The answer plz quickkkk

The tub of a washer goes into its spin cycle, starting from rest and gaining angular speed steadily for 7.00 s, at which time it

Schach [20]

Answer:

The tub of the washer does 37.5 revolutions while it is in motion.

Explanation:

At first we assume that the tub accelerates and decelerates uniformly. In this case, it is required to determine the number of revolutions done by the tub by means of the following equations of motion:

Acceleration

$\ddot n_{1} = \frac{\dot n_{1}-\dot n_{1,o}}{t_{1}}$ (Eq. 1)

Where:

$\ddot n_{1}$ - Angular acceleration experimented by the tub, measured in revolutions per square second.

$\dot n_{1,o}$ , $\dot n_{1}$ - Initial and final angular velocities of the tub, measured in revolutions per second.

$t_{1}$ - Acceleration time, measured in seconds.

$\Delta n_{1} = \dot n_{1,o}\cdot t_{1}+\frac{1}{2}\cdot \ddot n_{1}\cdot t_{1}^{2}$ (Eq. 2)

Where $\Delta n_{1}$ is the change in angular position of the tub during acceleration, measured in revolutions.

Deceleration

$\ddot n_{2} = \frac{\dot n_{2}-\dot n_{1}}{t_{2}}$ (Eq. 3)

Where:

$\ddot n_{2}$ - Angular acceleration experimented by the tub, measured in revolutions per square second.

$\dot n_{1}$ , $\dot n_{2}$ - Initial and final angular velocities of the tub, measured in revolutions per second.

$t_{2}$ - Deceleration time, measured in seconds.

$\Delta n_{2} = \dot n_{1}\cdot t_{2}+\frac{1}{2}\cdot \ddot n_{2}\cdot t_{2}^{2}$ (Eq. 4)

If we know that $\dot n_{1,o} =0\,\frac{rev}{s}$ , $\dot n_{1} = 5\,\frac{rev}{s}$ , $t_{1} = 7\,s$ , $\dot n_{2} = 0\,\frac{rev}{s}$ and $t_{2} = 8\,s$ , then the amount of revolutions done by the tub during each phase is, respectively:

(Eq. 1)

$\ddot n_{1} = \frac{5\,\frac{rev}{s}-0\,\frac{rev}{s} }{7\,s}$

$\ddot n_{1} = \frac{5}{7}\,\frac{rev}{s^{2}}$

(Eq. 3)

$\ddot n_{2} = \frac{0\,\frac{rev}{s}-5\,\frac{rev}{s} }{8\,s}$

$\ddot n_{2} = -\frac{5}{8}\,\frac{rev}{s^{2}}$

(Eq. 2)

$\Delta n_{1} = \left(0\,\frac{rev}{s} \right)\cdot (7\,s)+\frac{1}{2}\cdot \left(\frac{5}{7}\,\frac{rev}{s^{2}}\right) \cdot (7\,s)^{2}$

$\Delta n_{1} = \frac{35}{2}\,rev$

(Eq. 4)

$\Delta n_{2} = \left(5\,\frac{rev}{s} \right)\cdot (8\,s)+\frac{1}{2}\cdot \left(-\frac{5}{8}\,\frac{rev}{s^{2}}\right) \cdot (8\,s)^{2}$

$\Delta n_{2} = 20\,rev$

The total amount of revolutions done by the tub while it is in motion is:

$\Delta n_{T} = \Delta n_{1}+\Delta n_{2}$

$\Delta n_{T} = 37.5\,rev$

The tub of the washer does 37.5 revolutions while it is in motion.

4 0

3 years ago

a ball of mass 0.5 kg moving at 10 m/s collides with another ball of equal mass at rest. if the two balls move off together afte

nexus9112 [7]

Answer:

5 m/s

Explanation:

Here we can see there is no external force acted on a two masses when we consider the motion. If there is no external forces then momentum is conserved.

Initial momentum = Final momentum

0.5 × 10 = 1 × V

V = 5 m/s

7 0

4 years ago

True or false. in science, a hypothesis must be true?

pishuonlain [190]

No.A hypothesis is an educated guess but it will not always be right

5 0

3 years ago

A graph titled Position versus time for with horizontal axis time (seconds) and vertical axis position (meters). A straight blue

kondor19780726 [428]

Answer:

A, D, E

Explanation:

I answered it on Edgenuity.

4 0

4 years ago

Read 2 more answers

Can you pleas help me!

Sergeeva-Olga [200]

Answer:

T

Explanation:

4 0

4 years ago