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3 years ago

The density of gold is 19 300kg/m cube. what is the mass of gold cube with the length 0.2015m?

Physics

1 answer:

Sergeeva-Olga [200]3 years ago

4 0

Answer:

157.9 kg

Explanation:

Density: This can be defined as the ratio of the mass of a body and it's volume.

The S.I unit of density is kg/m³.

From the question,

Density = Mass/volume

D = m/v............................ Equation 1

Where D = Density of gold, m = mass of gold, v = volume of gold.

make m the subject of the equation

m = Dv.................... Equation 2

Since the gold is a cube,

v = l³................... Equation 3

Where l = length of the gold cube.

Substitute equation 3 into equation 2

m = Dl³............... Equation 4

Given: D = 19300 kg/m³, l = 0.2015 m

Substitute into equation 4

m = 19300(0.2015)³

m = 157.9 kg.

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A particle has a charge of q = +4.9 μC and is located at the origin. As the drawing shows, an electric field of Ex = +242 N/C ex

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$F_{E_x}=1.19\cdot 10^{-3}N$ (+x axis)

$F_{B_x}=0$

$F_{B_y}=0$

$F_{E_x}=1.19\cdot 10^{-3} N$ (+x axis)

$F_{B_x}=0$

$F_{B_y}=3.21\cdot 10^{-3}N$ (+z axis)

$F_{E_x}=1.19\cdot 10^{-3} N$ (+x axis)

$F_{B_x}=3.21\cdot 10^{-3} N$ (+y axis)

$F_{B_y}=3.21\cdot 10^{-3}N$ (-x axis)

Explanation:

The electric force exerted on a charged particle is given by

$F=qE$

where

q is the charge

E is the electric field

For a positive charge, the direction of the force is the same as the electric field.

In this problem:

$q=+4.9\mu C=+4.9\cdot 10^{-6}C$ is the charge

$E_x=+242 N/C$ is the electric field, along the x-direction

So the electric force (along the x-direction) is:

$F_{E_x}=(4.9\cdot 10^{-6})(242)=1.19\cdot 10^{-3} N$

towards positive x-direction.

The magnetic force instead is given by

$F=qvB sin \theta$

where

q is the charge

v is the velocity of the charge

B is the magnetic field

$\theta$ is the angle between the directions of v and B

Here the charge is stationary: this means $v=0$ , therefore the magnetic force due to each component of the magnetic field is zero.

In this case, the particle is moving along the +x axis.

The magnitude of the electric force does not depend on the speed: therefore, the electric force on the particle here is the same as in part a,

$F_{E_x}=1.19\cdot 10^{-3} N$ (towards positive x-direction)

Concerning the magnetic force, we have to analyze the two different fields:

- $B_x$ : this field is parallel to the velocity of the particle, which is moving along the +x axis. Therefore, $\theta=0^{\circ}$ , so the force due to this field is zero.

$- B_y$ : this field is perpendicular to the velocity of the particle, which is moving along the +x axis. Therefore, $\theta=90^{\circ}$ . Therefore, $\theta=90^{\circ}$ , so the force due to this field is: