio of magnitudes of their moments is : (a) 1:6 (b) 1:2 (c) 4:1 (d) 1:8
Answer» Answer: 1:2 Explanation: KINETIC energy =K.E $momentum = \sqrt{2m \:k.e}$ M(1) = (2×1×K.E)^1/2 M(2)= (2×4×K.E)^1/2 $\frac{m1}{m2} = \frac{1}{2}$ THEREFORE RATIO of their momenum is 1:2 Hope it helps to clear your concept . ☺✌

12.

Two masses of 1g and 4g are moving with equal kinetic energies. The ratio of magnitudes of their moments is : (a) 1:6 (b) 1:2 (c) 4:1 (d) 1:8

Answer»

Answer:

1:2

Explanation:

KINETIC energy =K.E

$momentum = \sqrt{2m \:k.e}$

M(1) = (2×1×K.E)^1/2

M(2)= (2×4×K.E)^1/2

$\frac{m1}{m2} = \frac{1}{2}$

THEREFORE RATIO of their momenum is 1:2

Hope it helps to clear your concept . ☺✌

Discussion

13.	Is square of distance is called acceleration??and if possible than please give reasons
Answer» no, it CANT be POSSIBLE. ..................€€€

13.

Is square of distance is called acceleration??and if possible than please give reasons

Answer»

no, it CANT be POSSIBLE.

..................€€€

Discussion

14.	Name the basic property of electric charges in Rn86 gives Po84+He2
Answer» GIVEN that, The equation is $Rn_{86}\Rightarrow Po_{84}+He_{2}$ We know that, Property of BALANCE equation, An equation is balance when the sum of charges and sum of masses are equal on both SIDES of the equation. In this equation, The charges on left side is equal to the sum of the charges on the right side. The name of ELECTRIC CHARGE property is subscript. In mathematically , $86=84+2$ L.H.S = R.H.S Hence, The name of the basic property of electric charges in equation is subscript.

14.

Name the basic property of electric charges in Rn86 gives Po84+He2

Answer»

GIVEN that,

The equation is

$Rn_{86}\Rightarrow Po_{84}+He_{2}$

We know that,

Property of BALANCE equation,

An equation is balance when the sum of charges and sum of masses are equal on both SIDES of the equation.

In this equation,

The charges on left side is equal to the sum of the charges on the right side.

The name of ELECTRIC CHARGE property is subscript.

In mathematically ,

$86=84+2$

L.H.S = R.H.S

Hence, The name of the basic property of electric charges in equation is subscript.

Discussion

15.	Two waves of the same pitch have amplitudes in the ratio 1:3 what will be the ratio of their intensities
Answer» Answer: 1:9 Explanation:

15.

Two waves of the same pitch have amplitudes in the ratio 1:3 what will be the ratio of their intensities

Answer»

Answer:

1:9

Explanation:

Discussion

16.	Calculate the pressure when a force of 20 n is applied over an area of hundred metre square
Answer» Answer: 0.2 PASCALS(N/metre SQUARED)

16.

Calculate the pressure when a force of 20 n is applied over an area of hundred metre square

Answer»

Answer:

0.2 PASCALS(N/metre SQUARED)

Discussion

17.	Which conductor having maximum resistance
Answer» EXPLANATION: The CONDUCTOR which has the HIGHEST RESISTANCE are ALLOYS

17.

Which conductor having maximum resistance

Answer»

EXPLANATION:

The CONDUCTOR which has the HIGHEST RESISTANCE are ALLOYS

Discussion

18.	What are Hydrides ?Discuss saline hydrides and covalent Hydrides
Answer» ┏_________∞◆∞_________┓ ✭✮ӇЄƦЄ ƖƧ ƳƠƲƦ ƛƝƧƜЄƦ✮✭ ┗_________∞◆∞_________┛ Hydrides are compounds of hydrogen where hydrogen is covalently bonded to the other atom For EXAMPLE H₂O is a hydride of oxygen, NH₃ is a ahydride of nitrogen etc Saline hydrides are compounds formed between hydrogen and the ALKALI and alkaline-earth metals of GROUP one and TWO elements For example NaH, CaH₂ etc Covalent hydrides are hydrides formed when hydrogen reacts with p-block elements. For example B₂H₆ , H₂O, CH₄ etc

18.

