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Ierofanga [76]
3 years ago
11

Zona onde não incide luz visível devido à interposição de um objeto opaco.

Physics
1 answer:
m_a_m_a [10]3 years ago
8 0

Answer:

can you please translate to english?

Explanation:

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How do you know the speed of an electromagnetic wave in a vacuum?
ycow [4]
Electromagnetic waves need no matter to travel - they can travel through empty space (a vacuum). In a vacuum, all electromagnetic waves travel at approximately 3 x 108 m/s - which is the fastest speed possible. ...
Light traveling value through an optical Fibre is, 2 x 108 m/s. Hope that helps.
5 0
3 years ago
In mechanics, massless strings are often assumed. Why is that not a good assumption when discussing waves on strings?
Marizza181 [45]

In mechanics, massless strings are often assumed. but this is not a good assumption when discussing waves on strings because the speed of a wave on a massless string would be infinite.

<h3>How to explain the information?</h3>

It should be noted that waves simply means the dynamic disturbance of a quantity.

It should be noted that in mechanics, massless strings are often assumed. but this is not a good assumption when discussing waves on strings because the speed of a wave on a massless string would be infinite.

Learn more about waves in:

brainly.com/question/15663649

#SPJ4

6 0
1 year ago
A planet exerts a gravitational force of magnitude 9e22 N on a star. If the planet were 2 times closer to the star (that is, if
Dmitrij [34]

To solve this problem we will use the related concepts in Newtonian laws that describe the force of gravitational attraction. We will use the given value and then we will obtain the proportion of the new force depending on the Radius. From there we will observe how much the force of attraction increases in the new distance.

Planet gravitational force

F_p = 6*10^{22}N

F_p = \frac{GMm}{R^2}

F_p = 9*10^{22}N

Distance between planet and star

r = \frac{R}{2}

Gravitational force is

F = \frac{GMm}{r^2}

Applying the new distance,

F = \frac{GMm}{(\frac{R}{2})^2}

F =  4\frac{GMm}{R^2}

Replacing with the previous force,

F = 4F_p

Replacing our values

F= 4(9*10^{22}N)

F = 36*10^{22}N

Therefore the magnitude of the force on the star due to the planet is  36*10^{22}N

5 0
3 years ago
Mass of 23.94kg and volume 2.10 cubic centimeters. Calculate the density of the copper wire in si units (ų/mm cubic)
Kamila [148]

Answer:

Density = 1.14 kg/mm³

Explanation:

Given the following data;

Volume = 2.10 cm³ to mm = 2.10 * 10 = 21 mm³

Mass = 23.94kg

To find the density;

Density can be defined as mass all over the volume of an object.

Simply stated, density is mass per unit volume of an object.

Mathematically, density is given by the equation;

Density = \frac{mass}{volume}

Substituting into the equation, we have;

Density = \frac{23.94}{21}

Density = 1.14 kg/mm³

7 0
3 years ago
Consider a 2 m wide and 5 m long slab resting on flat, fixed, earthen bed. The slab is 20 cm thick with uniform properties. You
4vir4ik [10]

Answer:

Explanation:

The coordinate sketch for the system is shown in the attached file below. Also, in the cartesian coordinate system, since the height is less than the length and width, we did neglect the height. Thus, we eliminate the height and converted it to a two-dimension.

8 0
3 years ago
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