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2 years ago

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Your chances of getting into a collision when talking on a cell phone _________: A. Double B. Triple C. Quadruple D. Remain the

same

1 answer:

Anit [1.1K]2 years ago

3 0

Answer:

C. Quadruple

Explanation:

¨Drivers who are talking on the phone, even on a hands-free device, are up to four times more likely to be involved in a crash.¨

I hope this helps! Have a great day!

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7. Two children of mass 20 kg and 30 kg sit balanced on a seesaw with the pivot point located at the center of the seesaw. If th

Ainat [17]

Answer:

Explanation:

Given

mass of children $m_1=20\ kg$

$m_2=30\ kg$

distance between two children $L=3\ m$

suppose small child is at a distance of x m from pivot point

so torque of small child and heavier child must be equal

$20\times (x)=30\times (3-x)$

$2x=9-3x$

$5x=9$

$x=1.8\ m$

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3 years ago

Two objects with a mass of 10,000-kg each and a distance of 1 meter between them.

Fofino [41]

Answer:

d) What is the force if we doubled both the masses AND we doubled the distance

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2 years ago

In which of these examples is the greatest movement occurring?

Elenna [48]

Answer:

not clear pic...but it's definitely not A)

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2 years ago

If the speed of an object doubles, how does that affect its kinetic energy? A. Halves B. Doubles C. Quarters D. Quadruples

kvasek [131]

Answer is :

D. Quadruples

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2 years ago

Read 2 more answers

A 120 g, 8.0-cm-diameter gyroscope is spun at 1000 rpm and allowed to precess. What is the precession period?

dolphi86 [110]

To solve this problem we will derive the expression of the precession period from the moment of inertia of the given object. We will convert the units that are not in SI, and finally we will find the precession period with the variables found. Let's start defining the moment of inertia.

$I = MR^2$

Here,

M = Mass

R = Radius of the hoop

The precession frequency is given as

$\Omega = \frac{Mgd}{I\omega}$

Here,

M = Mass

g= Acceleration due to gravity

d = Distance of center of mass from pivot

I = Moment of inertia

$\omega$ = Angular velocity

Replacing the value for moment of inertia

$\Omega= \frac{MgR}{MR^2 \omega}$

$\Omega = \frac{g}{R\omega}$

The value for our angular velocity is not in SI, then

$\omega = 1000rpm (\frac{2\pi rad}{1 rev})(\frac{1min}{60s})$

$\omega = 104.7rad/s$

Replacing our values we have that

$\Omega = \frac{9.8m/s^2}{(8*10^{-2}m)(104.7rad)}$

$\Omega = 1.17rad/s$

The precession frequency is

$\Omega = \frac{2\pi rad}{T}$

$T = \frac{2\pi rad}{\Omega}$

$T = \frac{2\pi}{1.17}$

$T = 5.4 s$

Therefore the precession period is 5.4s

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2 years ago