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g100num [7]
2 years ago
12

The global positioning system (GPS) uses electromagnetic waves sent between computer devices on Earth and satellites in space to

automatically determine the exact position of the computer device on Earth. In order to create this technology, engineers had to apply Einstein’s theory of relativity, which explains how electromagnetic waves change direction in space.
During which stage of developing the global positioning system did engineers first apply the theory of relativity to design the product?

first
second
third
fourth
Physics
2 answers:
zhenek [66]2 years ago
5 0

Answer:

first plz mark brarinlest

Explanation:

ANTONII [103]2 years ago
3 0

Answer:

SECOND

Explanation:

I got it wrong when I put in First, and Edge told me the answer was b. SECOND

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A single-phase 60-Hz overhead power line is symmetrically supported on a horizontal cross arm. Spacing between the centers of th
pentagon [3]

Complete question is;

A single-phase 60-Hz overhead power line is symmetrically supported on a horizontal cross arm. Spacing between the centers of the conductors acing between the centers of the conductors (say, a and b) is 2.5 m. A telephone line is also symmetrically supported on a horizontal cross arm 1.8 m directly below the power line. Spacing between the centers of these conductors (say, c and d) is 1.0 m.

The mutual inductance per unit length between circuit a-b and circuit c-d is given as 4 x 10^(-7) ln √((D_ad × D_bc)/(D_ac × D_bd)) H/m

where, for example, D_ad denotes the distance in meters between conductors a and d.

a. Hence, compute the mutual inductance per kilometer between the power line and the telephone line.

b. Find the 60-Hz voltage per kilometer induced in the telephone line when the power line carries 150 A

Answer:

A) M = 1.01 × 10^(-4) H/km

B) v_cd = 5.712 V/km

Explanation:

A) From the distances given in the question, we can deduce that;

D_ac = √(((2.5/2) - (1/2))² + 1.8²)

D_ac = 1.95 m

Also;

D_ad = √(((2.5/2) + (1/2))² + 1.8²)

D_ad = 2.51 m

I_a and I_b are put of phase by 180°. Thus, due to a and b, the flux linkages to c and d is given as;

φ_cd = 4 x 10^(-7)I_a( ln (2.51/1.95))

Mutual inductance per km is given as;

M = φ_cd/I_a

Thus;

M = 4 x 10^(-7)( ln (2.51/1.95))

M = 1.01 × 10^(-7) H/m

Per km;

M = 1.01 × 10^(-7) × 1000

M = 1.01 × 10^(-4) H/km

B) voltage per km is gotten by;

v_cd = ωMI

Now, ω = 2πf = 2π × 60 = 377 rad/s

Thus;

v_cd = 377 × 1.01 × 10^(-4) × 150

v_cd = 5.712 V/km

5 0
2 years ago
Determine the largest intensity w of the uniform loading that can be applied to the frame without causing either the average nor
jeka94

.........................|||||||..............................

8 0
3 years ago
A hockey puck on a frozen pond is given an initial speed of 20.0 m/s. If the puck always remains on the ice and slides 115 m bef
bekas [8.4K]

Answer:

μ_k = 0.1773

Explanation:

We are given;

Initial velocity;u = 20 m/s

Final velocity;v = 0 m/s (since it comes to rest)

Distance before coming to rest;s = 115 m

Let's find the acceleration using Newton's second law of motion;

v² = u² + 2as

Making a the subject, we have;

a = (v² - u²)/2s

Plugging relevant values;

a = (0² - 20²)/(2 × 115)

a = -400/230

a = -1.739 m/s²

From the question, the only force acting on the puck in the x direction is the force of friction. Since friction always opposes motion, we see that:

F_k = −ma - - - (1)

We also know that F_k is defined by;

F_k = μ_k•N

Where;

μ_k is coefficient of kinetic friction

N is normal force which is (mg)

Since gravity acts in the negative direction, the normal force will be positive.

Thus;

F_k = μ_k•mg - - - (2)

where g is acceleration due to gravity.

