1.Two stones are dropped from the top of a tower at an interval of 3 seconds.the relative velocity of the first Stone with respect to the stone at any instant during their fall is??

2.if density, acceleration and force are taken as basic quantities, then time period has dimension??
3.two forces of equal magnitude act as a point. The square of their resultant is three times their product.the angle between them is??
4.a point P of cylinder of radius R is in contact with a horizontal surface. The displacement of point P if the cylinder rolls formed by half resolution is??

Assalamuilikum to start off helping you list what you think you have to do for each question.

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1. The relative velocity between the two stones can be calculated by subtracting the velocity of one stone from the velocity of the other stone at any given instant.

Let's assume that the first stone is dropped from rest, so its velocity at any time t is v1 = gt, where g is the acceleration due to gravity.

The second stone is dropped 3 seconds later, so its velocity at any time t is v2 = g(t - 3).

The relative velocity of the first stone with respect to the second stone at any instant during their fall is given by:
relative velocity = v1 - v2 = gt - g(t - 3) = gt - gt + 3g = 3g

Therefore, the relative velocity of the first stone with respect to the second stone is 3g.

2. If density, acceleration, and force are taken as basic quantities, then time period has dimensions of [T].

A time period is defined as the time it takes for a complete cycle or oscillation of an event or motion to occur. It is usually represented by the symbol T.

Since density, acceleration, and force are taken as basic quantities, none of them have a dimension of time [T]. Therefore, the time period cannot be expressed in terms of these basic quantities. Instead, time period is considered an independent quantity and is represented by the dimension [T].

3. Let's assume the magnitude of each force is F.

The square of the resultant of the two forces is three times their product, so (resultant)^2 = 3(F * F).

Mathematically, (resultant)^2 = 3F^2.

By taking the square root of both sides, we get: resultant = sqrt(3F^2) = sqrt(3)F.

Now, let's assume the angle between the two forces is θ. The resultant of the two forces can be calculated using the law of cosines:

(resultant)^2 = (force1)^2 + (force2)^2 - 2(force1)(force2)cosθ.

Substituting the values, we have: (sqrt(3)F)^2 = F^2 + F^2 - 2(F)(F)cosθ.

Simplifying, we get: 3F^2 = 2F^2 - 2F^2cosθ.

Cancelling out the F^2 terms, we have: 3 = 2 - 2cosθ.

Rearranging, we get: 2cosθ = -1.

Dividing by 2, we have: cosθ = -1/2.

The angle θ where cosθ = -1/2 is 120 degrees or 2π/3 radians.

Therefore, the angle between the two forces is 120 degrees or 2π/3 radians.

4. The displacement of point P, assuming the cylinder rolls without slipping on a horizontal surface, can be calculated using the formula:

displacement = (angle of rotation in radians) * (radius of the cylinder).

Since the cylinder rolls formed by half resolution, we need to consider that it has made half a revolution.

Therefore, the angle of rotation in radians would be half of a complete revolution, which is π radians.

Substituting the angle of rotation (π radians) and the radius of the cylinder (R) into the displacement formula, we get:

displacement = π * R.

So, the displacement of point P would be π times the radius of the cylinder.

1. To find the relative velocity of the first stone with respect to the second stone at any instant during their fall, we need to consider their individual velocities and the direction in which they are falling.

Let's assume the velocity of the first stone is v1 and the velocity of the second stone is v2. Since they are dropped from the same height at an interval of 3 seconds, both stones experience the same downward acceleration due to gravity (which we'll call g).

For any instant during their fall, the relative velocity of the first stone with respect to the second stone is given by the formula:

Relative velocity = v1 - v2

Now, to calculate v1 and v2, we use the kinematic equation:

v = u + at

where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.

From the given information, we know that the time interval between the two stones is 3 seconds. Therefore, at any instant, the time for the second stone would be t - 3 (where t is the time in seconds since the first stone was dropped).

Using the kinematic equation for both stones:

For the first stone:
v1 = 0 + g*t

For the second stone:
v2 = 0 + g*(t - 3)

Plugging these values into the relative velocity equation:
Relative velocity = v1 - v2
= g*t - g*(t - 3)
= 3g

Therefore, the relative velocity of the first stone with respect to the second stone at any instant during their fall is 3 times the acceleration due to gravity (3g).

2. If density, acceleration, and force are taken as basic quantities, then the time period would have the dimension of time (T).

Time period refers to the time it takes for a complete cycle or oscillation to occur in a periodic motion. It is typically represented by the letter T and is measured in seconds.

Since density (mass/volume), acceleration (distance/time^2), and force (mass x acceleration) are the basic quantities, they will not have any direct relation to the dimension of time. Time period is independent of these quantities and is solely related to the concept of time.

3. To find the angle between two forces of equal magnitude whose resultant is three times their product, we can use the trigonometric relation between the magnitude of the resultant and the individual forces.

Let's assume the magnitude of each force is F. According to the problem, the square of their resultant is three times their product, which can be written as:

(Resultant)^2 = 3(F * F)
(Resultant)^2 = 3F^2

Taking the square root of both sides:
Resultant = √(3F^2)
Resultant = F√3

Now, we can use the formula for the magnitude of the resultant of two forces acting at an angle θ:

(Resultant)^2 = (Force1)^2 + (Force2)^2 + 2(Force1)(Force2)cos(θ)

Substituting the given values:
(F√3)^2 = F^2 + F^2 + 2(F)(F)cos(θ)
3F^2 = 2F^2 + 2F^2cos(θ)

Simplifying further:
3 = 4 + 2cos(θ)

Rearranging the equation:
cos(θ) = (3 - 4) / 2
cos(θ) = -1/2

The angle θ whose cosine is -1/2 can be evaluated using trigonometric ratios. We find that θ is 120 degrees or 2π/3 radians.

Therefore, the angle between the two forces is either 120 degrees or 2π/3 radians.

4. The displacement of point P when a half resolution is performed by a cylinder in contact with a horizontal surface can be calculated by considering the properties of rolling motion.

When a cylinder rolls without slipping, the point of contact between the cylinder and the surface has zero velocity. This means that the displacement of point P is equal to the distance it would have traveled if the cylinder had translated (moved without rotation) instead of rolling.

The distance traveled by a point on a rotating object can be calculated by multiplying the radius of the object by the angle through which it rotates. In this case, the radius of the cylinder is R.

Since the cylinder has performed a half resolution, it means that it has rotated by π radians (180 degrees).

Therefore, the displacement of point P would be:
Displacement = R x π
Displacement = πR

So, the displacement of point P when the cylinder rolls with half resolution is πR.