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vovikov84 [41]
3 years ago
9

Find the GCF 12, 18, and 84

Physics
2 answers:
ahrayia [7]3 years ago
5 0
To find the answer, plot down the factors for every number.

12:   1, 2 ,3 ,4, 6, 12

18:   1, 2, 3, 6, 9, 18

84:  1, 2, 3, 4, 6, 7, 12

If you noticed, the number that was common to the 3 numbers, were 1, 2, 3, and 6
And 6 is the bigger number
So 6 is your GCF
Paha777 [63]3 years ago
5 0

Answer: 6 is you GCF

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Consider a father pushing a child on a playground merry-go-round. The system has a moment of inertia of 84.4 kg.m^2. The father
Sophie [7]

Answer:

Explanation:

Given that:

the initial angular velocity \omega_o = 0

angular acceleration \alpha = 4.44 rad/s²

Using the formula:

\omega = \omega_o+ \alpha t

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t= \dfrac{\omega- \omega_o}{ \alpha }

where;

\omega = 1.53 \ rad/s^2

∴

t= \dfrac{1.53-0}{4.44 }

t = 0.345 s

b)

Using the formula:

\omega ^2 = \omega _o^2 + 2 \alpha \theta

here;

\theta = angular displacement

∴

\theta = \dfrac{\omega^2 - \omega_o^2}{2 \alpha }

\theta = \dfrac{(1.53)^2 -0^2}{2 (4.44) }

\theta =0.264 \ rad

Recall that:

2π rad = 1 revolution

Then;

0.264 rad = (x) revolution

x = \dfrac{0.264 \times 1}{2 \pi}

x = 0.042 revolutions

c)

Here; force = 270 N

radius = 1.20 m

The torque = F * r

\tau = 270 \times 1.20 \\ \\  \tau = 324 \ Nm

However;

From the moment of inertia;

Torque( \tau) = I \alpha \\ \\  Since( I \alpha) = 324 \ Nm. \\ \\  Then; \\ \\  \alpha= \dfrac{324}{I}

given that;

I = 84.4 kg.m²

\alpha= \dfrac{324}{84.4} \\ \\  \alpha=3.84 \ rad/s^2

For re-tardation; \alpha=-3.84 \ rad/s^2

Using the equation

t= \dfrac{\omega- \omega_o}{ \alpha }

t= \dfrac{0-1.53}{ -3.84 }

t= \dfrac{1.53}{ 3.84 }

t = 0.398s

The required time it takes= 0.398s

5 0
3 years ago
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