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Eduardwww [97]
1 year ago
14

If a car drives 10 miles NE and then due East for 10 minutes traveling 8

Physics
2 answers:
Nimfa-mama [501]1 year ago
7 0
I think the answer is 28 miles
abruzzese [7]1 year ago
4 0

Answer:

<h2>12.8 miles</h2>

Explanation:

d=\sqrt{a^{2}+b^{2}  } \\where;\\a=10\\b=8

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speed of sound is 343 Ms at 20 degrees Celsius. The frequency heard from the sound is 256 Hz. what is the sounds wavelength?
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S= 343m/s
F=256Hz

WL= 343ms/256-1
WL=V/F

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What us meant by Velocity ratio of simple machine is 4?<br><br><br>​
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Which is the BEST definition of refraction? A) Light or sound waves change direction. B) Light or sound waves bounce off a mediu
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4 0
3 years ago
Read 2 more answers
What is the most abundant gas in the atmosphere?
Snowcat [4.5K]

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8 0
3 years ago
A +71 nC charge is positioned 1.9 m from a +42 nC charge. What is the magnitude of the electric field at the midpoint of these c
viva [34]

Answer:

The net Electric field at the mid point is 289.19 N/C

Given:

Q = + 71 nC = 71\times 10^{- 9} C

Q' = + 42 nC = 42\times 10^{- 9} C

Separation distance, d = 1.9 m

Solution:

To find the magnitude of electric field at the mid point,

Electric field at the mid-point due to charge Q is given by:

\vec{E} = \frac{Q}{4\pi\epsilon_{o}(\frac{d}{2})^{2}}

\vec{E} = \frac{71\times 10^{- 9}}{4\pi\8.85\times 10^{- 12}(\frac{1.9}{2})^{2}}

\vec{E} = 708.03 N/C

Now,

Electric field at the mid-point due to charge Q' is given by:

\vec{E'} = \frac{Q'}{4\pi\epsilon_{o}(\frac{d}{2})^{2}}

\vec{E'} = \frac{42\times 10^{- 9}}{4\pi\8.85\times 10^{- 12}(\frac{1.9}{2})^{2}}

\vec{E'} = 418.84 N/C

Now,

The net Electric field is given by:

\vec{E_{net}} = \vec{E} - \vec{E'}

\vec{E_{net}} = 708.03 - 418.84 = 289.19 N/C

5 0
2 years ago
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