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not a satisfactory ans

y = 6 sin10 t + 8 cos10 t

Let 6 = R sin theta8 = R cos theta

y = R sin theta.sin 10 t + R cos theta. cos10 t

y = R sin(10 t + theta)

Where R is the amplitude of wave and theta is initial phase.

R^2 = (6)^2 + (8)^2 R^2 = 36 + 64R = sqrt(100)R = 10

Buoyant force = volume × density × g

= 1.6× 1030× 10= 17992.8=16480 N

Inside the lift the frame of reference (attached to the floor or celing)is moving up with an acceleration of g/2 upwards. So for applying Newton's laws, we need to add a pseudo force of mass*g/2 in the downward direction for each mass. In otherwords, we can say the effective g' = g + g/2 = 3g/2 downwards.

Since M1, the4 kilo mass is heavier, it will accelerate downwards with in the lift. M2, the 2kg mass will move upwards with the same acceleration as that of M1. We assume that the string is tight and so it has the same tension Tthrough the string. equations of motions for M1 and M2 are : M1 g' - T = M1 a => a = ( g'- T/M1) M2 a = T - M2 g' => T = M2 (a+g') = M2 (2g' - T/M1) T = 2 M2 g' - TM2 / M1 => T= 2 M1 M2 g'/ (M1+M2) T = 3 M1 M2 g / (M1+M2) = 3 *2 * 4 * 10 / (2+4) = 40 Newtons

option 1 is the answer

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According to second law of motion

F=ma mass =1kg force =1Nif one Newton of force applied on the body of mass 1kg and it moves with an acceleration of 1 m/s^2.

The body receives 1 m/s^2.

Standard unit of measurements provide a general reference point,

The concept here is that Area under acceleration time graph gives you the change in velocity.

From, 0 to 6s, Acceleration is positive. So, change in velocity = Area of upper triangle = 1/2 X Base X height= 1/2 X 6 X 10 = + 30 m/s

+ sign indicates that velocity is increasing.

From 6 to 8s, acceleration is negative. So, change in velocity= Area of lower triangle= 1/2 X Base X height= 1/2 X 2 X -10 = -10 m/s

- sign indicates that velocity is decreasing.

Total Area under Graph= Upper Triangle Area + Lower Triangle Area.= 30 - 10= 20 m/s

So, final velocity= Initial velocity + Total Area under Graph= -5 + 20= 15 m/s (Ans)

uniform motion is the motion in which speed is constant..

so, for speed to be constant.. the acceleration should be zero..

and the path is from A to B ..

so, option A.

1. AND logic gate performs multiplication. NOT logic gate performs inversion operation.

tuje dikta nhi hai Kya jab bol diya wrong answer

wrong answer

wrong answer Pehle bhi bola tha na

modules of velocity = √4+25= √29now kinetic energy = 1/2mv*v1/2*120*2917401.7Kgm^2/s^2

260 hai please like kaar

considering the motion to be projectile

maximum height is u²sin²∅/2g

so, if maximum height is same for all , and if ∅ is increased , the velocity should be reduced , to maintain the same height

so, 4 has less angle.. and 1 has maximum angle

=> velocity order is 4>3>2>1

option d

i think 2nd option is correct

plz inform me whether i am right or not

yes u r right

minimum force needed to overcome friction = k(m1+m2)g

now that force = F = at

=> at = k(m1+m2)g=> t = k(m1+m2)g/a

2+2+2 {series R} =66 ll 26*2/6+2=1.52+1.5+2=5.55.5*2/5.5+2=1.4662+1.466+2=5.466

Decreasing from 16 g to 2g is a decrease by a factor of 8 = 2^3 This requires three half-lives.

1

the correct answer is option (d) because nucleolus is present in nucleus which is only present in eukaryotes

thank you

Using Kirchoff'sVoltage Law the following eqn results220 = 20x1.8 + 15x20x0.1 + 15xR (where R is the external resistance required)or 15R = 220 - 36 -30.or 15R = 154or R = 10.3 ohms.

Work done = Q(potential difference between the two points).W = 3(125-115)W = 30 J

Maximum potential energy is obtained with the brick vertical along side a=mga/2

Minimum potential energy is obtained with the brick vertical along side c=mgc/2

:. Work done=mg(a/2-c/2)

Hence, option (B) is correct.

(i) For minimum work, the angle should be 90°. (ii) For maximum work, the angle should be zero.

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Aprojectileis an object that is given an initial velocity, and is acted on by gravity. The amount oftimeit spends in the air is called thetime of flight. The unit for thetime of flightis seconds (s).