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4 years ago

What is the purpose of the overrunning clutch in the starter drive?

Physics

1 answer:

nadya68 [22]4 years ago

5 0

<span>"prevent the engine from over speeding the armature"
hopes this help :) :D :)</span>

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Which statement is false?

podryga [215]

Answer:

D. Current with not flow through a wire if the magnet is outside the coil.

Explanation:

have a good day

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3 years ago

What happens to rocks as water combines

insens350 [35]

The answers is
D. The acid creates cracks in the rocks, which
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3 years ago

Which of the following can be an organic cause of chemical weathering?

Arturiano [62]

The answer would be B i just did this 3 days ago

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4 years ago

Read 2 more answers

How fast (in rpm) must a centrifuge rotate if a particle 6.00 cm from the axis of rotation is to experience an acceleration of 1

astra-53 [7]

The acceleration experienced by the particle is given by
$a=113000 g=113000 \cdot 9.81 m/s^2$
This corresponds to the centripetal acceleration of the motion, which is related to the angular speed $\omega$ of the particle and its distance r from the axis by the relationship
$a= \omega ^2 r$
In our problem, $r=6 cm=0.06 m$ , so we can solve for $\omega$ :
$\omega = \sqrt{ \frac{a}{r} } = \sqrt{ \frac{113000 \cdot 9.81 m/s^2}{0.06 m} }=4298 rad/s$
However, we must convert it into rpm (revolution per minute).
We know that 1 rad corresponds to $( \frac{1}{2 \pi} )$ revolutions, while $1 s = \frac{1}{60} min$ . So we the conversion is $\omega = 4298 rad/s \cdot ( \frac{1}{2\pi} rev/rad )( 60 s/min)=41067 rpm$

4 0

4 years ago

1. An asphalt block has a mass of 90 kg and a volume of 0.075 m. Determines the density of the asphalt.

Kisachek [45]

Answer:

1. Density = 1200[kg/m^3]; 2. Volume= 0.005775[m^3], mass= 15.59[kg]

Explanation:

1. We know that the density is defined by the following expression.

$Density = \frac{mass}{volume} \\where:\\mass=90[kg]\\volume=0.075[m^{3} ]\\density=\frac{90}{0.075} \\density=1200[\frac{kg}{m^{3} }]$

2. First we need to convert the units to meters.

wide = 35[cm] = 35/100 = 0.35[m]

long = 11 [dm] = 11 decimeters = 11/10 = 1.1[m]

Thick = 15[mm] = 15/1000 = 0.015[m]

Now we can find the density using the expression for the density.

$density= \frac{mass}{volume} \\where:\\volume = wide*long*thick\\volume=0.35*1.1*0.015 = 0.005775[m^3]\\\\mass= density*volume = 2700*0.005775 = 15.59[kg]$

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4 years ago