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4 years ago

Describe the motionof an object that gas acceleration of 0 m/s

Physics

1 answer:

Maru [420]4 years ago

3 0

If an object has acceleration of zero meters per second², then the object is traveling at a constant speed.

Note: The constant speed doesn't have to be zero, so the object is not necessarily at rest. It can be moving at any speed, as long as the speed doesn't change. (If the speed changed, then the acceleration wouldn't be zero.)

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Answer it fast and no spamming....! !

ExtremeBDS [4]

Answer:

When the resistances are connected in the parallel the equivalent resistance will be always less than the value of the lowest resistance in the parallel circuit.

The equivalent resistance in the parallel circuit can be calculated using the following formula.

1/R=1/R1+1/R2+1/R3

1/R= 1/10+1/20+1/30

1/R=(6+3+2)/60

1/R=11/60

R=60/11

R=5.45 Ohms

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3 years ago

What is the wavelength of a wave?

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4 years ago

A block is given a very brief push up a 20.0 degree frictionless incline to give it an initial speed of 12.0 m/s.(a) How far alo

Orlov [11]

Explanation:

(a) Net force acting on the block is as follows.

$F_{net} = -mg Sin (\theta)$

or, ma = -mg Sin (\theta)[/tex]

a = $-g Sin (\theta)$

= $-9.8 \times Sin (20^{o})$

= -3.35 $m/s^{2}$

According to the kinematic equation of motion,

$v^{2} - v^{2}_{o} = 2as$

Distance traveled by the block before stopping is as follows.

s = $\frac{v^{2} - v^{2}_{o}}{2a}$

= $\frac{(0)^{2} - (12.0)^{2}_{o}}{2 \times -3.35}$

= 21.5 m

According to the kinematic equation of motion,

v = $v_{o} + at$

0 = $12.0 m/s + \frac{1}{2} \times -3.35 m/s^{2} \times t$

$t_{1}$ = 7.16 sec

Therefore, before coming to rest the surface of the plane will slide the box till 7.16 sec.

(b) When the block is moving down the inline then net force acting on the block is as follows.

$F_{net} = -mg Sin (\theta)$

ma = $mg Sin (\theta)$

a = $g Sin (\theta)$

= $9.8 m/s^{2} \times Sin (20^{o})$

= 3.35 $m/s^{2}$

Kinematics equation of the motion is as follows.

s = $v_{o}t + \frac{1}{2}at^{2}$

21.5 m = $0 + \frac{1}{2} \times 3.35 m/s^{2} \times t^{2}$

$t_{2}$ = $\sqrt{\frac{2 \times 21.5 m}{3.35 m/s^{2}}}$

= 3.58 sec

Hence, total time taken by the block to return to its starting position is as follows.

t = $t_{1} + t_{2}$

= 7.16 sec + 3.58 sec

= 10.7 sec

Thus, we can conclude that 10.7 sec time it take to return to its starting position.

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Zanzabum

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