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slava [35]
3 years ago
8

.A particle starts on the origin. It is pushed back to -5.7 m in 2.1 s. Then it is pushed

Physics
2 answers:
harina [27]3 years ago
5 0

Answer: The average velocity is -0.965m/s

Explanation: The first step is to calculate the two velocities is both directions. A velocity is a distance per unit time.

V=d/ t

=-5.7/2.1

=-2.7m/s

For the other direction the velocity is

V=7.3/9.5

=0.77m/s

The average velocity the add the velocities and divide them by 2.

V=-2.7+0.77/2

V= 0.965m/s

GREYUIT [131]3 years ago
3 0

Answer:

The average velocity of the particle is 1.76 m/s.

Explanation:

Given that,

Distance = -5.7 m

Time t = 2.1 s

Forward distance = 7.3 m

Total time = 9.5 sec

We need to calculate the average velocity of the particle

Average velocity :

Average velocity is equal to the displacement divided by change in time.

v = \dfrac{\Delta D}{\Delta t}

Where, D = displacement

t = change in time

v=\dfrac{7.3-(-5.7)}{9.5-2.1}

v=1.76\ m/s

Hence, The average velocity of the particle is 1.76 m/s.

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You have a summer job at a company that developed systems to safely lower large loads down ramps. Your team is investigating a m
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Answer:

Note that the emf induced is

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v = [R m g sin (A) / B^2 d^2 cos^2 (A)]   [ANSWER, the constant speed, PART A]

******************************

v = [R m g sin (A) / B^2 d^2 cos^2 (A)]

Plugging in the units,

m/s = [ [ohm * kg * m/s^2] / [T^2 m^2] ]

Note that T = kg / (s * C), and ohm = J * s/C^2

Thus,

m/s = [ [J * s/C^2 * kg * m/s^2] / [(kg / (s * C))^2 m^2] ]

= [ [J * s/C^2 * kg * m/s^2] / [(kg^2 m^2) / (s^2 C^2)]

As J = kg*m^2/s^2, cancelling C^2,,

= [ [kg*m^2/s^2 * s * kg * m/s^2] / [(kg^2 m^2) / (s^2)]

Cancelling kg^2,

= [ [m^2/s^2 * s * m/s^2] / [(m^2) / (s^2)]

Cancelling m^2/s^2,

= [s * m/s^2]

Cancelling s,

=m/s   [DONE! WE SHOWED THE UNITS ARE CORRECT! ]

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3 years ago
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Answer:

18 N/C

Explanation:

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