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SIZIF [17.4K]
3 years ago
15

Chemistry The iron atom (Fe) has 26 protons, 30 neutrons, and 26 electrons. The diameter of the atom is approximately 1.0 × 10 −

10 m , while the diameter of its nucleus is about 9.2 × 10 − 15 m . (You can reasonably model the nucleus as a uniform sphere of charge.) What are the magnitude and direction of the electric field that the nucleus produces (a) just outside the surface of the nucleus and (b) at the distance of the outermost electron? (c) What would be the magnitude and direction of the acceleration of the outermost electron due only to the nucleus, neglecting any force due to the other electrons? SSM
Physics
1 answer:
Alisiya [41]3 years ago
4 0
Don’t mind this i just need to answer under something because i just signed up !
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Answer:

B. Gravitational force

Explanation:

The gravitational force of an object (also called weight) is given by

F=mg

where m is the mass of the object and g is the acceleration of gravity, and it always points in the downward direction.

In this problem, the arrow labelled with 4 is the only force pointing downward: therefore, it must be the gravitational force.

The other arrows represent:

1 --> component of the gravitational force parallel to the plane

2 --> force of friction

3 --> component of the gravitational force perpendicular to the plane

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What is the type of intermolecular force in methylated spirits
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3 years ago
A 1.0-in.-diameter hole is drilled on the centerline of a long, flat steel bar that is 1 2 thick and 4 in. wide. The bar is subj
Dominik [7]

Answer:

The answers are

The average stress = 20000 lb/in²

The maximum tensile stress immediately adjacent to the hole

= 31076.92 lb/in²

Explanation:

To solve the question we have

Weight of tensile load = 30,000 lb

Width of steel bar = 4 in

Thickness of steel bar = 1/2 in

Average Stress = Force/Area  

Size of hole drilled = 1.0 in diameter

Available width at cross section where the 1.00 in diameter hole is drilled =

(4 - 1) in = 3 inches

Cross sectional area at the point of reduced cross section due to the drilled hole = Width × Thickness (Since the item is a flat bar)

= 3 in × 1/2 in = 1.5 in²

Therefore Stress = (30000 lb)/(1.5 in²) = 20000 lb/in²

the maximum tensile stress immediately adjacent to the hole.

Bending stress = \sigma_B= \frac{M_y}{I} where I = \frac{(0.5^2 + 4^2)}{12}

0.5*30000/I = 11076.92 lb/in²

Max stress = 31076.92 lb/in²

8 0
3 years ago
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an athlete whirls a 7.00 kg hammer tied to the end of a 1.3 m chain in a horizontal circle the hammer moves at the rate of 1.0 r
ElenaW [278]

Answer:

The answer to your question is a = 1.3 m/s²

Explanation:

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Data

mass = 7 kg

radius = r = 1.3 m

angular rate = w = 1.0 rev/s

centripetal acceleration = a = ?

Formula

a = rw²

Substitution

a = (1.3)(1)²

Simplification and result

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