Thus,equating equation 1 and 2,we have;

−ma = μ_k•mg

m will cancel out to give;

-a = μ_k•g

μ_k = -a/g

g has a constant value of 9.81 m/s², so;

μ_k = - (-1.739/9.81)

μ_k = 0.1773

3 0
3 years ago
How many hydrogen and carbon atoms in a diamond
Wewaii [24]

Answer:

Explanation:

Thus, total 4+4=8 C atoms are present per unit cell of diamond. Carbon has an electronic arrangement of 2,4. In diamond, each carbon shares electrons with four other carbon atoms - forming four single bonds.

4 0
2 years ago
Given the isotope 2Fes, which has an actual mass of 55.934939 u: a) b) Determine the mass defect of the nucleus in atomic mass u
SSSSS [86.1K]

Answer:

Mass defect of each iron-56 nuclei:

The binding energy per nucleon of Iron-56 is approximately 8.6 MeV.

Explanation:

According to the physics constants table on Chemistry Libretexts:

  • Proton rest mass: \rm 1.0072765\;amu;
  • Neutron rest mass: \rm 1.0086649\; amu.
  • Speed of light in vacuum: \rm 2.99792458\times 10^{8}\;m\cdot s^{-1}.
  • Charge on an electron: \rm 1.6021765\times 10^{-19}\;C.

<h3>a)</h3>

The mass defect of a nucleus is equal to the sum of the mass of its parts (protons and, in most cases, neutrons) minus the mass of the nucleus.

The atomic number of iron is 26. There are 26 protons in each iron-56 nucleus. The mass number 56 indicates that there are 56 nucleons (neutrons and protons) in each iron-56 nucleus. The other 56 - 26 = 30 particles are neutrons.

The mass of protons and neutrons in each iron-56 nucleus will be:

\rm 26 \times 1.0072765 + 30 \times 1.0086649 = 56.464736\;amu.

According to this question, the mass of an iron-56 nucleus is equal to 55.934939 amu. The mass defect will be

\rm 56.464736 - 55.934939 = 0.514197\;amu.

<h3>b)</h3>

By the mass-energy equivalence,

E = m\cdot c^{2}.

Refer to this equation, the speed of light in vacuum c^{2} is the conversion factor between mass m and energy E. The value of c is usually given only in SI units \rm m\cdot s^{-1}. Accordingly, the value of c^{2} will be in the SI unit \rm m^{2}\cdot s^{-2} = J\cdot kg^{-1}.

Convert million electron-volts to joules.

One electron-volt is equal to the electrical work done moving an electron across a potential difference of one volt.  

\begin{aligned}\rm 1 MeV&= \rm 10^{6}\; eV\\ &= \rm (10^{6}\times 1.6021765\times 10^{-19}\;C)\times 1\; V\\&=\rm 1.6021765\times 10^{-19}\;J\end{aligned}.

Convert the unit of c^{2} from \rm m^{2}\cdot s^{-2} = J\cdot kg^{-1} to the desired \rm MeV \cdot amu^{-1}:

\begin{aligned}c^{2} &= \rm {\left(2.99792458\times 10^{8}\;m\cdot s^{-1}\right)}^{2}\\&=\rm 8.987551787\times 10^{16}\; m^{2}\cdot s^{-2}\\ &= \rm 8.987551787\times 10^{16}\; J\cdot kg^{-1}\\&= \rm 8.987551787\times 10^{16}\; J\cdot kg^{-1}\times \frac{1\;MeV}{1.6021765\times 10^{-13}\;J}\times \frac{1\times 10^{-3}\;kg}{6.022142\times 10^{23}\;amu}\\&\approx \rm 931.602164\;MeV\cdot amu^{-1}\end{aligned}.

Total binding energy in each iron-56 nucleus:

\begin{aligned}E &= m\cdot c^{2}\\&= \rm 0.514197\;amu \times 9.31602164\;MeV\cdot amu^{-1} \\&=\rm 479.027038\; MeV \end{aligned}.

Again, the mass number 56 indicates that there are 56 nucleons in each iron-56 nucleus. The binding energy per nucleon of iron-56 \mathrm{^{56}Fe} will be:

\displaystyle \rm \frac{479.027038\; MeV}{56} \approx 8.6\; MeV.

6 0
3 years ago
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