What are Hydrides ?Discuss saline hydrides and covalent Hydrides

Answer»

┏_________∞◆∞_________┓

✭✮ӇЄƦЄ ƖƧ ƳƠƲƦ ƛƝƧƜЄƦ✮✭

┗_________∞◆∞_________┛

Hydrides are compounds of hydrogen where hydrogen is covalently bonded to the other atom

For EXAMPLE H₂O is a hydride of oxygen, NH₃ is a ahydride of nitrogen etc

Saline hydrides are compounds formed between hydrogen and the ALKALI and alkaline-earth metals of GROUP one and TWO elements

For example NaH, CaH₂ etc

Covalent hydrides are hydrides formed when hydrogen reacts with p-block elements.

For example B₂H₆ , H₂O, CH₄ etc

Discussion

19.	The minimum value of y = x2 – 4x will exist at x is equal to
Answer» Answer: Evaluate y at x (PLUG x = 2 into the function) to find y-coordinate: The minimum VALUE of an upward curving parabola is the y-coordinate of the vertex. Hence the minimum value of the function is -9.

19.

The minimum value of y = x2 – 4x will exist at x is equal to

Answer»

Answer:

Evaluate y at x (PLUG x = 2 into the function) to find y-coordinate: The minimum VALUE of an upward curving parabola is the y-coordinate of the vertex. Hence the minimum value of the function is -9.

Discussion

20.	Action and reaction are equal in magnitude and opposite in direction.then why do they not balance each other?
Answer» ANSWER: They dont BALANCE each other out because they ACT on different objects so they don't cancel out. Explanation:

20.

Action and reaction are equal in magnitude and opposite in direction.then why do they not balance each other?

Answer»

ANSWER:

They dont BALANCE each other out because they ACT on different objects so they don't cancel out.

Explanation:

Discussion

21.	What is inertia write it's advantages
Answer» EXPLANATION: here is your ANSWER DEAR ✍️

21.

What is inertia write it's advantages

Answer»

EXPLANATION:

here is your ANSWER DEAR ✍️

Discussion

22.	76.6% of maximum torque
Answer» ANSWER: The performance of an engine is usually quoted by QUOTING the MAXIMUM power and torque and the engine rotation speeds at which they occur. $10 ❤️thanks = inbox$ $I hope this will helps you.$ Please LIKE my answer..........

22.

76.6% of maximum torque

Answer»

ANSWER:

The performance of an engine is usually quoted by QUOTING the MAXIMUM power and torque and the engine rotation speeds at which they occur.

$10 ❤️thanks = inbox$

$I hope this will helps you.$

Please LIKE my answer..........

Discussion

23.	Does NCERT has reduced class 11th and 12th physics syllabus ??
Answer» ANSWER: ASTER Q11 me KNOW if you Explanation: know when you're FREE

23.

Does NCERT has reduced class 11th and 12th physics syllabus ??

Answer»

ANSWER:

ASTER Q11 me KNOW if you

Explanation:

know when you're FREE

Discussion

24.	- একটি কাচের জানালার ক্ষেত্রফল 4 বর্গমিটার। কাচটির ক্ষেত্রফল প্রসারণেরজন্য জানালাটিতে 25 বর্গসেমি জায়গা রাখা আছে। কাচের দৈর্ঘ্য প্রসারণগৃণাঙ্ক 9x106°C-1 হলে, এটি সর্বোচ্চ কত উন্নতা পরিবর্তন সহ্য করতেপারবে ?[Ans. 34.72°C]
Answer» KRIPYA KAR iss prashan KO ya to English ya to Hindi m daale hm aapko uttar jarur denge gustakhi MAAF

24.

- একটি কাচের জানালার ক্ষেত্রফল 4 বর্গমিটার। কাচটির ক্ষেত্রফল প্রসারণেরজন্য জানালাটিতে 25 বর্গসেমি জায়গা রাখা আছে। কাচের দৈর্ঘ্য প্রসারণগৃণাঙ্ক 9x106°C-1 হলে, এটি সর্বোচ্চ কত উন্নতা পরিবর্তন সহ্য করতেপারবে ?[Ans. 34.72°C]

Answer»

KRIPYA KAR iss prashan KO ya to English ya to Hindi m daale hm aapko uttar jarur denge

gustakhi MAAF

Discussion

25.	Q:Find relation between E1 and E2.Please explain what did you do with the sinxsiny thing where you have two sine functions in one equation. NO SPAMMING
Answer» PLEASE PREFER PIC above ...

