All categories

3 years ago

V2 = v2. + 2 as, dimensional expression

Physics

2 answers:

laila [671]3 years ago

6 0

Answer:

Yirmigfhkjftiyjff6ftffyi

Monica [59]3 years ago

5 0

Answer:

This is the third equation of motion.

${ \rm{ {v}^{2} = {u}^{2} + 2as }}$

• Dimensions of velocity

${ \rm{velocity = \frac{distance}{time} }} \\ \\ { \rm{[velocity] = \frac{L}{T} }} \\ \\ { \rm{{[velocity] =LT {}^{ - 1} }}}$

• Dimensions of acceleration

${ \rm{[acceleration] = LT {}^{ - 2} }}$

Therefore, Answer is;

${ \rm{LT {}^{ - 1} = LT {}^{ - 1} + (2 \times LT {}^{ - 2} \times L) }} \\ \\ { \rm{LT {}^{ - 1} = 2LT {}^{ - 2}(T + L)}} \\ \\ { \boxed{ \rm{ \: \frac{T}{2} = L }}}$

You might be interested in

What happens when the objects submerged in a fluid at rest?

Talja [164]

It will act upon a buoyant force on the magnitude of which is equal to weight of the fluid

3 0

3 years ago

During a game the same batter swings at a ball thrown by the pitcher and hits a line drive. Just before the ball is hit it is tr

sergeinik [125]

Answer:

The total work on the ball is 36.25 Joules

Explanation:

There is an important principle on classical mechanics that is the work-energy principle it states that the total work on an object is equal the change on its kinetic energy, mathematically expressed as:

$W_{net}=\Delta K = K_f -K_i$ (1)

With W net the total work, Kf the final kinetic energy and Ki the initial kinetic energy. We're going to use this principle to calculate the total work on the baseball by the force exerted by the bat.

Kinetic energy is the energy related with the movement of an object and every classical object with velocity has some kinetic energy, it is defined as:

$K=\frac{mv^2}{2}$

With m the mass of the object and v its velocity, knowing this we can use on:

$W_{net}= \frac{mv_f^2 -mv_i^2}{2}=\frac{m(v_f^2 -v_i^2)}{2}$

In our case vf is the velocity just after the hit and vi the velocity just before the hit. For an average baseball its mass is 145g that is 0.145 kg, then

$W_{net}=\frac{0.145*(30.0^2 -20.0^2)}{2}$

$W_{net}=36.25 J$

8 0

3 years ago

Consider the model above. It represents the electrical force. As r increases, the attractive force decreases. How would this mod

aivan3 [116]

Answer:

As we keep on increasing the radius the value of the gravitation force of attraction decreases and as we decrease the radius the gravitation force increases.

Explanation:

Like the coulombs law of electrostatics, the law of gravitation also depends inversely on the square of the value of r. Therefore, as we keep on increasing the value of r the value of the gravitation force decreases and as we decrease the value of the r the value of gravitation force increases.

Gravitation Force= $\frac{Gm_{1}m_{2} }{r^{2}}$