All categories

2 years ago

The radius of a lead atom is 175 pm. how many lead atoms would have to be laid side by side to span a distance of 6.11 mm?

Physics

1 answer:

podryga [215]2 years ago

7 0

The number of atoms needed to be laid side by side to span a distance of 6.11 mm is 17,457,143.

<h3>What is Unit conversion?</h3>

Unit conversion is a way of converting some common units into another without changing their real value. for, example, 1 centimeter is equal to 10 mm, though the real measurement is still the same the units and numerical values have been changed.

Given that the radius of a lead atom is 175 picometers. Therefore, the radius of the lead atom in mm can be written as,

1 picometer = 10⁻⁹ millimeter

175 picometer = 175×10⁻⁹ millimeter

The radius of the lead atom = 175×10⁻⁹ millimeter

Since the atoms are needed to be arranged side by side, therefore, the diameter of one lead atom will join with the other, thus, the diameter of the lead atom is,

Diameter of lead atom = 2 × 175 × 10⁻⁹ millimeter = 350 ×10⁻⁹ millimeter

Now, let the number of lead atoms be represented by n. Therefore, the number of atoms needed to be laid side by side to span a distance of 6.11 mm is,

6.11 mm = Number of atoms × Diameter of atoms

6.11 mm = n × 350 ×10⁻⁹ mm

Solving for n,

n = 6.11 mm / 350 ×10⁻⁹ mm

n = 17,457,142.86 ≈ 17,457,143

Hence, the number of atoms needed to be laid side by side to span a distance of 6.11 mm is 17,457,143.

Learn more about Units conversion here:

brainly.com/question/4736731

#SPJ1

You might be interested in

When two or more forces act together on an object what do they combine to form

Daniel [21]

Without knowing anything about their magnitudes or directions, the only thing you can always say about thier combination is that it's the "net force" on the object.

7 0

3 years ago

Determine the magnitude of the average friction force exerted on the collar when the velocity of the collar at c is 3.39 m/s and

djverab [1.8K]

The magnitude of the average friction force exerted on the collar (F)= 8.641 N

<h3>How can we calculate the magnitude of the average friction force exerted on the collar?</h3>

To calculate the magnitude of the average friction force exerted on the collar we are using the formula,

$\frac{1}{2} k(x^2_f - x^2_i ) + F\times y + \frac{1}{2} m v^{2} _{c} = mgy$

Here we are given,

k = The spring has a spring constant.

= 25.5 N/m.

$x_f$ = Final length of the spring .

= $\sqrt{1.25^2+1.8^2} -0.60$

= 1.591 m

$x_i$ = The initial length of the spring.

= 1.25−0.60

=0.65 m

y=The collar then travels downward a distance.

= 1.80 m.

m= The mass of the collar.

=3.55 kg

$v_c$ = the velocity of the collar.

= 3.39 m/s.

g = The acceleration due to gravity.

= 9.81 m/s²

We have to calculate the magnitude of the average friction force exerted on the collar = F

Now we put the known values in the above equation, we get;

$\frac{1}{2} k(x^2_f - x^2_i ) + F\times y + \frac{1}{2} m v^{2} _{c} = mgy$

Or, $\frac{1}{2} \times 25.5 \times((1.591)^2 - (0.65)^2 ) + F\times 1.80 + \frac{1}{2}\times 3.55\times (3.39)^{2} = 3.55\times 9.81\times 1.80$

Or, F= 8.641 N

From the above calculation we can conclude that,

The magnitude of the average friction force exerted on the collar (F)= 8.641 N

Learn more about friction:

brainly.com/question/24338873

#SPJ4

Disclaimer: This question is incomplete in the portal. Here is the complete question.

Question:

The 3.55 kg collar shown below is attached to a spring and released from rest at A. The collar then travels downward a distance of y = 1.80 m. The spring has a spring constant of k = 25.5 N/m. The distance a is given as 1.25 m. The datum for gravitational potential energy is set at the horizontal line through A and B. Determine the magnitude of the average friction force exerted on the collar when the velocity of the collar at c is 3.39 m/s and the spring has an unstretched length of 0.60 m .