25.

Q:Find relation between E1 and E2.Please explain what did you do with the sinxsiny thing where you have two sine functions in one equation. NO SPAMMING

Answer»

PLEASE PREFER PIC above ...

Discussion

26.	Please answer the question with explanation.I will mark you as brainliest answer.
Answer» ANSWER: C÷n Explanation: because it is only given as air so we take the absolute refractive index and the more correct answer will be C÷Va In which C is the SPEED of light in vaccum and Va is the speed of light in air Hope it HELPS you and if so PLEASE mark it as the brainliest answer

26.

Please answer the question with explanation.I will mark you as brainliest answer.

Answer»

ANSWER: C÷n

Explanation: because it is only given as air so we take the absolute refractive index and the more correct answer will be C÷Va

In which C is the SPEED of light in vaccum and Va is the speed of light in air

Hope it HELPS you and if so PLEASE mark it as the brainliest answer

Discussion

27.	a truck ccelerates from 24m/s at constant rate 2.1m/s2 for 15sec. what is the final speed of the car at the end of the 15seconds.
Answer» Answer: 55.55 m/s Explanation: Given : INITIAL velocity = u = 24 m/s Acceleration of the truck = a = 2.1 m/s² Time taken = t = 15 seconds To find : Final velocity of the truck after 15 seconds USING the first EQUATION of motion : V=u+at V=24+(2.1×15) V=24+31.5 V=55.5 m/s The final velocity of the truck after 15 seconds is equal to 55.55 m/s More : First equation of motion = v = u+at Second equation of motion = s = ut+½×at² Third equation of motion = v²-u²=2as

27.

a truck ccelerates from 24m/s at constant rate 2.1m/s2 for 15sec. what is the final speed of the car at the end of the 15seconds.

Answer»

Answer:

55.55 m/s

Explanation:

Given :

INITIAL velocity = u = 24 m/s

Acceleration of the truck = a = 2.1 m/s²

Time taken = t = 15 seconds

To find :

Final velocity of the truck after 15 seconds

USING the first EQUATION of motion :

V=u+at

V=24+(2.1×15)

V=24+31.5

V=55.5 m/s

The final velocity of the truck after 15 seconds is equal to 55.55 m/s

More :

First equation of motion = v = u+at

Second equation of motion = s = ut+½×at²

Third equation of motion = v²-u²=2as

Discussion

28.	A person crossing a road with a certain velocity due north sees a car moving towards east the relative velocity of car with respect to the person is root 2 x that of the velocity of the person the angle made by the relative velocity with the East isanswer is 45° I just need the answer
Answer» I sᴇᴇᴍᴇᴅ ᴛʜᴀᴛ ʙʀᴀɪɴʟʏ ᴍᴏᴅᴇʀᴀᴛᴏʀs ᴅᴏᴇsɴ'ᴛ ʟɪᴋᴇ ᴍʏ ᴀɴsᴡᴇʀ sᴏ ᴛʜᴀᴛ ᴛʜᴇʏ sᴀɪᴅ ʀᴜᴅᴇ ! ᴡᴇ ᴅᴏɴ'ᴛ ᴜsᴇ sᴜᴄʜ ᴡᴏʀᴅs ʜᴇʀᴇ sᴏ ᴛʜᴀᴛ I ʜᴀᴅ ɢɪᴠᴇɴ ʏᴏᴜ sᴄʀᴇᴇɴsʜᴏᴛ

Discussion

29.	. An R - C circuit has R = 50 and C = 2000 μF. The time constant of the circuit is
Answer» ANSWER: i NEED your HELP GIVE EXAMPLES

29.

. An R - C circuit has R = 50 and C = 2000 μF. The time constant of the circuit is

Answer»

ANSWER:

i NEED your HELP GIVE EXAMPLES

Discussion

30.	State the nature and the lens and the focal of its power +2.5
Answer» EXPLANATION: PLEASE MARK me as BRANLIEST

30.

