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3 years ago

Which type of modulation is used for GPS? amplitude frequency phase pulse

Physics

2 answers:

Alenkinab [10]3 years ago

4 0

Frequency - transferred through satellites and picked up through their receivers...

lisov135 [29]3 years ago

3 0

Answer:

phase

Explanation:Answer for E2020

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In scientific notation, (6.2 x 10^4) x (3.3 x 10^2 ) equals ____________________.

castortr0y [4]

You will get 20460000 as your answer which is broken down into, 2.046 x 10^7 as your number has to be between 1-10.

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2 years ago

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What is the meaning of viscosity?

Vera_Pavlovna [14]

the state of being thick, sticky, and semifluid in consistency, due to internal friction.

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2 years ago

Why are there only two elements in the first period of the periodic table?(1 point)

Katen [24]

Answer:

because only two electrons can fit in the first orbit around the nucleus, and each period on the table is organized by number of orbits

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2 years ago

calculate the mass of potassium chlorate (kcio3) required to obtain 10g of oxygen in the following reaction:kclO3-kcl+O2

igor_vitrenko [27]

First, balance the reaction:

_ KClO₃ ==> _ KCl + _ O₂

As is, there are 3 O's on the left and 2 O's on the right, so there needs to be a 2:3 ratio of KClO₃ to O₂. Then there are 2 K's and 2 Cl's among the reactants, so we have a 1:1 ratio of KClO₃ to KCl :

2 KClO₃ ==> 2 KCl + 3 O₂

Since we start with a known quantity of O₂, let's divide each coefficient by 3.

2/3 KClO₃ ==> 2/3 KCl + O₂

Next, look up the molar masses of each element involved:

• K: 39.0983 g/mol

• Cl: 35.453 g/mol

• O: 15.999 g/mol

Convert 10 g of O₂ to moles:

(10 g) / (31.998 g/mol) ≈ 0.31252 mol

The balanced reaction shows that we need 2/3 mol KClO₃ for every mole of O₂. So to produce 10 g of O₂, we need

(2/3 (mol KClO₃)/(mol O₂)) × (0.31252 mol O₂) ≈ 0.20835 mol KClO₃

KClO₃ has a total molar mass of about 122.549 g/mol. Then the reaction requires a mass of

(0.20835 mol) × (122.549 g/mol) ≈ 25.532 g

of KClO₃.

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2 years ago

A uniform electric field exists in the region between two oppositely charged plane parallel plates. A proton is released from re

spayn [35]

The magnitude of the electric field can be calculated using the equation

$E = \frac{F}{q}$ , where F is the Force acting on the proton and q is the charge of the proton.

We know that the charge of the proton is $q = 1.6 x 10^{-19} C$

We have to calculate the Force first.

We know that F = ma from Newton's 2nd law, where m is the mass of the proton and a is the acceleration.

We know that the mass of the proton $m = (1.67) X 10^{-27} kg$

So it turns out that we have to calculate acceleration before anything else.

In order to calculate the acceleration, we make use of the following data from the question:

Initial Velocity of the proton $V_{i} = 0$

Distance traveled $D = 1.70 cm$ = 0.017 m

Time taken for the travel between the plates $t = (1.48) X 10^{-6} s$

Acceleration a = ?

Using the equation, $D = V_{i}t + \frac{1}{2} at^{2}$ , we get

Knowing that initial velocity is 0, the equation reduces to $D = \frac{1}{2}at^{2}$

Rearranging the equation so as to make a the subject of the formula, we have

$a = \frac{2D}{t^{2} }$

Plugging in the numbers and simplifying gives us a = 1.5 x $10^{10} m/s^{2}$

We can now calculate the Force using F = ma

Plugging in the known values, we get F = 2.5 x $10^{-17} N$

Using this, we can calculate E through the equation $E = \frac{F}{q}$

Plugging the numbers and simplifying gets us E = 156.25 N/C

Thus, the magnitude of the electric field between the plates of the capacitor is 156.25 N/C

B) To calculate the Final Velocity of the proton, we can make use of the equation

$V_{f} = V_{i} + at$

Plugging the numbers in and simplifying gets us $V_{f} = (2.22) * 10^{4} m/s$

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3 years ago

Read 2 more answers