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3 years ago

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In orbit, gravity:

1 answer:

ZanzabumX [31]3 years ago

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AA this is only similarity

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Ian walks 2 km to his best friend's house, then walks 0.5 km to the library. He then makes a 2.5 km walk home. The entire walk t

earnstyle [38]

Average speed = (total distance covered) / (time to cover the distance)

Ian's total distance covered = (2km + 0.5km + 2.5km) = 5 km.

His time to cover the distance = 3 hours.

Average speed = (5 km) / (3 hrs)

Average speed = (5/3) (km/hr)

<em>Average speed = 1.67 km/hr</em>

5 0

2 years ago

Please help me ,create a slogan that promotes the role of internet in education

aleksklad [387]

Answer:

¨Facts you didn´t know¨ or ¨unknown facts¨

Explanation:

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3 years ago

The following are possible ways to express the quantity 0.391 (Give ALL correct answers, i.e., B, AC, BCD...) Note: 3.45E-8 is a

laiz [17]

To answer, evaluate the power of 10 in the given choices. If it is positve, move the decimal n places to the right. If it is negative, move the decimal n corresponding places to the left. From all the choices given, only the choices D, E, and F will give us the correct answer.

4 0

3 years ago

An electron is initially at rest in a uniform electric field having a strength of 1.85 × 106 V/m. It is then released and accele

kirza4 [7]

Answer:

$W = 462.5 keV$

Explanation:

As we know that when electron moved in electric field then work done by electric field must be equal to the change in kinetic energy of the electron

So here we have to find the work done by electric field on moving electron

So we have

$F = qE$

$F = (1.6 \times 10^{-19})(1.85 \times 10^6)$

$F = 2.96 \times 10^{-13} N$

now the distance moved by the electron is given as

$d = 0.25 m$

so we have

$W = F.d$

$W = (1.6 \times 10^{-19})(1.85 \times 10^6)(0.25)$

$W = 7.4 \times 10^{-14} J$

now we have to convert it into keV units

so we have

$1 keV = 1.6 \times 10^{-16} J$

$W = 462.5 keV$

5 0

3 years ago

A 3874-kg rollercoaster is brought to the top of a 42m hill in 40 seconds,then drops 28m before the next hill.

Ede4ka [16]

(a) The work required to get the coaster to the top of the first hill is 1,594,538.4 J.

(b) The power required to bring the train to the top of the first hill is 39,863.46 W.

(c) The energy lost when the coaster drops is 531,512.8 J.

(d) The left at the bottom is determined as 1,063,025.6 J.

<h3>Work done to bring the rollercoaster top of the hill</h3>

W = Fn x d = mgh

W = 3874 x 9.8 x 42

W = 1,594,538.4 J

<h3>Power dissipated in bringing the rollercoaster on top hill</h3>

P = Fv

P = Fd/t

P = W/t

P = 1,594,538.4 /40 = 39,863.46 W

<h3>Energy lost when the coaster drops</h3>

E = 1,594,538.4 - (3874 x 9.8 x 28)

E = 531,512.8 J

<h3>Energy left at the bottom</h3>

E = 3874 x 9.8 x 28 = 1,063,025.6 J

Learn more about energy here: brainly.com/question/13881533

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6 0

2 years ago