State the nature and the lens and the focal of its power +2.5

Answer»

EXPLANATION:

PLEASE MARK me as BRANLIEST

Discussion

31.	Give odd one out and please give me the reason also
Answer» TUMHARI CHOICE HAI KUCH BHI KRO

Discussion

32.	PLEASE EXPLAIN THE ANSWER FIRST AND BE THE BRAINLIEST FOR SURE Find the displacement and distance traveled by a body in 10sec, using the v-t graph given below (Diagram)
Answer» ANSWER: displacement=5m distance=25m

32.

PLEASE EXPLAIN THE ANSWER FIRST AND BE THE BRAINLIEST FOR SURE Find the displacement and distance traveled by a body in 10sec, using the v-t graph given below (Diagram)

Answer»

ANSWER:

displacement=5m

distance=25m

Discussion

33.	A body moves a distance of 6m towards east and then 8m towards north. Calculate the total distance travelled by the body and the net displacement of the body.
Answer» $\bigstar$ Answer $\bigstar$ $\checkmark$ Given: A body MOVES a distance of 6 m towards east and then 8 m towards NORTH $\checkmark$ To Find: Total Distance travelled Net Displacement $\checkmark$ SOLUTION: Given that, A body moves a distance of 6 m towards east and then 8 m towards north Therefore, Distance travelled is 6 m towards east and then 8 m towards north Hence, Total Distance travelled (D) D = 6 m+ 8 m = 14 m Therefore, $\star$ Total Distance travelled = 14 m According to the Question, We are ALSO ASKED to find the Net Displacement Therefore, Net Displacement (d) $\red{\implies \sf d=\sqrt{6^{2}+8^{2}} \ m}$ $\blue{\implies \sf d=\sqrt{36+64} \ m}$ $\green{\implies \sf d=\sqrt{100} \ m}$ $\pink{\implies \sf d=10 \ m}$ Hence, $\star$ The Net Displacement = 10 m

33.

A body moves a distance of 6m towards east and then 8m towards north. Calculate the total distance travelled by the body and the net displacement of the body.

Answer»

$\bigstar$ Answer $\bigstar$

$\checkmark$ Given:

A body MOVES a distance of 6 m towards east and then 8 m towards NORTH

$\checkmark$ To Find:

Total Distance travelled
Net Displacement

$\checkmark$ SOLUTION:

Given that,

A body moves a distance of 6 m towards east and then 8 m towards north

Therefore,

Distance travelled is 6 m towards east and then 8 m towards north

Hence,

Total Distance travelled (D)

D = 6 m+ 8 m = 14 m

Therefore,

$\star$ Total Distance travelled = 14 m

According to the Question,

We are ALSO ASKED to find the Net Displacement

Therefore,

Net Displacement (d)

$\red{\implies \sf d=\sqrt{6^{2}+8^{2}} \ m}$

$\blue{\implies \sf d=\sqrt{36+64} \ m}$

$\green{\implies \sf d=\sqrt{100} \ m}$

$\pink{\implies \sf d=10 \ m}$

Hence,

$\star$ The Net Displacement = 10 m

Discussion

34.	A body moves from a point a to b under the action of a force in figure force f is in newton and distance x in meter . what is the amount of work done ???
Answer» ┏_________∞◆∞_________┓ ✭✮ӇЄƦЄ ƖƧ ƳƠƲƦ ƛƝƧƜЄƦ✮✭ ┗_________∞◆∞_________┛ We know that, *WORK done = Force DISPLACEMENT W = FS* Given that, Force = F = f N Displacement = S = X m Hence, Work done = FX W = fx Joules

34.

A body moves from a point a to b under the action of a force in figure force f is in newton and distance x in meter . what is the amount of work done ???

Answer»

┏_________∞◆∞_________┓

✭✮ӇЄƦЄ ƖƧ ƳƠƲƦ ƛƝƧƜЄƦ✮✭

┗_________∞◆∞_________┛

We know that,

WORK done = Force * DISPLACEMENT

W = F*S

Given that,

Force = F = f N

Displacement = S = X m

Hence,

Work done = FX

W = fx Joules

Discussion

35.	If a,b,c are three vectors such that a vector.b vector
Answer» Answer: PLEASE MAKE SURE THAT YOUR QUESTION IS RIGHT.

35.

If a,b,c are three vectors such that a vector.b vector

Answer»

Answer:

PLEASE MAKE SURE THAT YOUR QUESTION IS RIGHT.

