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2 years ago

what is the difference in the charges on a balloon rubbed in your hair and a glass rod rubbed with silk? why?

1 answer:

8 0

The balloon steals electrons from your hair, leaving the hair positively charged, and the balloon negatively charged. It causes the hair to be apart from each other, because they have the same charge. Glass has a weaker hold on electrons, and silk absorbs the lost electrons.

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What is the equation for finding the amount of inertia in forces and motion? Explain how you solve that equation.

jolli1 [7]

Answer:

Inertia = angular momentum / angular velocity

4 0

2 years ago

I need help in this pls i really need a answer

IrinaK [193]

Answer:

In terms of distance, average speed is 20 km/h

In terms of displacement, average speed is 0 km/h.

Explanation:

Total distance:

$= (40.0 - 10.0) + (20.0 - 10.0) + (40.0 - 20.0) \\ = 30.0 + 10.0 + 20.0 \\ = 60.0 \: km$

Total time is 3.0 hours

but:

$average \: speed = \frac{total \: distance}{total \: time} \\$

In terms of distance.

substitute:

$average \: speed = \frac{60.0}{3.0} \\ \\ = 20 \: {kmh}^{ - 1}$

$displacement = ( - 30.0) + 10 .0+ 20.0 \\ = 0$

In terms of displacement:

$speed = \frac{0}{3} \\ \\ = 0 \: {kmh}^{ - 1}$

3 0

3 years ago

A box weighing 52.4 N is sliding on a rough horizontal floor with a constant friction force of magnitude LaTeX: ff. The box's in

german

Answer:

The magnitude of the friction force exerted on the box is 2.614 newtons.

Explanation:

Since the box is sliding on a rough horizontal floor, then it is decelerated solely by friction force due to the contact of the box with floor. The free body diagram of the box is presented herein as attachment. The equation of equilbrium for the box is:

$\Sigma F = -f = m\cdot a$ (Eq. 1)

Where:

$f$ - Kinetic friction force, measured in newtons.

$m$ - Mass of the box, measured in kilograms.

$a$ - Acceleration experimented by the box, measured in meters per square second.

By applying definitions of weight ( $W = m\cdot g$ ) and uniform accelerated motion ( $v = v_{o}+a\cdot t$ ), we expand the previous expression:

$-f = \left(\frac{W}{g} \right)\cdot \left(\frac{v-v_{o}}{t}\right)$

And the magnitude of the friction force exerted on the box is calculated by this formula:

$f = -\left(\frac{W}{g} \right)\cdot \left(\frac{v-v_{o}}{t}\right)$ (Eq. 1b)

Where:

$W$ - Weight, measured in newtons.

$g$ - Gravitational acceleration, measured in meters per square second.

$v_{o}$ - Initial speed, measured in meters per second.

$v$ - Final speed, measured in meters per second.

$t$ - Time, measured in seconds.

If we know that $W = 52.4\,N$ , $g = 9.807\,\frac{m}{s^{2}}$ , $v_{o} = 1.37\,\frac{m}{s}$ , $v = 0\,\frac{m}{s}$ and $t = 2.8\,s$ , the magnitud of the kinetic friction force exerted on the box is: