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2 years ago

The earth has a radius of 6.38 × 106 m and turns on its axis once every 23.9 h.

Physics

1 answer:

Vladimir [108]2 years ago

8 0

Answer:

a) V = 465.9 m/s

b) θ = 70.529°

Explanation:

Let's first calculate angular velocity of earth:

$\omega=\frac{2\pi}{23.9h}*1h/3600s$

Velocity of a person on Ecuador will be:

$V_E = \omega*R$

$V_E = 465.9 m/s$

For part b, since angular velocity is the same:

$\frac{\omega*R}{3}=\omega*(R*cos\theta )$

Solving for θ:

$\theta=acos(1/3)$

$\theta=70.529\°$

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2 years ago

(Another tomato/skyscraper problem.) You are looking out your window in a skyscraper, and again your window is at a height of 45

Ivan

Answer:

1027.2 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 32.2 ft/s

$s=ut+\frac{1}{2}at^2\\\Rightarrow u=\frac{s-\frac{1}{2}at^2}{t}\\\Rightarrow u=\frac{450-\frac{1}{2}\times 32.2\times 2^2}{2}\\\Rightarrow u=192.8\ ft/s$

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The height the tomato would fall is 450+577.2 = 1027.2 m

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3 years ago

A biker travels at an average speed of 18 km/hr along a 0.30 km straight segment of a bike path. How much time (in hours) does t

lawyer [7]

Answer: 0.016 h

Explanation:

$\text{Average speed} = \frac{\text {Total Distance}}{\text {total time taken}}$

It is given that, biker has an average speed = 18 km/h

Total distance traveled = 0.30 km

Therefore, time taken by biker to travel this distance:

$\Rightarrow \text{total time taken} = \frac{0.30 km}{18 km/h}=0.016 h$

Thus, the biker takes 0.016 hours to travel the segment of 0.30 km at an average speed of 18 km/h.

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2 years ago

A piano string sounds a middle A by vibrating primarily at 220 Hz.a)Calculate its period.b)Calculate its angular frequency.c)Cal

chubhunter [2.5K]

a) 4.5 ms

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$T=\frac{1}{f}$

where f is the frequency.

For the note in this problem, f = 220 Hz, so the period of the wave is

$T=\frac{1}{f}=\frac{1}{220 Hz}=4.5\cdot 10^{-3} s = 4.5 ms$

b) 1381.6 rad/s

The angular frequency is given by:

$\omega=2 \pi f$

where f is the frequency.

In this problem, f = 220 Hz, so the angular frequency is

$\omega=2 \pi (220 Hz)=1381.6 rad/s$

c) 1.1 ms

The frequency of the "high A" is four times the frequency of the piano string, so

$f=4 \cdot 220 Hz=880 Hz$

And so, its period is

$T=\frac{1}{f}=\frac{1}{880 Hz}=1.1\cdot 10^{-3} s=1.1 ms$

d) 5526.4 rad/s

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where f is the frequency.

For this note, f = 880 Hz, so the angular frequency is

$\omega=2 \pi (880 Hz)=5526.4 rad/s$

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2 years ago

Recall that the essential matrix E describing the stereo geometry of two calibrated cameras is a function only of the relative r

Mashcka [7]

Answer:

The essential matrix in this case is given by

$E=[T_{x}]R=\left[\begin{array}{ccc}0&0&0\\0&0&d\\0&-d&0\end{array}\right]$

the values of the 3x3 essential matrix E for this stereo system

$E=[T_{x}]R=\left[\begin{array}{ccc}0&0&0\\0&0&0.12\\0&-0.12&0\end{array}\right]$

Explanation:

The explanation is shown on the first uploaded image

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2 years ago