Discussion

36.	A dragster accelerates to a speed of 11m/s over a distance of 398m . Determine the acceleration (assumed uniform) of the dragster
Answer» Given : ▪ Initial velocity = zero ▪ Final velocity = 11m/s ▪ Distance travelled = 398m To Find : ▪ ACCELERATION of dragster. Concept : ☞ In this question, Acceleration has said to be uniform throughout the motion, we can EASILY APPLY third equation of KINEMATICS to solve this question. ✴ Third equation of kinematics : ↗ v² - u² = 2as where, ▪ v denotes final velocity ▪ u denotes initial velocity ▪ a denotes acceleration ▪ s denotes distance Calculation : → v² - u² = 2as → (11)² - (0)² = 2a(398) → 121 = 796a → a = 121/796 → a = 0.15m/s²

36.

A dragster accelerates to a speed of 11m/s over a distance of 398m . Determine the acceleration (assumed uniform) of the dragster

Answer»

Given :

▪ Initial velocity = zero

▪ Final velocity = 11m/s

▪ Distance travelled = 398m

To Find :

▪ ACCELERATION of dragster.

Concept :

☞ In this question, Acceleration has said to be uniform throughout the motion, we can EASILY APPLY third equation of KINEMATICS to solve this question.

✴ Third equation of kinematics :

↗ v² - u² = 2as

where,

▪ v denotes final velocity

▪ u denotes initial velocity

▪ a denotes acceleration

▪ s denotes distance

Calculation :

→ v² - u² = 2as

→ (11)² - (0)² = 2a(398)

→ 121 = 796a

→ a = 121/796

→ a = 0.15m/s²

Discussion

37.	What is magnet and magnetic feild
Answer» Answer: A MAGNETIC field is a vector field that describes the magnetic influence of electric charges in relative MOTION and magnetized materials. A charge that is moving parallel to a CURRENT of other charges experiences a force perpendicular to its own velocity. I hope it helps you plz MARK me as BRAINLIEST

37.

What is magnet and magnetic feild

Answer»

Answer:

A MAGNETIC field is a vector field that describes the magnetic influence of electric charges in relative MOTION and magnetized materials. A charge that is moving parallel to a CURRENT of other charges experiences a force perpendicular to its own velocity.

I hope it helps you

plz MARK me as BRAINLIEST

Discussion

38.	(3) A car hitting a motorcycleTwo objects of masses 10 kg and 15 kg are moving in the same direction with different velocities5 m/min and 10 m/min respectively. If they collide and stick together and move with a common velocity,then the magnitude of their common velocity is
Answer» ANSWER: Answer: 8m/min Explanation: M1u + m2v = ( M1+m2)V 10(5) +15(10)= (15+10)v 150+50= 25v 200=25v 8m/min = v

38.

(3) A car hitting a motorcycleTwo objects of masses 10 kg and 15 kg are moving in the same direction with different velocities5 m/min and 10 m/min respectively. If they collide and stick together and move with a common velocity,then the magnitude of their common velocity is

Answer»

ANSWER:

Answer: 8m/min

Explanation:

M1u + m2v = ( M1+m2)V

10(5) +15(10)= (15+10)v

150+50= 25v

200=25v

8m/min = v

Discussion

39.	Can anyone give me examples of gravitational force
Answer» ┏_________∞◆∞_________┓ ✭✮ӇЄƦЄ ƖƧ ƳƠƲƦ ƛƝƧƜЄƦ✮✭ ┗_________∞◆∞_________┛ Gravitational force is the attractive force between 2 objects or bodies having mass This force is given by the equation F = Gm₁m₂/r² where, m₁ and m₂ are the MASSES of the body G is the universal gravitational constant r is the distance between the 2 masses Examples of gravitational FORCES are: Force that HOLD the moon towards the earth Force which makes a car to move downhill on a slope even without ACCELERATING Force which makes a BALL to fall downwards when dropped

39.

