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3 years ago

On a distance vs. time graph, how do you know when the object is moving away from its starting position?

Physics

2 answers:

Ainat [17]3 years ago

8 0

U know by if they are in first place

den301095 [7]3 years ago

8 0

Answer:

u look at em .<.

Explanation:

lol sorry im just bored

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Xavier is roller skating at 14 km/h and tosses a set of keys forward on the ground at 8 km/h. The speed of the keys relative to

JulijaS [17]

Answer:

22km/h

Explanation:

The speed of the keys relative to the ground will be given by the sum of speed of keys with respect to the ground and the speed of keys with respect to the skater

Given that the speed of keys with respect to the ground is 8 m/s while the speed of keys with respect to skater is 14 km/h then the relative speed which combines both speeds will be 8+14=22 km/hr

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4 years ago

Question 3 (1 point) The nucleus occupies most of the space of an atom True O False

Marizza181 [45]

Answer:

the correct answer is false

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4 years ago

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: