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Gelneren [198K]
3 years ago
15

An airplane starts at rest and accelerates at 6.8 m/s2 at an angle of 39° south of west. After 9 s, how far in the westerly dire

ction has the airplane traveled?
Physics
1 answer:
cestrela7 [59]3 years ago
7 0

Answer:36.8

Explanation:

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E=<br> (500.0lm)<br> 4 (100)
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Answer:

didn't understand your question

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3 years ago
Find the speed for Barkely, the dog, if he runs 5.5 miles in 3 hours
amm1812

Answer: B

Explanation:

5.5/3

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If you shine a single light source on a screen you will see that the entire screen is lit up. Assume this light source is of a s
kompoz [17]

Answer:

An interference pattern.

Explanation:

When we have two light source of the same frequency turned on close to each other, the light emitted by them will interfere since light is also a wave. This means that an interference pattern will appear in a screen put ahead of them, that is, bands of light and darkness where the waves are interfering constructively and destructively.

8 0
3 years ago
The big bang theory is one of the most accepted theories on the origin of the universe because of scientific evidence, such as _
solmaris [256]
<span>The big bang theory is one of the most accepted theories on the origin of the universe because of scientific evidence, such as d: distant, exploding quasars were found   </span>
3 0
3 years ago
Read 2 more answers
A tube of mercury with resistivity 9.84 × 10 -7 Ω ∙ m has an electric field inside the column of mercury of magnitude 23 N/C tha
slava [35]

Answer:

The current through the tube is 73.39A.

Explanation:

The relationship between the resistivity \rho, the electric field E, and the current density J is given by

\rho = \dfrac{E}{J}

This equation can be solved for J to get:

J = \dfrac{E}{\rho}

Since the current is I = J\cdot A

I= J\cdot A  = \dfrac{E}{\rho} \cdot A

Now, for the tube of mercury \rho = 9.84*10^{-7}\: \Omega \cdot m, E = 23N/C, and the area is A = \pi r^2 = \pi (1.0*10^{-3}m)^2 = 3.14*10^{-6}m^2; therefore,

I= \dfrac{23N/C}{9.84*10^{-7}\Omega\cdot m } *3.14*10^{-6}m^2

\boxed{I = 73.39A.}

Hence, the current through the mercury tube is 73.39A.

5 0
3 years ago
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