Can anyone give me examples of gravitational force

Answer»

┏_________∞◆∞_________┓

✭✮ӇЄƦЄ ƖƧ ƳƠƲƦ ƛƝƧƜЄƦ✮✭

┗_________∞◆∞_________┛

Gravitational force is the attractive force between 2 objects or bodies having mass

This force is given by the equation

F = Gm₁m₂/r²

where,

m₁ and m₂ are the MASSES of the body

G is the universal gravitational constant

r is the distance between the 2 masses

Examples of gravitational FORCES are:

Force that HOLD the moon towards the earth
Force which makes a car to move downhill on a slope even without ACCELERATING
Force which makes a BALL to fall downwards when dropped

Discussion

40.	If a car, starting from point P. goes to point (seefigure 1.2 b) and then returns to point P. how muchdistance has it travelled and what is its displacement
Answer» Answer: 1 irafathima8642 02.07.2018 Physics Secondary School +15 pts Answered If a car,starting from point P,goes to point Q and then returns to point P ,how much distance has it travelled and what is its DISPLACEMENT 1 SEE ANSWER Log in to add comment Answers Me · Beginner Know the answer? Add it here! amitnrw Genius Answer: no displacement Explanation: If a car,starting from point P,goes to point Q and then returns to point P ,how much distance has it travelled and what is its displacement distance travelled = TWICE the distance between TWO POINTS displacement = 0 No displacement as car return to original point and displacement is shortest distance between two points. here both points are same so no displacement

40.

If a car, starting from point P. goes to point (seefigure 1.2 b) and then returns to point P. how muchdistance has it travelled and what is its displacement

Answer»

Answer:

1

irafathima8642

02.07.2018

Physics

Secondary School

+15 pts

Answered

If a car,starting from point P,goes to point Q and then returns to point P ,how much distance has it travelled and what is its DISPLACEMENT

1

SEE ANSWER

Log in to add comment

Answers

Me · Beginner

Know the answer? Add it here!

amitnrw

Genius

Answer:

no displacement

Explanation:

If a car,starting from point P,goes to point Q and then returns to point P ,how much distance has it travelled and what is its displacement

distance travelled = TWICE the distance between TWO POINTS

displacement = 0

No displacement

as car return to original point and displacement is shortest distance between two points.

here both points are same so no displacement

Discussion

41.	A train starts from a station and stops in the next station in 4 min.The distance is 2 km between two stations. If the train goes the first part of the journey with a uniform acceleration a1 and the second part with a retardation a2 then show that 1/a1+1/a2=4
Answer» Answer: The distance between the station P and Q is 4KM and the train TAKES 4mins to complete JOURNEY, then 1/A1 + 1/A2 = ? Motion.

41.

A train starts from a station and stops in the next station in 4 min.The distance is 2 km between two stations. If the train goes the first part of the journey with a uniform acceleration a1 and the second part with a retardation a2 then show that 1/a1+1/a2=4

Answer»

Answer:

The distance between the station P and Q is 4KM and the train TAKES 4mins to complete JOURNEY, then 1/A1 + 1/A2 = ? Motion.

Discussion

42.	A cube of sides 5cm is immersed in water. What is the difference in pressure at the bottom and top faces of the cube?Given, the depth of the top face from the water surface is 10cm, density of water is 1000 kg/m³ and g=9.8m/s².
Answer» ANSWER: nothinkskdksomwmwkwkwkowoksnendbebbdbebebeb

42.

A cube of sides 5cm is immersed in water. What is the difference in pressure at the bottom and top faces of the cube?Given, the depth of the top face from the water surface is 10cm, density of water is 1000 kg/m³ and g=9.8m/s².

Answer»

ANSWER:

nothinkskdksomwmwkwkwkowoksnendbebbdbebebeb

Discussion

43.	Select the odd one out
Answer» ANSWER: Sit Resultant force grovoved tyres Hope it's helpful Need more HELP 25 ❣ = INBOX

43.

Select the odd one out

Answer»

ANSWER:

Sit
Resultant force
grovoved tyres

Hope it's helpful

Need more HELP 25 ❣ = INBOX

Discussion

44.	Prove that force of fiction depends on the action is two surfaces in contact
Answer» ANSWER: I am INTERESTED in the office Explanation: jjdjjd me PLEASE

44.

