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Bess [88]
3 years ago
5

Atoms are spherical in shape. Therefore, the Pt atoms in the cube cannot fill all the available space. If only 74.0 percent of t

he space inside the cube is taken up by Pt atoms, calculate the radius in picometers of a Pt atom. The mass of a single Pt atom is 3.240 × 10−22 g. [The volume of a sphere of radius r is (4/3) πr3. The volume of a cube is l3, where l is the length of a side. Avogadro's number is 6.022 × 1023.]
Physics
1 answer:
quester [9]3 years ago
5 0

Answer:

A)6.6×10^22atoms of Pt in the cube

B)1.4×10^-8m

Explanation:

(a) Calculate the number of Pt atoms in the cube.

an edge length of platinum (Pt) = 1.0 cm.

Then Volume= 1.0 cm×1.0 cm×1.0 cm=1cm^3

Then we have volume of the cube as 1cm^3

Given:

The density Pt = 21.45 g/cm3

the mass of a single Pt atom =3.240 x 10^-22 g

Then with 1atom of the platinum element, we can calculate the number of Pt atoms in the cube as

Density of pt/mass of a single Pt atom

=(21.45 /=3.240 x 10^-22)

=6.6×10^22atoms of Pt in the cube

B)Volume of cube V=4/3πr^3

V= 4/3 ×π×r^3

V= 4.19067r^3

r^3= V/4.19067

But volume is not total volume but just 74% of it, then With 74% of the space inside the cube is taken up by Pt atoms, then we need to find 74% of volume of the cube which is 1cm^3

74/100 ×1= 0.74cm^3

Then our new volume V is 0.74cm^3

r^3=0.74/4.19067×6.620 x 10^22

r^3=2.6674×10^-24

r= 3√2.6674×10^-24

r=1.4×10^-8m

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3 years ago
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1. How far can a person run if they go 6m/s for 15 seconds?
Ray Of Light [21]

Answer:

90m

Explanation:

Speed = <u>distance</u>

Time

Distance = Speed ×Time

Distance = 6×15

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3 years ago
A negative test charge experiences a force to the right as a result of an electric field. What is the best conclusion to draw ba
sergiy2304 [10]

Answer:

The electric field points to the left because the force on a negative charge is opposite to the direction of the field.

Explanation:

Normally,in an electric field, the positive charge has tendency to move from higher to lower potential and therefore, experience the electric force in the direction of electric field since electric field lines are directed from high to low potential whike a negative charge has tendency to move from low to high potential and hence experience the electric force in the direction opposite to  electric field since electric field lines are directed from high to low potential.

Thus, we can say that; The electric field points to the left because the force on a negative charge is opposite to the direction of the field.

6 0
3 years ago
An MRI technician moves his hand from a region of very low (B=0) magnetic field strength into an MRI scanner’s 3.50 T field with
sattari [20]

Answer:

The change in the magnetic flux in the ring is 10.99 Wb.

Explanation:

Given that,

An MRI technician moves his hand from a region of very low (B=0) magnetic field strength into an MRI scanner’s 3.50 T field with his fingers pointing in the direction of the field.

Diameter of the ring, d = 2 cm

Radius, r = 1 cm

It takes 0.4 s to move it into the field. We need to find the change in magnetic flux in the ring. Magnetic flux is given by :

\phi=BA\\\\\phi=B\pi r^2\\\\\phi=3.5\pi \times (1)^2\\\\\phi=10.99\ Wb

So, the change in the magnetic flux in the ring is 10.99 Wb. Hence, this is the required solution.

4 0
3 years ago
Activity Problem 8.3 This problem asks you to compare the finite current element approximation to the infinite straight wire app
Tresset [83]

Answer: (a) B = 2 x 10⁻⁵T

               (b) B = 1.94 x 10⁻⁵T

               (c) B = 1.8 x 10⁻⁴T

Explanation: A magnetic field due to a current passing through a straight wire is calculated using the <u>Biot-Savart</u> <u>Law</u>:

dB=\frac{\mu_{0}}{4.\pi} \frac{IdLXR}{r^{3}}

where

dL is current length element

\mu_{0} is permeability of free space (4.\pi.10^{-7}T.m/A)

(a) For a infinite straight wire:

B=\frac{\mu_{0}I}{2.\pi.R}

B=\frac{4.\pi.10^{-7}.2}{2.\pi.2.10^{-2}}

B = 2x10⁻⁵T

For an infinite, long and straight wire, magnetic field is 2x10⁻⁵T.

(b) For a finite wire:

B=\frac{\mu_{0}I}{2.\pi.R}\frac{L}{\sqrt{L^{2}+R^{2}} }

B=\frac{4.\pi.10^{-7}.2}{2.\pi.2.10^{-2}} \frac{8.10^{-2}}{\sqrt{(8.10^{-2})^{2}+(2.10^{-2})^{2}} }

B = 1.94x10⁻⁵T

The magnetic field for a finite wire in the same conditionsas infinite wire is 1.94x10⁻⁵T.

(c) For a finite wire at a point distant from the end of the wire:

B=\frac{\mu_{0}I}{4.\pi.L\sqrt{2} }

B=\frac{4.\pi.10^{-7}.2}{4.\pi.8.10^{-2}\sqrt{2} }

B = 0.18x10⁻⁵T

At a point at the end, magnetic field is 1.8x10⁻⁴T.

4 0
3 years ago
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