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3 years ago

An atom gains an additional electron. What is the overall charge of the ion that is formed?

2 answers:

8 0

An is formed when an atom loses or gains one or more electrons. Because the number of electrons in an ion is different from the number of protons, an ion does have an overall electric charge. Consider how a positive ion can form from an atom. The left side of the illustration below represents a sodium (Na) atom

german3 years ago

7 0

It is a negative ion

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What Mechanisms give rise to magnetism

MrRa [10]

Answer:

Explanation:Magnetism is a class of physical attributes that are mediated by magnetic fields. Electric currents and the magnetic moments of elementary particles give rise to a magnetic field, which acts on other currents and magnetic moments. Magnetism is one aspect of the combined phenomenon of electromagnetism.

4 0

3 years ago

The rigid beam is supported by the three suspender bars. bars ab and ef are made of aluminum and bar cd is made of steel. if eac

faltersainse [42]

Answer:

Pmax = 67.5 KN

Explanation:

We need to calculate the maximum allowable value of P for both aluminum and steel bars.

FOR STEEL BARS:

Since,

(σallow)st = (Pmax)st/A

where,

(σallow)st = maximum allowable stress of steel bar = 200 MPa = 2 x 10⁸ Pa

A = Cross-sectional area of steel bar = 450 mm² = 0.45 x 10⁻³ m²

(Pmax)st = Maximum allowable force for steel bar = ?

Therefore,

2 x 10⁸ Pa = (Pmax)st/0.45 x 10⁻³ m²

(Pmax)st = (2 x 10⁸ Pa)(0.45 x 10⁻³ m²)

(Pmax)st = 9 x 10⁴ N = 90 KN

FOR Aluminum BARS:

Since,

(σallow)al = (Pmax)al/A

where,

(σallow)al = maximum allowable stress of Al bar = 150 MPa = 1.5 x 10⁸ Pa

A = Cross-sectional area of Aluminum bar = 450 mm² = 0.45 x 10⁻³ m²

(Pmax)al = Maximum allowable force for Aluminum bar = ?

Therefore,

1.5 x 10⁸ Pa = (Pmax)al/0.45 x 10⁻³ m²

(Pmax)al = (1.5 x 10⁸ Pa)(0.45 x 10⁻³ m²)

(Pmax)al = 6.75 x 10⁴ N = 67.5 KN

Since,

(Pmax)al < (Pmax)st

Therefore,

The maximum allowable force will be:

Pmax = (Pmax)al

Pmax = 67.5 KN

3 0

4 years ago

A less-dense liquid of density rho1 floats on top of a more-dense liquid of density rho2. A uniform cylinder of length l and den

Umnica [9.8K]

Answer:

Explanation:

Given

$\rho=density\ of\ cylinder\\\rho_1=less\ dense\ cylinder\\\rho_2=more\ dense\ cylinder$

Suppose V is the volume of a cylinder

Also $L_1=length\ in\ less\ dense\ part\\L_2=length\ in\ dense\ part\\L=length\ of\ cylinder\\V=A\times L\\where\ A=area\ of\ cross-section\\$

Now, we can write

The weight of the cylinder is supported by the buoyant forces of two liquids

$\rho Vg=\rho_1 V_1g+\rho_2 V_2g\\\rho ALg=\rho_1 A\times L_1g+\rho_2 A\times L_2g\\\\\rho L=\rho_1 L_1+\rho_2 L_2\\Also, L=L_1+L_2$

using the above equation we can write

$L_2=\frac{\rho-\rho_1}{\rho_2-\rho_1}\cdot L\\\frac{L_2}{L}=\frac{\rho-\rho_1}{\rho_2-\rho_1}$

5 0

3 years ago

A slender rod is 80.0 cm long and has mass 0.390 kg . A small 0.0200-kg sphere is welded to one end of the rod, and a small 0.05

Lilit [14]

Answer:

v = 1.08 m/s

Explanation:

What is the linear speed of the 0.0500-kg sphere as its passes through its lowest point?

The decrease in PE is

d = 80.0cm * 1 / 1000m = 0.80m

h = 0.80 m /2 = 0.40 m

ΔPE = m*g*h

ΔPE = (0.0500 - 0.0200)kg * 9.8m/s² * 0.400 m

ΔPE = 0.1176 J

The moment of inertia of the assembly is

I = 1/12*m*L² + (m1 + m2)*(L/2)²

I = 1/12*0.390kg*(0.800m)² + 0.0700kg*(0.400m)²

I = 0.032 kg·m²

KE = ½Iω²

0.1176 J = ½ * 0.032kg·m² * ω²

ω = 2.71 rad/s

v = ωr = 2.71 rad/s * 0.400m

The linear velocity

v = 1.08 m/s

3 0

3 years ago

Two particles, each with charge Q, and a third charge q, are placed at the vertices of an equilateral triangle as shown. The tot

Gelneren [198K]

Answer:

D. The total force on the particle with charge q is perpendicular to the bottom of the triangle.

Explanation:

The image is shown below.

The force on the particle with charge q due to each charge Q = $\frac{kQq}{r^{2} }$

we designate this force as N

Since the charges form an equilateral triangle, then, the forces due to each particle with charge Q on the particle with charge q act at an angle of 60° below the horizontal x-axis.

Resolving the forces on the particle, we have

for the x-component

$N_{x}$ = N cosine 60° + (-N cosine 60°) = 0

for the y-component

$N_{y}$ = -f sine 60° + (-f sine 60) = -2N sine 60° = -2N(0.866) = -1.732N

The above indicates that there is no resultant force in the x-axis, since it is equal to zero ( $N_{x}$ = 0).

The total force is seen to act only in the y-axis, since it only has a y-component equivalent to 1.732 times the force due to each of the Q particles on q.

The total force on the particle with charge q is therefore perpendicular to the bottom of the triangle.

5 0

4 years ago