Prove that force of fiction depends on the action is two surfaces in contact

Answer»

ANSWER:

I am INTERESTED in the office

Explanation:

jjdjjd me PLEASE

Discussion

45.	An artificial satellite is moving uniformly in a circular orbit of radius 42250 km. Find it's speed if it takes 24hrs to revolve around the Earth.
Answer» Answer: 11,055.41 km/hr Explanation: GIVEN : Radius of the CIRCULAR orbit = 42,250 km Time taken to revolve AROUND the earth = 24 hrs To find : Speed it takes to revolve around the earth Speed = distance / time Distance = 2×3.14×r Distance = 6.28×42,250 Distance = 2,65,330 kilo metres The Circumference or the distance COVERED by the SATELLITE is equal to 2,65,330 kilometres Time taken = 24 hrs Speed = 265330/24 Speed = 11,055.41 km/hr The speed of the satellite is equal to 11,055.41 km/hr

45.

An artificial satellite is moving uniformly in a circular orbit of radius 42250 km. Find it's speed if it takes 24hrs to revolve around the Earth.

Answer»

Answer:

11,055.41 km/hr

Explanation:

GIVEN :

Radius of the CIRCULAR orbit = 42,250 km

Time taken to revolve AROUND the earth = 24 hrs

To find :

Speed it takes to revolve around the earth

Speed = distance / time

Distance = 2×3.14×r

Distance = 6.28×42,250

Distance = 2,65,330 kilo metres

The Circumference or the distance COVERED by the SATELLITE is equal to 2,65,330 kilometres

Time taken = 24 hrs

Speed = 265330/24

Speed = 11,055.41 km/hr

The speed of the satellite is equal to 11,055.41 km/hr

Discussion

46.	एक कार ट्रैवल 30 किलोमीटर ऐट यूनिफार्म स्पीड ऑफ 40 किलोमीटर पर आवर इंटरनेट 30 किलोमीटर ऐट यूनिफार्म स्पीड ऑफ 20 किलोमीटर पर आवर फाइंड इट्स एवरेज स्पीड
Answer» Explanation: for half DISTANCE or same distance and TWO DIFFERENT velocities the SHORT formula is $average \: speed = 2 \times v1 \times v2 \div v1 + v2$ v1=40 v2=20 $average \: speed = 2 \times 40 \times 20 \div 40 + 20 \\ average \: speed = 1600 \div 60 \\ average \: speed = 80 \div 3$ hope you like

46.

एक कार ट्रैवल 30 किलोमीटर ऐट यूनिफार्म स्पीड ऑफ 40 किलोमीटर पर आवर इंटरनेट 30 किलोमीटर ऐट यूनिफार्म स्पीड ऑफ 20 किलोमीटर पर आवर फाइंड इट्स एवरेज स्पीड

Answer»

Explanation:

for half DISTANCE or same distance and TWO DIFFERENT velocities

the SHORT formula is

$average \: speed = 2 \times v1 \times v2 \div v1 + v2$

v1=40

v2=20

$average \: speed = 2 \times 40 \times 20 \div 40 + 20 \\ average \: speed = 1600 \div 60 \\ average \: speed = 80 \div 3$

hope you like

Discussion

47.	Find net magnetic field at p l =10cm
Answer» the RADIUS of the magnetic field is 10 CM and the diameter = 20 cm therefore the net magnetic field =2×d =40cm ${\huge{\underbrace{\overbrace{\white{hope \: it \: helps \: you \: plz \: mark \: me \: brainliest \: .-}}}}}$

47.

Find net magnetic field at p l =10cm

Answer»

the RADIUS of the magnetic field is 10 CM

and the diameter = 20 cm

therefore the net magnetic field =2×d

=40cm

${\huge{\underbrace{\overbrace{\white{hope \: it \: helps \: you \: plz \: mark \: me \: brainliest \: .-}}}}}$

Discussion

48.	A man walks on a straight road from his home to market 2.5km away with a speed of 5km\h. finding the market closed, he instntly turns and backs home with a speed of 7.5km .the average speed of the man over the interval of time 0 to 40 min. is equal to
Answer» Explanation: Time taken by the MAN to reach the market from home,t1 = 2.5/5 = 1/2 h = 30 min Time taken by the man to reach home from the market, t2 = 2.5/7.5 = 1/3 h = 20 min Total time taken in the whole journey = 30 + 20 = 50 min (i) 0 to 30 min Average velocity = Displacement/Time = 2.5/(1/2) = 5 km/h Average speed = Distance/Time = 2.5/(1/2) = 5 km/h (II) 0 to 50 min Time = 50 min = 50/60 = 5/6 h Net displacement = 0 Total distance = 2.5 + 2.5 = 5 km Average velocity = Displacement / Time = 0 Average speed = Distance / Time = 5/(5/6) = 6 km/h (iii) 0 to 40 min Speed of the man = 7.5 km/h Distance travelled in first 30 min = 2.5 km Distance travelled by the man (from market to home) in the next 10 min = 7.5 × 10/60 = 1.25 km Net displacement = 2.5 – 1.25 = 1.25 km Total distance travelled = 2.5 + 1.25 = 3.75 km Average velocity = Displacement / Time = 1.25 / (40/60) = 1.875 km/h Average speed = Distance / Time = 3.75 / (40/60) = 5.625 km/h

