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2 years ago

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A car is traveling at 30 m/s ( approximately 67 mph) on I-95 highway. The driver applies the brakes and the car comes to a stop

in 4.0 s . What is the car's acceleration, in m/s2

1 answer:

o-na [289]2 years ago

7 0

Answer:

step 1 the initial velocity of the car is u = 30 m/s.the final velocity of the car is v = 0 m/s. the time taken by

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In which condition the acceleration of a moving vehicle become zero

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Explanation:

When,the vehicle has uniform velocity, it's acceleration becomes zero

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In a science museum, a 140 kg brass pendulum bob swings at the end of a 16.8 m -long wire.

Mice21 [21]

The period of the pendulum is 8.2 s

Explanation:

The period of a simple pendulum is given by the equation:

$T=2\pi \sqrt{\frac{L}{g}}$

where

L is the length of the pendulum

g is the acceleration of gravity

T is the period

We notice that the period of a pendulum does not depend at all on its mass, but only on its length.

For the pendulum in this problem, we have

L = 16.8 m

and

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Therefore the period of this pendulum is

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¿ cuales son las especies q mas características tienes en común ?

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3 years ago

A meter stick is suspended vertically at a pivot point 22 cm from the top end. It is rotated on the pivot until it is horizontal

suter [353]

Answer:

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$\omega$ = Angular velocity

Moment of inertia of the system is given by

$I=I_c+mr^2\\\Rightarrow I=\frac{mL^2}{12}+mr^2\\\Rightarrow I=\frac{m1^2}{12}+m0.28^2\\\Rightarrow I=m(\frac{1}{12}+0.0784)$

As the energy in the system is conserved

$mgh=I\frac{\omega^2}{2}\\\Rightarrow mgh=m(\frac{1}{12}+0.0784)\frac{\omega^2}{2}\\\Rightarrow gh=(\frac{1}{12}+0.0784)\frac{\omega^2}{2}\\\Rightarrow \omega=\sqrt{\frac{2gh}{\frac{1}{12}+0.0784}}\\\Rightarrow \omega=\sqrt{\frac{2\times 9.81\times 0.28}{\frac{1}{12}+0.0784}}\\\Rightarrow \omega=5.82812\ rad/s$

The maximum angular velocity is 5.82812 rad/s

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3 years ago