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noname [10]
3 years ago
7

A woman flies 300 kilometers north, 200 kilometers west, and 500 kilometers south. How far is she from her starting point?

Physics
1 answer:
maria [59]3 years ago
5 0
She is
\sqrt{200^2+200^2}=  ~282.84 kilometers away from her starting point
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What is the magnification of an astronomical telescope whose objective lens has a focal length of 71 cm and whose eyepiece has a
Rus_ich [418]

To solve this problem it is necessary to apply the concepts related to optical magnification (is the process of enlarging the apparent size, not physical size, of something.). Specifically the angular magnification of an optical telescope is given by

M = -\frac{f_o}{f_e}

Where,

f_o = Focal length of the objective lens in a refractor

f_e = Focal length of the eyepiece

Our values are given as

f_o = 71cm

f_e =  2.1cm

Replacing we have

M = -\frac{f_o}{f_e}

M = -\frac{71}{2.1}

M = - 33.81

Therefore the magnification of this astronomical telescope is -33.81

7 0
3 years ago
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zhuklara [117]

Answer:

In a column of fluid, pressure increases with depth as a result of the weight of the overlying fluid. Thus a column of fluid, or an object submerged in the fluid, experiences greater pressure at the bottom of the column than at the top. This difference in pressure results in a net force that tends to accelerate an object upwards.

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Granite: 2.70 × 10 32.70 × 10 3

Lead: 1.13 × 10 41.13 × 10 4

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4 0
3 years ago
State a situation in which force is applied on a body, but no work is done​
LenaWriter [7]
If you press on your arm force is applied work done is if it moves.
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8 0
3 years ago
Read 2 more answers
What are the magnitude and direction of the acceleration of an electron at a point where the electric field has magnitude 7400 N
bezimeni [28]

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Explanation: The magnitude of the force required to move the electron is given as

F = ma

The force exerted on the charge by the electric field of intensity (E) is given by

F = Eq

Thus

Eq = ma

a = E * q/ m

Where a = acceleration of charge

E = strength of electric field = 7400N/c

q = magnitude of electronic charge = 1.609 * 10^-6c

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a = 7400 * 1.609 * 10^-16/ 9.109 * 10^-31

a = 11906.6 * 10^-16 / 9.019 * 10^-31

a = 1.19 * 10^-12 / 9.019 * 10^-31

a = 0.132 * 10^19

a = 1.32 * 10^18m/s²

As stated in the question, the direction of the electric field is due north hence, the direction of it force will also be north thus making the electron experience a force due north ( according to Newton second law of motion)

5 0
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What statements correctly describe magnetism?
insens350 [35]
A best describes magnetism
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