48.

A man walks on a straight road from his home to market 2.5km away with a speed of 5km\h. finding the market closed, he instntly turns and backs home with a speed of 7.5km .the average speed of the man over the interval of time 0 to 40 min. is equal to

Answer»

Explanation:

Time taken by the MAN to reach the market from home,t1 = 2.5/5 = 1/2 h = 30 min

Time taken by the man to reach home from the market, t2 = 2.5/7.5 = 1/3 h = 20 min

Total time taken in the whole journey = 30 + 20 = 50 min

(i) 0 to 30 min

Average velocity = Displacement/Time = 2.5/(1/2) = 5 km/h

Average speed = Distance/Time = 2.5/(1/2) = 5 km/h

(II) 0 to 50 min

Time = 50 min = 50/60 = 5/6 h

Net displacement = 0

Total distance = 2.5 + 2.5 = 5 km

Average velocity = Displacement / Time = 0

Average speed = Distance / Time = 5/(5/6) = 6 km/h

(iii) 0 to 40 min

Speed of the man = 7.5 km/h

Distance travelled in first 30 min = 2.5 km

Distance travelled by the man (from market to home) in the next 10 min

= 7.5 × 10/60 = 1.25 km

Net displacement = 2.5 – 1.25 = 1.25 km

Total distance travelled = 2.5 + 1.25 = 3.75 km

Average velocity = Displacement / Time = 1.25 / (40/60) = 1.875 km/h

Average speed = Distance / Time = 3.75 / (40/60) = 5.625 km/h

Discussion

49.	Rahul swims in a 120 m long swimming pool. He covers 240 m in 1 minute by swimming from one end to the other and back along the same straight path. What is his average velocity?
Answer» Rahul swims in a 120 m long pool. (Length of pool = 120 m). He covers 240m in 1 min (60 SEC) by swimming from one end to the other and BACK along the same straight path. We have to find the average velocity of Rahul. Average SPEED is defined as the total DISTANCE covered with respect to total time taken. Whereas average velocity is defined as the ratio of total displacement with respect to the total time taken. As given in QUESTION that, the Rahul covers a distance of 240 m in 60 sec. (Total distance covered is 240 m and time taken is 60 sec). Average speed = 240/60 = 4 m/s Therefore, the average speed of the Rahul is 4 m/s. Since, he is swimming from one end to the other and back along the same straight path. So, the total displacement of the Rahul is zero. Therefore, the average velocity of the Rahul is 0 m/s.

49.

Rahul swims in a 120 m long swimming pool. He covers 240 m in 1 minute by swimming from one end to the other and back along the same straight path. What is his average velocity?

Answer»

Rahul swims in a 120 m long pool. (Length of pool = 120 m). He covers 240m in 1 min (60 SEC) by swimming from one end to the other and BACK along the same straight path.

We have to find the average velocity of Rahul.

Average SPEED is defined as the total DISTANCE covered with respect to total time taken. Whereas average velocity is defined as the ratio of total displacement with respect to the total time taken.

As given in QUESTION that, the Rahul covers a distance of 240 m in 60 sec.

(Total distance covered is 240 m and time taken is 60 sec).

Average speed = 240/60 = 4 m/s

Therefore, the average speed of the Rahul is 4 m/s.

Since, he is swimming from one end to the other and back along the same straight path. So, the total displacement of the Rahul is zero.

Therefore, the average velocity of the Rahul is 0 m/s.

Discussion

50.	When have you lost your pen? spot error
Answer» ANSWER: I THINK , you have STOLE it

50.

When have you lost your pen? spot error

Answer»

ANSWER:

I THINK , you have STOLE it

Discussion

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