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ra1l [238]
4 years ago
7

Sarah, who lives in Australia notices that the length of the shadow of the flagpole is getting shorter every day when measured a

t noon. Paul, who lives in Canada, notices that the length of the shadow of the flagpole is getting longer everyday when measured at noon. Draw a model to explain how both students can be observing this at the same time. Include these in your model: Sun; earth; tilted axis; rotation and revolution.
Physics
1 answer:
Allisa [31]4 years ago
3 0

Answer:

Time zone is one important factor in difference in location and this in turn affects the result of the resolution and rotation of shadow produced from the sun or other illumination.

Therefore someone at a place might see a clear large shadow due to shinny sun reflection and another a small or dull Shadow at same time if the intensity of the sun or lighting source is going down.

Explanation:

The closer a body/object is to a lighting source the larger the shadow it produces, and the farther the body the smaller the shadow produced.

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2) Two railway tracks are parallel to west east direction. Along one track, train A moves with a speed of 45 m/s from
Natalija [7]

Answer:

105 m/s

Explanation:

Given that the speed of train A, V_A = 45 m/s from west to east.

Speed of train B, V_B = 60 m/s from east to west.

Train B is moving in the opposite direction with respect to the speed of train A. Assuming that the speed from east to west direction is positive.

So, the speed of train A from east to west= - 45 m/s

The speed of train B w.r.t train A = V_B - V_A=60-(-45)=60+45=105 m/s

Hence, the speed of train B w.r.t train A is 105 m/s from east to west.

5 0
3 years ago
Newtons first law stateS that object will move With a constant velocity if nothing acts on it. Does our every day experience con
bazaltina [42]

Answer:

hi

Explanation:

hi

4 0
3 years ago
On a cold winter day, a penny (mass 2.50 g) and a nickel (mass 5.00 g) are lying on the smooth (frictionless) surface of a froze
aleksklad [387]
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8 0
4 years ago
Starting from rest, a disk rotates about its central axis with constant angular acceleration. In 1.00 s, it rotates 21.0 rad. Du
ELEN [110]

With constant angular acceleration \alpha, the disk achieves an angular velocity \omega at time t according to

\omega=\alpha t

and angular displacement \theta according to

\theta=\dfrac12\alpha t^2

a. So after 1.00 s, having rotated 21.0 rad, it must have undergone an acceleration of

21.0\,\mathrm{rad}=\dfrac12\alpha(1.00\,\mathrm s)^2\implies\alpha=42.0\dfrac{\rm rad}{\mathrm s^2}

b. Under constant acceleration, the average angular velocity is equivalent to

\omega_{\rm avg}=\dfrac{\omega_f+\omega_i}2

where \omega_f and \omega_i are the final and initial angular velocities, respectively. Then

\omega_{\rm avg}=\dfrac{\left(42.0\frac{\rm rad}{\mathrm s^2}\right)(1.00\,\mathrm s)}2=42.0\dfrac{\rm rad}{\rm s}

c. After 1.00 s, the disk has instantaneous angular velocity

\omega=\left(42.0\dfrac{\rm rad}{\mathrm s^2}\right)(1.00\,\mathrm s)=42.0\dfrac{\rm rad}{\rm s}

d. During the next 1.00 s, the disk will start moving with the angular velocity \omega_0 equal to the one found in part (c). Ignoring the 21.0 rad it had rotated in the first 1.00 s interval, the disk will rotate by angle \theta according to

\theta=\omega_0t+\dfrac12\alpha t^2

which would be equal to

\theta=\left(42.0\dfrac{\rm rad}{\rm s}\right)(1.00\,\mathrm s)+\dfrac12\left(42.0\dfrac{\rm rad}{\mathrm s^2}\right)(1.00\,\mathrm s)^2=63.0\,\mathrm{rad}

5 0
4 years ago
The crest of one wave with an amplitude of 4 m meets up with the crest of a second
trapecia [35]

Answer:

Constructive

Explanation:

3 0
3 years ago
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