Explore topic-wise InterviewSolutions in .

thank you so much

The question is incomplete...please complete the question first..

1. Constant2. non uniform3. Uniform

the first option is the ans

uniform motion in first casenon uniform in second as well as in third case

uniform motion non uniform motionnon uniform motion

1 is the correct answer of this question

frequencyh/I = ML²T⁻¹ / ML²

=T⁻¹

=Frequency

Sincekinetic energyis based on motion, it is always a positive value. If it is not in motion, thekinetic energyof that object is zero.

Kinetic energy cannever be anegativevalue.

Kinetic energy canbe quantified as one half of the mass times the velocity squared (KE =1/2*m*v²).

Brewster's law states that:

If the light is incident on the surface of a transparent medium with an angle of incidence given by,

μ = tan ip,

μ is the refractive index of the medium.

The reflected light is completely polarised along one direction. This is known asBrewster’s Law. The refracted light is partly polarised.

f== 20cm

u = -30cm

v = ?

using the lens formula : 1/f = 1/v - 1/u

1/20 = 1/v - 1/-30

1/v = 1/20 - 1/30

1/v = (30 - 20)/600

1/v = 10/600

v = 60cm

therefore the image is formed at 60cm from the lens.

Magnification = v/u

= 60/-30

= -2 D

hence, the image formed is real and inverted.

On September 20,1932 Mahatma Gandhi began a fast unto death in Yervada Jail against Communal award of Ramsay MacDonald

please say in detail

On September 16th of 1932, Mahatma Gandhi began a “fast unto death” to protest British support of a new Indian Constitution that would separate the Indian electorate by caste, thereby segregating the “untouchables” & ensuring that the social classes would remain unfairly divided.

After just six days, the British accepted an alternate proposal & he was able to break the fast.

However, Gandhi’s non-violent & steadfast commitment to social justice continued until “The Great Soul” was assassinated in 1948.

thanks a lot

2734.2=distanceDisplacment=0

During sunriseand sunset, the rays have to travel a larger part of the atmosphere because they are very close to the horizon. Therefore, light other thanredis mostly scattered away. Most of theredlight, which is the least scattered, enters our eyes. Hence, thesunand the skyappear red.

thankyou

S=125=UT+1/2AT^2as U is =0therefore S=1/2AT^2here A is 10m/s^2therefore 125=1/2AT^2250=10T^2therefore T =√25=5sec

V^2-U^2=2AS

here A is -10 and S is -125m as its falling from a height and U =0 therefore

V^2=2(-10)(-125)V=√2500V=50m/s

half time by it reaches =5/2=2.5secS=UT+1/2AT^2S=1/2x10xT^2therefore S=31.25m

but H=125-31.25therefore H=93.75m

a)The time by which the ball reaches the ground is=5secondsb) The velocity of the ball on reaching the ground=50m/sc) The height of the ball at half time it takes to reach the ground=93.75m

m = 0.2 kg

h = 45m

v^2 = u^2 +2as

v^2 = 2gs

v^2 = 2*10*45

v^2=900m/s

v= 30m/s

rebound velocity =2/3v

20m/s

since it is opp direction speed =-20m/s

change in momentum = m(v-u)

=0.2(30+20)

=10Nsec

average force = rate of change in momentum

=10/0.1

100N

v = u + gt ( acceleration = acc. due to gravity ) v = 0 + 10 x 1v = 10 m / s the ball will reach the ground at a speed of10 m/s .

acc. to second equation v^2 = u^2 + 2gs100 = 0 + 2 x 10 x s100 /20 = s s = 5 m.

The wall is ¼ of the range away.The maximum height is ½ of the range away.So it takes as long to get from the throw point to the wall as it takes to get from the wall to maximum height.If the total flight time is T, then the vertical velocity Vv isVv at t = 0,½*Vv at t = T/4,0 at t = T/2,-½Vv at t = 3T/4, and-Vv at t = T.Then the average velocity during the first interval is 3Vv/4, and the average velocity during the second interval is Vv/4.Then the vertical distance traveled during the first interval is 3 times the vertical distance during the second interval.Since the vertical distance traveled during the first interval was 3 m, then the vertical distance traveled during the second interval was 1 m, and the maximum height is 4 m.

range = (V²sin(2Θ))/g → 24m = V²sin(2Θ) / gmax height = (V·sinΘ)² / (2g) → 4m = V²sin²Θ / 2gDivide range by max height:24m / 4m = 6 = 2sin(2Θ) / sin²ΘUse trig identity sin(2Θ) = 2sinΘcosΘ:6 = 4sinΘcosΘ / sin²Θ = 4cosΘ / sinΘ = 4 / tanΘtanΘ = 4/6 = 2/3Θ = tan^-1(2/3) = assuming the ball just clears the wall.

1. The diameter of a circular track is200 m200 mand one round is completed in40 s40 s.

Circumference of a circular track is given as,

C=2πrC=2πr.

Where, the radius of circular track isrrand the circumference isCC.

By substituting the given value in the above equation, we get

C=2π×100= 200π mC=2π×100= 200π m

So, the athlete covers a distance of200π m200π min40 s40 s.

Total distance covered in2 min 20s2 min 20sor140 s140 sis,

s=200π40×140≈2200 ms=200π40×140≈2200 m

Since, every40 s40 sathlete covers one circular track therefore, athlete will take 3 round in120 s120 s. In the next20 s20 ssecond, he will cover half of the circumference. So, his displacement will be the diameter of the circular track that is200 m200 m.

Thus, the distance covered in the given time is2200 m2200 mand the displacement is200 m200 m.

Option (4) is correct.

Ans: Since the slope of the curve is changing continuously, thus the body is accelerating.

The maximum speed of the particle corresponds to the area of the a→t graph.

Maximum speed = Area of the graph

= 1/2 ×10 × 11

= 55 m/s

4 volts is the answer

potential difference between 24ohm is 4volts.

4/3

bhai koi bhi to answer do

2√7 the correct answer of the given question