Motion in A Plane Test - 12

A constant vector in uniform circular motion the magnitude of centripetal acceleration remain constant but its direction always get changed due to change in position of object.

So centripetal acceleration is not a constant vector.

(i) The work done on a body moving in a circular path is also zero. This is because, when a body moves in a circular path, then the centripetal force acts along the radius of the circle, and it is at right angles to the motion of the body.

(ii) Thus, the work done in the case of moon moving round the earth is also zero.

Electrostatic force provides necessary centripetal force for circular motion of electron.

In a uniform circular motion, the acceleration is directed towards the centre while velocity is acting tangentially.

Using,

\(v = r \omega \) = 0.5 x 70 = 35 m/s

Given,

Radius of circle, r = 20 cm = 0.2m

Angular velocity, \(\omega\) = 10 rad/s

Linear velocity, v = r x \(\omega\)

= 0.2 x 10

= 2 m/s

(i) A tachometer (revolution-counter, tach, rev-counter, RPM gauge) is an instrument measuring the rotation speed of a shaft or disk, as in a motor or other machine.

(ii) The device usually displays the revolutions per minute (RPM) on a calibrated analogue dial, but digital displays are increasingly common.

Centripetal force = F = ma = \(\frac{mv^2}{r}\)

Hence if radius is doubled the force will reduce to half when r is replaced by 2r.

The velocity with which a bullet or shell leaves the muzzle of a gun.

We  know that in projection of the particle for maximum range, \(\theta\) = 45°

Now maximum range

\(R = \frac{\mu^2 sin 2 \theta}{g}\)

\(R = \frac{\mu^2 sin 90^o}{g}\) (sin 90° = 1)

µ = \(\sqrt {Rg}\)     .......(1)

Given:

\(R_{max} = 16 km = 16 \times 10^3 m\)

g = \(10 m/s^2\)

Now from equation (1)

µ = \(\sqrt{16 \times 10^3 \times 10} \)

= 400 m/s

In projectile motion vertical component of velocity changes, but horizontal component of velocity remains always constant.

As the multiple of \(\hat j\) in the given vector is zero therefore this vector lies in XZ plane and projection of this vector on y-axis is zero.

\(\vec{A} = 2 \hat i + 4 \hat j - 5 \hat k\)

\(|\vec A| = \sqrt{2^2 + 4^2 + (-5)^2}\) = \(\sqrt{45}\)

\(cos \alpha = \frac{2}{\sqrt{45}}\)

\(cos \beta = \frac{4}{\sqrt{45}}\)

\(cos \gamma = \frac{-5}{\sqrt{45}}\)

Diagonal of the hall = \(\sqrt{l^2 + b^2 + h^2}\)

= \(\sqrt{10^2 + 12^2 + 14^2}\)

= \(\sqrt{100 + 144 + 196}\)

= \(\sqrt {400}\) = 20 m

If the angle between all forces which are equal and lying in one plane are equal then resultant force will be zero.

Magnitude of unit vector = 1

\(\sqrt{(0.5)^2 + (0.8)^2 + c^2} = 1\)

c = \(\sqrt{0.11}\)

\( \vec F_1 + \vec F_2+\vec F_3 =0\)

There should be minimum three coplanar vectors having different magnitude which should be added to give zero resultant.

The cross product vector A x B it a vector, with its direction perpendicular to both vector A and B. A x B is area. If side B is zero, area is zero. vector A x 0 is a zero vector.

If in case 0 is a scalar, then also the product is zero. But a scalar x a vector is also a vector.

Hence one gets a zero vector in any case.

When velocity of projection of a body is made n times, and its time of flight becomes n times and range becomes  \(n^2 \) times.
so Assertion is correct but Reason is incorrect

Energy (E) = F x d

F = \(\frac{E}{d}\)

so Erg/metre can be the unit of force.