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2 years ago

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As an atom's nucleus gets larger the electric charges repelling the protons get larger. To compensate for this greater electric

charge the atom.
A.) needs more neutrons to gain enough nuclear binding forces to overcome
the repulsion.

B.) needs fewer neutrons to gain enough nuclear binding forces to overcome
the repulsion.

C.)needs more electrons to gain enough nuclear binding forces to overcome
the repulsion.

D.)needs fewer electrons to gain enough nuclear binding forces to overcome
the repulsion.

1 answer:

Nady [450]2 years ago

5 0

Answer:

b

Explanation:

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A dart is thrown at a dartboard 3.66 m away. When the dart is released at the same height as the center of the dartboard, it hit

lapo4ka [179]

Answer:

The angle is $\theta = 15.48^o$

Explanation:

From the question we are told that

The distance of the dartboard from the dart is $d = 3.66 \ m$

The time taken is $t = 0.455 \ s$

The horizontal component of the speed of the dart is mathematically represented as

$u_x = ucos \theta$

where u is the the velocity at dart is lunched

so

$distance = velocity \ in \ the\ x-direction * time$

substituting values

$3.66 = ucos \theta * (0.455)$

=> $ucos \theta = 8.04 \ m/s$

From projectile kinematics the time taken by the dart can be mathematically represented as

$t = \frac{2usin \theta }{g}$

=> $usin \theta = \frac{g * t}{2 }$

$usin \theta = \frac{9.8 * 0.455}{2 }$

$usin \theta = 2.23$

=> $tan \theta = \frac{usin\theta }{ucos \theta } = \frac{2.23}{8.04}$

$\theta = tan^{-1} [0.277]$

$\theta = 15.48^o$

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2 years ago

g a small smetal sphere, carrying a net charge is held stationarry. what is the speed are 0.4 m apart

weeeeeb [17]

Complete Question

A small metal sphere, carrying a net charge q1=−2μC, is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of q2= -8μC and mass 1.50g, is projected toward q1. When the two spheres are 0.80m apart, q2 is moving toward q1 with speed 20ms−1. Assume that the two spheres can be treated as point charges. You can ignore the force of gravity.The speed of q2 when the spheres are 0.400m apart is.

Answer:

The value $v_2 = 4 \sqrt{10} \ m/s$

Explanation:

From the question we are told that

The charge on the first sphere is $q_1 = 2\mu C = 2*10^{-6} \ C$

The charge on the second sphere is $q_2 = 8 \mu C = 8*10^{-6} \ C$

The mass of the second charge is $m = 1.50 \ g = 1.50 *10^{-3} \ kg$

The distance apart is $d = 0.4 \ m$

The speed of the second sphere is $v_1 = 20 \ ms^{-1}$

Generally the total energy possessed by when $q_2$ and $q_1$ are separated by $0.8 \ m$ is mathematically represented

$Q = KE + U$

Here KE is the kinetic energy which is mathematically represented as

$KE = \frac{1 }{2} m (v_1)^2$

substituting value

$KE = \frac{1 }{2} * ( 1.50 *10^{-3}) (20 )^2$

$KE = 0.3 \ J$

And U is the potential energy which is mathematically represented as

$U = \frac{k * q_1 * q_2 }{d }$

substituting values

$U = \frac{9*10^9 * 2*10^{-6} * 8*10^{-6} }{0.8 }$

$U = 0.18 \ J$

So

$Q = 0.3 + 0.18$

$Q = 0.48 \ J$

Generally the total energy possessed by when $q_2$ and $q_1$ are separated by $0.4 \ m$ is mathematically represented

$Q_f = KE_f + U_f$

Here $KE_f$ is the kinetic energy which is mathematically represented as

$KE_f = \frac{1 }{2} m (v_2^2$

substituting value

$KE_f = \frac{1 }{2} * ( 1.50 *10^{-3}) (v_2 )^2$

$KE_f = 7.50 *10^{ -4} (v_2 )^2$

And $U_f$ is the potential energy which is mathematically represented as

$U_f = \frac{k * q_1 * q_2 }{d }$

substituting values

$U_f = \frac{9*10^9 * 2*10^{-6} * 8*10^{-6} }{0.4 }$

$U_f = 0.36 \ J$

From the law of energy conservation

$Q = Q_f$

So

$0.48 = 0.36 +(7.50 *10^{-4} v_2^2)$

$v_2 = 4 \sqrt{10} \ m/s$

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2 years ago

What is the Octet rule

egoroff_w [7]

Answer:

the octate rule says that electron after completing all the valency of the element doesnot goes as 9 but continues next round after completing the next octat then the first shell completes the all electron valency .thanks if you dont understand then please comment l will try to solve your confusion

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3 years ago

Two stars orbit each other in a _ solar system

Brilliant_brown [7]

This is a binary star system

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3 years ago

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A 100kg crate slides along a floor with a starting velocity of 21ms If the force due to friction is 8N how long will it take for

GaryK [48]

A 100kg crate slides along a floor with a starting velocity of 21 m/s. If the force due to friction is 8N, then, it will take 262.5 s for the box to come to rest.

We'll begin by calculating the declaration of the box. This can be obtained as follow:

Force (F) = –8 N (opposition)

Mass (m) = 100 Kg

<h3>Deceleration (a) =? </h3>

<h3>F = ma</h3>

–8 = 100 × a

Divide both side by 1000

$a = \frac{-8}{100}$

<h3>a = –0.08 ms¯²</h3>

Therefore, the deceleration of the box is –0.08 ms¯²

Finally, we shall determine the time taken for the box to come to rest. This can be obtained as follow:

Deceleration (a) = –0.08 ms¯²

Initial velocity (u) = 21 ms¯¹

Final velocity (v) = 0 ms¯¹

<h3>Time (t) =.? </h3>

<h3>v = u + at</h3>

0 = 21 + (–0.08×t)

0 = 21 – 0.08t

Collect like terms

0 – 21 = –0.08t

–21 = –0.08t

Divide both side by –0.08

$t = \frac{-21}{-0.08}\\\\$

<h3>t = 262.5 s</h3>

Therefore, it will take 262.5 s for the box to come to rest.

Learn more: brainly.com/question/14446351

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2 years ago