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Hatshy [7]
3 years ago
8

Please help with this​

Physics
1 answer:
hichkok12 [17]3 years ago
5 0

Answer:

3/5i + 4/5J

Explanation:

use the commtative property to reorder the terms

3i + 4 j

to

4j + 3i

The conjugate of a complex number a=b is the number a-bi

4j -3i

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An astronaut has a momentum of 280 kg and travels 10 m/s. what is the mass of the astronaut?
Kamila [148]

Answer:

The answer is

<h2>28 kg</h2>

Explanation:

The mass of an object given it's momentum and velocity / speed can be found by using the formula

m =  \frac{p}{v}  \\

where

m is the mass

p is the momentum

v is the speed or velocity

From the question

p = 280 kg/ms

v = 10 m/s

The mass of the object is

m =  \frac{280}{10}  = 28 \\

We have the final answer as

<h3>28 kg</h3>

Hope this helps you

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3 years ago
When using the scientist method, which step comes last?
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Here, I hope this helps.

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Which atomic model proposed that electrons move in specific orbits around the nucleus of an atom
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Bohr's atomic model
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Suppose you wish to fabricate a uniform wire out of 1.10 g of copper. If the wire is to have a resistance R = 0.390 Ω, and if al
Citrus2011 [14]

To solve this problem we will apply the concepts related to volume, as a function of length and area, as of mass and density. Later we will take the same concept of resistance and resistivity, equal to the length per unit area. Once obtained from the known constants it will be possible to obtain the area by matching the two equations:

Mass of copper wire(m) = 1.10g = 1.10*10^{-3} kg

Density (\rho)= 8.92*10^3kg/m^3

Resistively of copper (\gamma) = 1.7*10^{-8}\Omega \cdot m

Resistance (R) = 0.390\Omega

Volume is defined as,

V= lA \text{ and }\frac{m}{\rho}

lA= \frac{1.10*10^{-3}}{8.92*10^3}

lA = 1.233*10^{-7} m^3 (1)

We know that,

\frac{l}{A} = \frac{R}{\gamma}

\frac{l}{A}= \frac{0.390\Omega}{1.7*10^{-8}\Omega m}

\frac{l}{A} = 2.2941*10^7 m^{-1} (2)

Multiplying equation we have

l^2 = (1.233*10^{-7})( 2.2941*10^7)

l^2 = 2.8286m^2

l =\sqrt{2.8286m^2}

l = 1.68m

Therefore the length of the wire is 1.68m

6 0
2 years ago
What distance is required for a train to stop if its initial velocity is 23 m/s and its
Varvara68 [4.7K]

-- The train starts at 23 m/s and slows down by 0.25 m/s every second.

So it'll take (23/0.25) = 92 seconds to stop.

-- Its average speed during that time will be (1/2)(23+0) = 11.5 m/s

-- Moving at an average speed of 11.5 m/s for 92 sec, the train will cover

(11.5 m/s) x (92 sec)  =  <em>1,058 meters</em> .

7 0
2 years ago
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