a) The speed of the drums is increased from 2 ft/s to 4 ft/s in 4 s. From the below kinematic equations the acceleration of the drums can be determined.

$v_1 = v_0 + at \\4 = 2 + 4a\\a = 0.5~ft/s^2$

This is the linear acceleration of the drums. Since the tape does not slip on the drums, by the rule of rolling without slipping,

$v = \omega R\\a = \alpha R$

where α is the angular acceleration.

In order to continue this question, the radius of the drums should be given.

Let us denote the radius of the drums as R, the angular acceleration of drum B is

α = 0.5/R.

b) The distance travelled by the drums can be found by the following kinematics formula:

$v_1^2 = v_0^2 + 2ax\\4^2 = 2^2 + 2(0.5)x\\x = 12 ft$

One revolution is equal to the circumference of the drum. So, total number of revolutions is

$x / (2\pi R) = 6/(\pi R)$

Question 2)

a) In a rocket propulsion question, the acceleration of the rocket can be found by the following formula:

$a = \frac{dv}{dt} = -\frac{v_{fuel}}{m}\frac{dm}{dt} = -\frac{13000}{2600}25 = 125~ft/s^2$

b) $a = -\frac{v_{fuel}}{m}\frac{dm}{dt} = - \frac{13000}{400}25 = 812.5~ft/s^2$

5 0

3 years ago

Kelli weighs 440 N, and she is sitting on a playground swing that hangs 0.41 m above the ground. Her mom pulls the swing back an

kifflom [539]

Answer:

Explanation:

Initial height from the ground = .41 m

Final height = 1m

Height by which Kelli was raised ( h )= .59 m

When she passes through the lowest point , she loses P E

= mgh

= 440 x .59

= 259.6 J

kinetic energy possessed by her

= 1/2 mv²

= .5 x (440/9.8) x 2²

= 89.8 J

Difference of energy is lost due to work by air friction

work done by friction = 89.8 - 259.6

= - 169.8 J

4 0

3 years ago