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3 years ago

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A girl is riding a bike with a velocity of 7 m/s. The bike undergoes an acceleration of 2 m/s2. What is the velocity of the bike

after 2 seconds

1 answer:

Dmitry_Shevchenko [17]3 years ago

4 0

From the information provided, the velocity after 2 seconds is 11 m/s.

From the equations of motion;

v = u + at

Where;

v = final velocity

a = acceleration

t = time taken

The initial velocity of the bike was 7 m/s and its acceleration is 2 m/s2. This change occurred within a time of 2 seconds.

Hence;

v = 7 + 2(2)

v = 11 m/s

The velocity after 2 seconds is 11 m/s.

Learn more: brainly.com/question/12550364

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A 0.5 m diameter wagon wheel consists of a thin rim having a mass of 7 kg and six spokes, each with a mass of 1.2 kg. 1.2 kg 7 k

Arte-miy333 [17]

Explanation:

It is given that,

Mass of the rim of wheel, m₁ = 7 kg

Mass of one spoke, m₂ = 1.2 kg

Diameter of the wagon, d = 0.5 m

Radius of the wagon, r = 0.25 m

Let I is the the moment of inertia of the wagon wheel for rotation about its axis.

We know that the moment of inertia of the ring is given by :

$I_1=m_1r^2$

$I_1=7\times (0.25)^2=0.437\ kgm^2$

The moment of inertia of the rod about one end is given by :

$I_2=\dfrac{m_2l^2}{3}$

l = r

$I_2=\dfrac{m_2r^2}{3}$

$I_2=\dfrac{1.2\times (0.25)^2}{3}=0.025\ kgm^2$

For 6 spokes, $I_2=0.025\times 6=0.15\ kgm^2$

So, the net moment of inertia of the wagon is :

$I=I_1+I_2$

$I=0.437+0.15=0.587\ kgm^2$

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4 years ago

Current is the movement of negative charges called protons.<br> A. True<br> B. False

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Explanation:

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3 years ago

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iren2701 [21]

Answer:

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If we got an uncertainty $\Delta r$ the formula for the uncertainty of V is:

$\Delta V(r) = \sqrt{ (\frac{dV}{dr} \Delta r)^2 }$

We can calculate this uncertainty, first we obtain the derivative:

$\frac{dV}{dr} = 3 * \frac{4}{3} \ \pi \ r^2$

$\frac{dV}{dr} = 4 \ \pi \ r^2$

And using it in the formula:

$\Delta V(r) = \sqrt{ (4 \ \pi \ r^2\Delta r)^2 }$

$\Delta V(r) = \sqrt{ 4^2 \ \pi^2 \ r^4 \Delta r^2 }$

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The relative uncertainty is:

$\frac{\Delta V(r)}{V(r)}$

$\frac{ 4 \ \pi \ r^2 \Delta r }{ \frac{4}{3} \ \pi \ r^3}$

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Using the values for the problem:

$\frac{ 3 * 0.09 m }{ 5.66 m} = 0.0477$

This is, a percent uncertainty of 4.77 %

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