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myrzilka [38]
3 years ago
13

A speed-time graph is a horizontal line with a y-value of 4. Which describes the objects motion?

Physics
1 answer:
taurus [48]3 years ago
5 0
It means the speed is constant with a value of 4 units.
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Which of the following is formed by constructive erosion? 1. delta 2. gully 3. rill 4. glacier
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1. Delta, is formed by constructive erosion.
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Most earthquakes occur along or near the edges of the earth's
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What is the main reason water from the ocean turns to water vapor , and then evaporates into the air
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C) heat from the sun

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Because the sun’s heat makes the water turn to steam—or what we call “water vapor.”

Hope this helped and I hope I answered in time!

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2 years ago
Read 2 more answers
As the launch force increase the launch velocity will
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As the launch force increase the launch velocity will

<em><u>Increase</u></em>

The reason for your answer to number six is because

<em><u>There is a direct relationship between force and acceleration.</u></em>

<em><u /></em>

Explanation:

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5 0
2 years ago
Each plate of a parallel‑plate capacitor is a square of side 4.19 cm, 4.19 cm, and the plates are separated by 0.407 mm. 0.407 m
alexandr1967 [171]

Answer:

The electric field strength inside the capacitor is 49880.77 N/C.

Explanation:

Given:

Side length of the capacitor plate (a) = 4.19 cm = 0.0419 m

Separation between the plates (d) = 0.407 mm = 0.407\times 10^{-3}\ m

Energy stored in the capacitor (U) = 7.87\ nJ=7.87\times 10^{-9}\ J

Assuming the medium to be air.

So, permittivity of space (ε) = 8.854\times 10^{-12}\ F/m

Area of the square plates is given as:

A=a^2=(0.0419\ m)^2=1.75561\times 10^{-3}\ m^2

Capacitance of the capacitor is given as:

C=\dfrac{\epsilon A}{d}\\\\C=\frac{8.854\times 10^{-12}\ F/m\times 1.75561\times 10^{-3}\ m^2 }{0.407\times 10^{-3}\ m}\\\\C=3.819\times 10^{-11}\ F

Now, we know that, the energy stored in a parallel plate capacitor is given as:

U=\frac{CE^2d^2}{2}

Rewriting in terms of 'E', we get:

E=\sqrt{\frac{2U}{Cd^2}}

Now, plug in the given values and solve for 'E'. This gives,

E=\sqrt{\frac{2\times 7.87\times 10^{-9}\ J}{3.819\times 10^{-11}\ F\times (0.407\times 10^{-3})^2\ m^2}}\\\\E=49880.77\ N/C

Therefore, the electric field strength inside the capacitor is 49880.77 N/C

8 0
3 years ago
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