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2 years ago

Which statement best describes metallic bonding

Physics

1 answer:

horrorfan [7]2 years ago

4 0

Answer:

It's a type of chemical bonding that rises from the electrostatic attractive force between conduction electrons and positively charged metal bars. It can also be described as the sharing of free electrons among a structure of positively charged ions

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Can someone explain to how to calculate this

Karo-lina-s [1.5K]

answer

option d is the correct answer

explanation

as we know frequency is equal to 1 /t

f= 457 Hz

t=1

SO, 1/457

=0.0022sev

3 0

3 years ago

A spring has a force constant k, and an object of mass m is suspended from it. The spring is cut in half and the same object is

kenny6666 [7]

Answer:

$f2/f1 = \sqrt{2}$

Explanation:

From frequency of oscillation

$f = 1/2pi *\sqrt{k/m}$

Initially with the suspended string, the above equation is correct for the relation, hence

$f1 = 1/2pi *\sqrt{k/m}$

where k is force constant and m is the mass

When the spring is cut into half, by physics, the force constant will be doubled as they are inversely proportional

$f2 = 1/2pi *\sqrt{2k/m}$

Employing f2/ f1, we have

$f2/f1 = \sqrt{2}$

3 0

2 years ago

Two flat 4.0 cm × 4.0 cm electrodes carrying equal but opposite charges are spaced 2.0 mm apart with their midpoints opposite ea

serious [3.7K]

Answer:

1.77 x 10^-8 C

Explanation:

Let the surface charge density of each of the plate is σ.

A = 4 x 4 = 16 cm^2 = 16 x 10^-4 m^2

d = 2 mm

E = 2.5 x 10^6 N/C

ε0 = 8.85 × 10-12 C2/N ∙ m2

Electric filed between the plates (two oppositively charged)

E = σ / ε0

σ = ε0 x E

σ = 8.85 x 10^-12 x 2.5 x 10^6 = 22.125 x 10^-6 C/m^2

The surface charge density of each plate is ± σ / 2

So, the surface charge density on each = ± 22.125 x 10^-6 / 2

= ± 11.0625 x 10^-6 C/m^2

Charge on each plate = Surface charge density on each plate x area of each plate

Charge on each plate = ± 11.0625 x 10^-6 x 16 x 10^-4 = ± 1.77 x 10^-8 C

7 0

3 years ago

1. A boat initially moving at 10 m/s accelerates at 2 m/s2 for 10 s. What is the velocity of the boat after 10 seconds?

DIA [1.3K]

Answer:

5 m/s is the answer so yeah

5 0

2 years ago

A box has a weight of 150 N and is being pulled across a horizontal floor by a force that has a magnitude of 110 N. The pulling

ivann1987 [24]

Answer:

42.99°

Explanation:

$F_h$ = Kinetic friction force

$F_{\theta}$ = Pulling force at angle $\theta$

$N_h$ = Weight of the box = 150 N

Kinetic friction force

$F_h=\muN_h$

Pulling force at angle $\theta$

$F_{\theta}=\muN_{\theta}$

N = Pulling force

According to question

$\frac{F_h}{F_{\theta}}=\frac{2}{1}\\\Rightarrow \frac{\muN_h}{\muN_{\theta}}=2\\\Rightarrow \frac{N_h}{N_{\theta}}=2\\\Rightarrow N_{\theta}=\frac{N_h}{2}\\\Rightarrow N_{\theta}=\frac{150}{2}\\\Rightarrow N_{\theta}=75\ N$

Applying Newton's second law in the vertical direction we get

$N_h-Nsin\theta=N_{\theta}\\\Rightarrow 150-110sin\theta=75\\\Rightarrow \theta=sin^{-1}\frac{75}{110}\\\Rightarrow \theta=42.99\ ^{\circ}$

The angle is 42.99°

8 0

3 years ago