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3 years ago

An Airplane taxis onto the runway going at 10 m/s. If it can accelerate steady at 3

Physics

1 answer:

kirza4 [7]3 years ago

4 0

Answer: 1333 m

the length of runway it will need is S = $\frac{90^{2}-10^{2} }{2.3}=1333$ (m)

Explanation:

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horsena [70]

Coke has more fizz than Pepsi, because Coke has more carbonation in it. Pepsi contains more sugar (2 more tablespoons) than Coke, so it tastes slightly sweeter to many people.

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One of the longer sides of a kite is 39 feet. one of the shorter sides is 25 feet. half the length of the shorter diagonal is 15

pychu [463]

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3 years ago

A proton and an electron traveling with the same velocity enter a uniform electric field. compared to the acceleration of the pr

kenny6666 [7]

The acceleration of the electron is larger than the acceleration of the proton.

The reason for this is that the mass of the electron is smaller (about 1000 times smaller) than the mass of the proton. The two particles have same charge (e), so they experience the same force under the same electric field E:
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2 years ago

Question 8

FinnZ [79.3K]

Answer:

A hope this helps

Explanation:

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2 years ago

Explain how mirrors can produce images that are larger or smaller than life size, as well as upright or inverted

galina1969 [7]

Answer:

1) When $d_{o}$ < $d_{i}$ (hence $d_{o}$ < f ) and they are both in front of the mirror (positive), the image will be larger and inverted

2) When $d_{o}$ > $d_{i}$ (and $d_{o}$ < f ) such that they are both positive (in front of the mirror), the image will be smaller and inverted

3) When the image is behind the mirror, for convex mirrors and the object is in front the image will be uptight. The magnification of the image will be the ratio of the image distance to the object distance from the mirror

Explanation:

The position of an object in front of a concave mirror of radius of curvature, R, determines the size and orientation of the image of the object as illustrated in the mirror equation

$\dfrac{1}{f}=\dfrac{1}{d_{o}} + \dfrac{1}{d_{i}}$

$Magnification, \, m = \dfrac{h_{i}}{h_{o}} = -\dfrac{d_{i}}{d_{o}}$

Where:

f = Focal length of the mirror = R/2

$d_{i}$ = Image distance from the mirror

$d_{o}$ = Object distance from the mirror

$h_{i}$ = Image height

$h_{o}$ = Object height

$d_{o}$ is positive for an object placed in front of the mirror and negative for an object placed behind the mirror

$d_{i}$ is positive for an image formed in front of the mirror and negative for an image formed behind the mirror

m is positive when the orientation of the image and the object is the same

m is negative when the orientation of the image and the object is inverted

f and R are positive in the situation where the center of curvature is located in front of the mirror (concave mirrors) and f and R are negative in the situation where the center of curvature is located behind the mirror (convex mirrors)

∴ When $d_{o}$ < $d_{i}$ (hence $d_{o}$ < f ) and they are both in front of the mirror (positive), the image will be larger and inverted

When $d_{o}$ > $d_{i}$ (and $d_{o}$ < f ) such that they are both positive (in front of the mirror), the image will be smaller and inverted

When the image is behind the mirror, for convex mirrors and the object is in front the image will be uptight. The magnification of the image will be the ratio of the image distance to the object distance from the mirror.

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3 years ago