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Which was more difficult to determine (latitude or longitude) and why?

Ivenika [448]

Longitude was. Determining longitude requires knowing the exact time of day, which was difficult prior to modern clocks. The source book below tells the story of Englishman John Harrison's life-long pursuit of building a reliable clock and its importance to navigation.

7 0

4 years ago

What is physics? In your own words too

denis23 [38]

The study of how the world works

4 0

3 years ago

50POINTS! Find the orbital speed of a satellite in a circular orbit 1700km above the surface of the Earth. M_earth 5.97e24kg, r_

ivolga24 [154]

<h2>Answer: 7020.117 m/s</h2>

Explanation:

The velocity of a satellite describing a circular orbit is<u> constant</u> and defined by the following expression:

$V=\sqrt{G\frac{M}{R}}$ (1)

Where:

$G=6.674({10}^{-11})\frac{N{m}^{2}}{{kg}^{2}}$ is the gravity constant

$M_{Earth}=5.97{10}^{24}kg$ the mass of the massive body around which the satellite is orbiting, in this case, the Earth .

$R=r_{Earth}+h=8080000m$ the radius of the orbit (measured from the center of the planet to the satellite).

This means the radius of the orbit is equal to <u>the sum</u> of the average radius of the Earth $r_{Earth}$ and the altitude of the satellite above the Earth's surface $h$ .

Note this orbital speed, as well as orbital period, does not depend on the mass of the satellite. It depends on the mass of the massive body (the Earth).

Now, rewriting equation (1) with the known values:

$V=\sqrt{(6.674({10}^{-11})\frac{N{m}^{2}}{{kg}^{2}})\frac{5.97{10}^{24}kg}{8080000m}}$

$V=7020.117\frac{m}{s}$

6 0

4 years ago

Read 2 more answers

A diffusion couple, made by welding a thin onecentimeter square slab of pure metal A to a similar slab of pure metal B, was give

bezimeni [28]

Answer:

The value is $H = 18*10^{2} \ Atom / sec$

Explanation:

From the question we are told that

The atom fraction of metal A at point G is $A = 0.30 \ m$

The atom fraction of metal A at a distance 5000nm from G is $A_2 = 0.35$

The number of atoms per $m^3$ is $N_h = 9 * 10^{28}$

The diffusion coefficient is $D = 2* 10^{-14 } m^2/s$

Generally of the concentration of atoms of metal A at G is

$N_A = A * N_h$

=> $N_A = 0.3 * 9 * 10^{28}$

=> $N_A = 2.7 * 10^{28} 2.7 atoms/m^3$

Generally of the concentration of atoms of metal A at a distance 5000nm from G is

$D = 0.35 *9 * 10^{28}$

=> $D = 3.15 * 10^{28} \ atoms / m^3$

The concentration gradient is mathematically represented as

$\frac{dN_A}{dx} = \frac{(3.15 - 2.7) * 10^{28} }{5000nm - 0 }$

=> $\frac{dN_A}{dx} = \frac{(3.15 - 2.7) * 10^{28} }{[5000 *10^{-9}] - 0 }$

=> $\frac{dN_A}{dx} = 9 * 10^{20} / m^4$

Generally the flux of the atoms per unit area according to Fick's Law is mathematically represented as

$J = -D* \frac{d N_A}{dx}$

=> $J = -2* 10^{-14 * 9 * 10^{20}$

=> $J = 18*10^{6}\ atoms\ crossing\ /m^2 s$

Generally if the cross-section area is $a = 1 cm^2 = 10^{-4} \ m^2$

Generally the number of atom crossing the above area per second is mathematically is

$H = 18*10^{6} * 10^{-4}$

=> $H = 18*10^{2} \ Atom / sec$

7 0

4 years ago

Professional baseball pitchers deliver pitches that can reach the blazing speed of 100 mph (miles per hour). A local team has dr

Wittaler [7]

Answer:

A. ) K =126. 7 J

B. ) h= 91.1 m.

Explanation:

A)

Assuming no air resistance, once released by the pitcher, the speed must keep constant through all the trajectory, so the kinetic energy of the ball can be expressed as follows:

$K = \frac{1}{2}*m*v^{2} = \frac{1}{2}*0.142 kg*(42.24m/s)^{2} = 126.7 J (1)$

B)

Neglecting air resistance, total mechanical energy must be the same at any point, so, if we choose the ground level as the zero reference level for the gravitational potential energy, and assuming that the ball attains this kinetic energy just before striking ground, this value must be equal to the gravitational potential energy just before be dropped, so we can write the following equality:

$U_{o} = K_{f} = 126. 7 J (2)$

⇒ m*g*h = 126. 7 J

Solving for h, we get:

$h = \frac{K_{f}}{m*g} = \frac{126.7J}{0.1420kg*9.8m/s2} = 91.1 m (3)$

4 0

3 years ago