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Igoryamba
3 years ago
10

IE is the energy required to remove an electron from an atom. As atomic radius increases, the valence electrons get farther from

the nucleus. How do you think an atom’s size will affect its ability to hold on to its valence electrons? Why?
Physics
1 answer:
Bond [772]3 years ago
4 0

Answer:

The bigger the atom the lesser the ability of the atom to hold on to its valence electrons.

Explanation:

Atomic radius can be looked at as the distance between the nucleus and the outermost energy level. As an atom gets bigger, the outer shell gets further and further from the positive nucleus. this means that electrons that are in the outer energy level become less held (attracted) by the nucleus because of distance and shielding of the attractive forces by the electrons in the lower energy levels. This means that as an atom becomes bigger, its ability to hold on to its outer electrons lessens.

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A 15 kg box is sliding down an incline of 35 degrees. The incline has a coefficient of friction of 0.25. If the box starts at re
valina [46]

The box has 3 forces acting on it:

• its own weight (magnitude <em>w</em>, pointing downward)

• the normal force of the incline on the box (mag. <em>n</em>, pointing upward perpendicular to the incline)

• friction (mag. <em>f</em>, opposing the box's slide down the incline and parallel to the incline)

Decompose each force into components acting parallel or perpendicular to the incline. (Consult the attached free body diagram.) The normal and friction forces are ready to be used, so that just leaves the weight. If we take the direction in which the box is sliding to be the positive parallel direction, then by Newton's second law, we have

• net parallel force:

∑ <em>F</em> = -<em>f</em> + <em>w</em> sin(35°) = <em>m a</em>

• net perpendicular force:

∑<em> F</em> = <em>n</em> - <em>w</em> cos(35°) = 0

Solve the net perpendicular force equation for the normal force:

<em>n</em> = <em>w</em> cos(35°)

<em>n</em> = (15 kg) (9.8 m/s²) cos(35°)

<em>n</em> ≈ 120 N

Solve for the mag. of friction:

<em>f</em> = <em>µ</em> <em>n</em>

<em>f</em> = 0.25 (120 N)

<em>f</em> ≈ 30 N

Solve the net parallel force equation for the acceleration:

-30 N + (15 kg) (9.8 m/s²) sin(35°) = (15 kg) <em>a</em>

<em>a</em> ≈ (54.3157 N) / (15 kg)

<em>a</em> ≈ 3.6 m/s²

Now solve for the block's speed <em>v</em> given that it starts at rest, with <em>v</em>₀ = 0, and slides down the incline a distance of ∆<em>x</em> = 3 m:

<em>v</em>² - <em>v</em>₀² = 2 <em>a</em> ∆<em>x</em>

<em>v</em>² = 2 (3.6 m/s²) (3 m)

<em>v</em> = √(21.7263 m²/s²)

<em>v</em> ≈ 4.7 m/s

4 0
2 years ago
How can icalculate 3 seconds and 5 meters into a speed?
topjm [15]
To find speed you have to divide distance by time. In this case:

5 meters➗3 seconds = about 1.66666666 and so on m/s.

You could round to 1.67 or 1.7 if you'd like.
8 0
3 years ago
d is at rest at the top of a 2 m high slope. The sled has a mass of 45 kg. The sled's potential energy is J?
Anna71 [15]
<span>D is at rest at the top of a 2 m high slope. The sled has a mass of 45 kg. The sled's potential energy is J?
</span>Answer: The sled's potential energy is 882 Joules
5 0
3 years ago
Read 2 more answers
The velocity of a car doubles over a short period of time. Which additional information is sufficient to determine the amount of
Rom4ik [11]

Answer:

Time and velocity

Explanation:

The time taken for the velocity to double is very important to find the amount of acceleration the car acquires.

Acceleration is the rate of change of velocity with time.

  Acceleration  = \frac{v -u}{t}

v is the initial velocity

u is the final velocity

t is the time taken

 So, the velocity and time is needed to calculate the value of the acceleration the car undergoes.

4 0
3 years ago
A horizontal insulating rod of length 11.8-cm and charge 19 nC is in a plane with a long straight vertical uniform line charge.
SCORPION-xisa [38]

Answer:

11.962337 × 10^-4 N

Explanation:

Given the following :

Length L = 11.8

Charge = 29nC = 29 × 10^-9 C

Linear charge density λ = 1.4 × 10^-7 C/m

Radius (r) = 2cm = 2/100 = 0.02 m

Using the relation:

E = 2kλ/r ; F =qE

F = 2kλq/L × ∫dr/r

F = 2*k*q*λ/L × (In(0.02 + L) - In(0.02))

2*k*q*λ/L = [2 × (9 * 10^9) * (29 * 10^9) * (1.4 * 10^-7)]/ 0.118] = 6193.2203 × 10^(9 - 9 - 7) = 6193.2203 × 10^-7 = 6.1932203 × 10^-4

In(0.02 + 0.118) - In(0.02) = In(0.138) - In(0.02) = 1.9315214

Hence,

(6.1932203 × 10^-4) × 1.9315214 = 11.962337 × 10^-4 N

3 0
3 years ago
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