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4 years ago

When light passes through an object unchanged, scientists call that process _____.

Physics

1 answer:

Nataly [62]4 years ago

5 0

<u>Answer</u>:

When light passes through an object unchanged, scientists call that process Transmission.

<u>Explanation</u>:

Transmission is the process where all the light that is passed through the material moves via the material without being absorbed. The Transmission depends on the affected radiation.The Transmittance of the medium is defined as the ratio between transmitted radiant power and incident radiant power. The light that is passed through the medium and not reflected will be either scattered or reflected. The light can be transmitted only through transparent or translucent material. Opaque object does not allows transmission of light.

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Suppose you have three objects of equal mass. The first object has a volume of 5 cm3 and is made of a material of density 2 g/cm

liubo4ka [24]

Answer:

2 g/cm³

Explanation:

m = Mass of object

v = Volume

ρ = Density

When ρ = 2 g/cm³ , v = 5 cm³

$\rho=\frac{m}{v}\\\Rightarrow m=\rho v\\\Rightarrow m=2\times 5\\\Rightarrow m=10\ g$

All the objects have same mass, which is 10 grams

When v = 5 cm³

$\rho=\frac{m}{v}\\\Rightarrow \rho=\frac{10}{5}\\\Rightarrow \rho=2\ g/cm^3$

Density of third object is 2 g/cm³

5 0

3 years ago

Read 2 more answers

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

irina1246 [14]

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

$F_{B_y}=qvB_y$

where:

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

$v=345 m/s$ is the velocity

$B_y = +1.9 T$ is the magnetic field

Substituting,

$F_{B_y}=(4.9\cdot 10^{-6})(345)(1.9)=3.21\cdot 10^{-3} N$

And the direction of this force can be found using the right-hand rule:

- Index finger: direction of the velocity (+x axis)

- Middle finger: direction of the magnetic field (+y axis)

- Thumb: direction of the force (+z axis)

As in part b), the electric force has not change, since it does not depend on the veocity of the particle:

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

For the field $B_x$ , the velocity (+z axis) is now perpendicular to the magnetic field (+x axis), so the force is

$F_{B_x}=qvB_x$

And by substituting,

$F_{B_x}=(4.9\cdot 10^{-6})(345)(1.9)=3.21\cdot 10^{-3} N$

And by using the right-hand rule:

- Index finger: velocity (+z axis)

- Middle finger: magnetic field (+x axis)

- Thumb: force (+y axis)

For the field $B_y$ , the velocity (+z axis) is also perpendicular to the magnetic field (+y axis), so the force is

$F_{B_y}=qvB_y$

And by substituting,

$F_{B_y}=(4.9\cdot 10^{-6})(345)(1.9)=3.21\cdot 10^{-3} N$

And by using the right-hand rule:

- Index finger: velocity (+z axis)

- Middle finger: magnetic field (+y axis)

- Thumb: force (-y axis)

3 0

4 years ago

What is the direction of the electric field at the dot??

Georgia [21]

What is the magnitude of the electric field at the dot in the figure? (E=? V/m) What is the direction of the electric field at the dot in the figure? 1. to the left 2. to the right 3. upward 4. downward Thanks!

8 0

4 years ago

How will you relate the distance of the planets with their period of revolution

My name is Ann [436]

Answer:

${t}^{2} \alpha {r}^{2}$

In other words, the time period squared is directly proportional to the radius of their orbit cubed

Explanation:

By combing equations for gravitational force and centripetal force, you can create a formula for t or r.

Feel free to message me if you want more help

3 0

4 years ago

An object accelerates from 9 m/s to 26 m/s in 8 seconds. What is the acceleration? Round your answer to

qwelly [4]

Explanation:

Acceleration is the change in velocity over time.

a = Δv / Δt

a = (26 m/s − 9 m/s) / 8 s

a ≈ 2.1 m/s²

3 0

3 years ago

Read 2 more answers