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3 years ago

Why do aeroplanes take longer to fly west than east?

1 answer:

8 0

Every planet/moon has global wind that are mostly determined by the way the planet/moon rotates and how evenly the Sun illuminates it. On the Earth the equator gets much more Sun than the poles. resulting in warmer air at the equator than the poles and creating circulation cells (or "Hadley Cells") which consist of warm air rising over the equator and then moving North and South from it and back round.

The Earth is also rotating. When any solid body rotates, bits of it that are nearer its axis move slower than those which are further away. As you move north (or south) from the equator, you are moving closer to the axis of the Earth and so the air which started at the equator and moved north (or south) will be moving faster than the ground it is over (it has the rotation speed of the ground at the equator, not the ground which is is now over). This results in winds which always move from the west to the east in the mid latitudes.

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What is the SI unit for speed?

Maksim231197 [3]

That is meters per second, same as velocity.

3 0

3 years ago

Read 2 more answers

A 86 kg human stands on the surface of Venus. The mass of Venus is 4.9 × 1024 kg and its radius is 6.1 × 106 m. Collapse questio

iris [78.8K]

Answer:

755.37 N

Explanation:

We are given that

Mass of venus= $m_1=4.9\times 10^{24}kg$

Radius= $r=6.1\times 10^6$ m

Mass of human= $m_2=86 kg$

We know that the gravitational force between two bodies

$F=G\frac{m_1m_2}{r^2}$

Where G=Gravitational constant= $6.67\times 10^{-11}Nm^2/kg^2$

Using the formula

The magnitude of the gravitational force exerted by Venus on the human= $F=\frac{6.67\times 10^{-11}\times 86\times 4.9\times 10^{24}}{(6.1\times 10^6)^2}$

The magnitude of the gravitational force exerted by Venus on the human=F= $755.37N$

7 0

3 years ago

During a very quick stop, a car decelerates at 6.8 m/s^2. Assume the forward motion of the car corresponds to a positive directi

geniusboy [140]

Answer:

-24.28571 rad/s²

29.57239 revolutions

3.91176 seconds

52.026478 m

Explanation:

$a_t$ = Tangential acceleration = -6.8 m/s²

r = Radius of wheel = 0.28

$\omega_i$ = Initial angular velocity = 95 rad/s

$\theta$ = Angle of rotation

$\omega_f$ = Final angular velocity

t = Time taken

Angular acceleration is given by

$\alpha=\frac{a_t}{t}\\\Rightarrow \alpha=\frac{-6.8}{0.28}\\\Rightarrow \alpha=-24.28571\ rad/s^2$

The angular acceleration is -24.28571 rad/s²

$\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \theta=\frac{\omega_f^2-\omega_i^2^2}{2\alpha}\\\Rightarrow \theta=\frac{0^2-95^2}{2\times -24.28571}\\\Rightarrow \theta=185.80885\ rad=185.80885\times \frac{1}{2\pi}\\\Rightarrow \theta=29.57239\ rev$