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soldi70 [24.7K]
3 years ago
10

if a cyclist is travelling a road due east at 12km/h and a wind is blowing from south-west at 5km/h. find the velocity of the wi

nd relative to the cyclist.​
Physics
1 answer:
slamgirl [31]3 years ago
7 0

Answer:

The velocity of wind with respect to cyclist is -15.5 \widehat{i} - 3.5 \widehat{j}.

Explanation:

speed of cyclist = 12 km/h east

speed of wind = 5 km/h south west

Write the speeds in the vector form

\overrightarrow{vc} = 12 \widehat{i}\\\\overrightarrow{vw} = - 5 (cos 45 \widehat{i} + sin 45 \widehat{j})\\\\\overrightarrow{vw} =-3.5 \widehat{i} - 3.5 \widehat{j}

The velocity of wind with respect to cyclist is

\overrightarrow{vw} =-3.5 \widehat{i} - 3.5 \widehat{j}\overrightarrow{v_(w/c)} = \overrightarrow{vw}-\overrightarrow{vc}\\\\\overrightarrow{v_(w/c)} = - 3.5 \widehat{i} - 3.5 \widehat{j}  - 12 \widehat{i}\\\\\overrightarrow{v_(w/c)} =-15.5 \widehat{i} - 3.5 \widehat{j}

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When the Earth and sun are perfectly lined up, then it will happen. They can tell when it's going to happen.

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Suppose that you are holding a pencil balanced on its point. If you release the pencil and it begins to fall, what will be the a
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Answer:

The angular acceleration of the pencil<em> α  = 17 rad·s⁻²</em>

Explanation:

Using Newton's second angular law or torque to find angular acceleration, we get the following expressions:

    τ = I α                              (1)

    W r = I α                          (2)

The weight is that the pencil has is,

   sin 10 = r / (L/2)

   r = L/2(sin(10))

 

The shape of the pencil can be approximated to be a cylinder that rotates on one end and therefore its moment of inertia will be:

    I = 1/3 M L²

Thus,

   mg(L / 2)sin(10) = (1/3 m L²)(α) 

   α(f) = 3/2(g) / Lsin(10)

   α  = 3/2(9.8) / 0.150sin(10)

  <em> α  = 17 rad·s⁻²</em>

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3 0
3 years ago
How many significant digits are in the number 0.010?
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There a two significant digits

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6 0
3 years ago
The density of mobile electrons in copper metal is 8.4 1028 m-3. Suppose that i = 4.6 1018 electrons/s are drifting through a co
Vesna [10]

Answer:

The time is 106.7 minute.

Explanation:

Given that,

Density = 8.4\times10^{28}\ m^3

Current i = 4.6\times10^{18}\ electron/s

Diameter of wire = 1.2 mm

Length = 31 cm

We need to calculate the drift velocity

Using formula of drift velocity

v_{d}=\dfrac{I}{neA}

v_{d}=\dfrac{Ne}{tne\times\pi r^2}

Put the value into the formula

v_{d}=\dfrac{4.6\times10^{18}}{8.4\times10^{28}\times\pi\times(0.6\times10^{-3})^2}

v_{d}=4.842\times10^{-5}\ m/s

We need to calculate the time

Using formula for time

v_{d}=\dfrac{l}{t}

t=\dfrac{l}{v_{d}}

Where, l = length

v_{d} = drift velocity

Put the value into the formula

t=\dfrac{31\times10^{-2}}{4.842\times10^{-5}}

t=6402.31\ sec

t=106.7\ minute

Hence, The time is 106.7 minute.

7 0
3 years ago
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