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3 years ago

8

DONT ANSWER IF YOU DONT KNOW. For a person who is farsighted, what would they need to have a correct focal point?

1 answer:

fiasKO [112]3 years ago

5 0

A) Convex Lens.... hope it helps :)

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How do we know Earth is spinning, not the Sun?

Tpy6a [65]

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Because it is said that the earth rotates and revolves around the sun and moon so it is impossible for the earth not to be spinning.

Explanation:

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3 years ago

Benjamina started her walk from the front door of her ground floor apartment. She walked 6 meters to the corner of the building

Cloud [144]

The distance is the total distance she walked which is 16 meters adding the 6 meters to the corner and 10 meters to her friend's apartment. Her displacement is the distance from her original starting point so you set up a triangle with side lengths of 6 and 10 and solve for the hypotenuse which gives you a displacement of 11.66 meters.

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4 years ago

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5 ways in which people can explore space without physically <br> going there

goblinko [34]

Telescope
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3 years ago

I need some help with my science homework about energy, work, and power. It would be greatly appreciated :)

fgiga [73]

Answer:

1. B

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Work = Force × Distance

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4 years ago

An unstable atomic nucleus of mass 1.58 10-26 kg initially at rest disintegrates into three particles. One of the particles, of

lara31 [8.8K]

Answer:

a The velocity of the third particle is $v_3 = (-15.629*10^{6}i -14.45*10^{6}j)\ m/s$

b The total kinetic energy increase $E = 6.50 *10^{-13} J$

Explanation:

To solve this question we are going to be using the concept of conservation of momentum which is mathematically represented as

$m_1v_1 +m_2v_2 +m_3v_3 = 0$

From the question

Mass of unstable atomic nucleus $m_0= 1.58 *10^{-26} kg$

Mass of x-axis particle $m_1 = 8.44* 10^{-27} kg$

Velocity of x-axis particle $v_ 1 = 4.00* 10^6 m/s$

Mass of y-axis particle $m_2 =5.20* 10^{-27} kg$

Velocity of y-axis particle $v_2 = 6.00 *10^6 m/s$

Since the initial mass is known we can obtain the mass of the third particle

Using the conservation of mass mathematical expression

$m_3 = m_0 - (m_1 +m_2)$

$m_3 = 1.58 *10^{-26} - ( 8.44* 10^{-27} + 5.20* 10^-27)$

$= 2.16 *10 ^{-27}kg$

Using the mathematical expression above for conservation of momentum

$(8.44 * 10^{-27})( 4.00* 10^6i) + (5.20 *10^{-27})(6.00* 10^6j) +(2.16*0^{-27})v_3 =0$

$v_3 = (-15.629*10^{6}i -14.45*10^{6}j)\ m/s$

The above is the vector solution now to obtain the value in numeric form we evaluate the magnitude of the vector

$Mag = \sqrt{(15.629*10^6)^2 +(14.45*10^6)^2}$

$= 21.285*10^6 \ m/s$

To obtain the total increase in kinetic energy which is mathematically represented as

$E = \frac{1}{2}m_1v_1 + \frac{1}{2}m_2v_2 + \frac{1}{2} m_3 v_3$

$E =\frac{1}{2} [(5.20*10^{-27})(6.0*0^{6})^2 + (8.44*10^{-27})(4.00*10^6)^2+\\\\(2.16*10^{-27})(21.285*10^6)^2]$

$= 6.50 *10^{-13} J$

5 0

3 years ago