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3 years ago

15

The voltage in the lines that carry electric power to homes is typically 2000 V. What is the required ratio of the loops in the

primary and secondary coils of the transformer to drop the voltage to 120 V?

2 answers:

Sav [38]3 years ago

7 0

Answer:

b

Explanation:

EDGE 2021

mariarad [96]3 years ago

6 0

Answer:

yurrrrr 221n54

Explanation:

yurrrrrrrrrrrrrrrrrrrrrr

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If a high jumper needs to make his center of gravity rise 1.50 m, how fast must he be able to sprint? Assume all of his kinetic

sergiy2304 [10]

Answer:

v = 5.42 m/s

Explanation:

given,

height of the jumper = 1.5 m

velocity of sprinter = ?

kinetic energy can be transformed into potential energy

$m g h = \dfrac{1}{2}mv^2$

$g h = \dfrac{1}{2}v^2$

$v =\sqrt{2gh}$

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v = 5.42 m/s

Speed of the sprinter is equal to v = 5.42 m/s

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Newton's first law of motion is the law of inertia. What does it state?

vladimir2022 [97]

Answer:

b is the right answer

Explanation:

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3 years ago

"What is the magnifying power of an astronomical telescope using a reflecting mirror whose radius of curvature is 5.9 m and an e

notka56 [123]

Answer:

The Magnifying power of a telescope is $M = 109.26$

Explanation:

Radius of curvature R = 5.9 m = 590 cm

focal length of objective $f_{objective}$ = $\frac{R}{2}$

⇒ $f_{objective}$ = $\frac{590}{2}$

⇒ $f_{objective}$ = 295 cm

Focal length of eyepiece $f_{eyepiece}$ = 2.7 cm

Magnifying power of a telescope is given by,

$M = \frac{f_{objective} }{f_{eyepiece} }$

$M = \frac{295}{2.7}$

$M = 109.26$

therefore the Magnifying power of a telescope is $M = 109.26$

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