Because the mechanical advantage of the machine is affected by friction and weight but velocity ratio is not. So, mechanical advantage is less than velocity rate. Thus, the machine's efficiency is less than 100% and can't be a perfect machine

3 0

3 years ago

A 12-kg dog jumps up in the air to catch a ball. The dog's center of mass is normally 0.20 m above the ground, and he is 0.50 m

ad-work [718]

Explanation:

Given Data:

mass of dog = 12 Kg

dog's center of mass = 0.20m

length of dog = 0.50m

height of dog's jump = ?

Solution:

Work done of gravitational force = Gain in Potential energy

2.1 × mgΔh = mg (h - 0.1)

2.1 × (0.3 - 0.1) = (h - 0.1)

h = 0.52 m

5 0

3 years ago

The equation Bold r (t )equals(8 t plus 9 )Bold i plus (2 t squared minus 8 )Bold j plus (6 t )Bold k is the position of a parti

KATRIN_1 [288]

Explanation:

It is given that, the position of a particle as as function of time t is given by :

$r(t)=(8t+9)i+(2t^2-8)j+6tk$

Let v is the velocity of the particle. Velocity of an object is given by :

$v=\dfrac{dr(t)}{dt}$

$v=\dfrac{d[(8t+9)i+(2t^2-8)j+6tk]}{dt}$

$v=(8i+4tj+6k)\ m/s$

So, the above equation is the velocity vector.

Let a is the acceleration of the particle. Acceleration of an object is given by :

$a=\dfrac{dv(t)}{dt}$

$a=\dfrac{d[8i+4tj+6k]}{dt}$

$a=(4j)\ m/s^2$

At t = 0, $v=(8i+0+6k)\ m/s$

$v(t)=\sqrt{8^2+6^2} =10\ m/s$

Hence, this is the required solution.

7 0

3 years ago

Does changing the height of point C affect the speed of the coaster car at point D?​

Does changing the height of point C affect the speed of the coaster car at point D?