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Irina18 [472]
2 years ago
12

If you wanted to know the location of a vehicle that ran out of gas after taking a zigzag route through the city, which quantity

would be
most useful?

a. the vehicle's speed

b. the vehicle's distance

c. the vehicle's direction

d. the vehicle's displacement
Physics
1 answer:
Papessa [141]2 years ago
8 0

Explanation:

the vehicles displacement, since displacement deals with position

You might be interested in
A helium nucleus is composed of two protons (positively charged) and two neutrons (charge neutral). The distance between the two
lora16 [44]

Answer:

Using the given values

F = K q^2 / r^2 = 9 * 10E9 * (1.6 * E-19)^2 / (5.18 * E-15)^2 N

E = 9 * 1.6^2 / 5.18^2 * 10 = 8.5 N

5 0
2 years ago
1mm is equal to .1 what
bulgar [2K]

Answer:

When you have to do an English-Metric (SI) length conversion, and you already know the English units of length (miles, yards, feet, inches, etc.), all you need to remember is one simple relationship, and you can readily convert any length in the SI system, to the equivalent length in the other.

1 foot (ft) = 0.3048 meters (m)

BIn this case you need your answer in inches. You (hopefully) know there are 12 inches in a foot, so you just do the following:

1 inch (in) = 1/12 ft = 0.3048/12 m = 0.0254 m

8 0
2 years ago
Chromatic aberration comes from the fact that different wavelengths of light travel at different speeds through the material of
gtnhenbr [62]

Answer:

 y_red / y_blue = 1.11

Explanation:

Let's use the constructor equation to find the image for each wavelength

         1 /f = 1 /o + 1 /i

Where f is the focal length, or the distance to the object and i the distance to the image

Red light

           1 / i = 1 / f - 1 / o

           1 / i_red = 1 / f_red - 1 / o

           1 / i_red = 1 / 19.57 - 1/30

           1 / i_red = 1,776 10-2

           i_red = 56.29 cm

Blue light

            1 / i_blue = 1 / f_blue - 1 / o

            1 / i_blue = 1 / 18.87 - 1/30

            1 / i_blue = 1,966 10-2

            i_blue = 50.863 cm

Now let's use the magnification ratio

             m = y ’/ h = - i / o

             y ’= - h i / o

Red Light

            y_red ’= - 5 56.29 / 30

            y_red ’= - 9.3816 cm

Light blue

            y_blue ’= 5 50,863 / 30

            y_blue ’= - 8.47716 cm

The ratio of the height of the two images is

            y_red ’/ y_blue’ = 9.3816 / 8.47716

            y_red / y_blue = 1,107

            y_red / y_blue = 1.11

5 0
3 years ago
A charge of 0.91 C is spread uniformly throughout a 25 cm rod of radius 4 mm. What are the volume and linear charge densities
Oxana [17]

The volume of the rod is 1.26×10⁻⁵ m³, and the linear charge density of the rod is 3.64 C/m

<h3>What is volume?:</h3>

This is the product of the height of a solid object and its crossectional area.

The Volume of the rod is can be calculated using the formula below.

Note: A rod has the shape of a cylinder.

Formula:

  • V = πr²h............... Equation 1

Where:

  • V = Volume of the rod
  • r = radius of the rod
  • h = height of the rod.

From the question,

Given:

  • r = 4mm = 0.004 m
  • h = 25 cm = 0.25 m
  • π = 3.14

Substitute these values into equation 1

  • V = 3.14(0.004²)(0.25)
  • V = 1.26×10⁻⁵ m³

<h3>What is linear charge density:</h3>

This is the ratio of the charge on an object to the length of the object.

The linear charge density of the rod can be calculated using the formula below.

  • D = Q/h.................... Equation 2

Where:

  • D = Linear charge density of the rod
  • Q = Charge on the rod.
  • h = height or length of the rod

From the question

Given:

  • Q = 0.91 C
  • h = 25 cm = 0.25 m

Substitute these values into equation 2

  • D = 0.91/0.25
  • D = 3.64 C/m

Hence, The volume of the rod is 1.26×10⁻⁵ m³, and the linear charge density of the rod is 3.64 C/m

Learn more about charge density here: brainly.com/question/14568868

4 0
2 years ago
A cube of side 5.0m is in a region where the electric field is directed outward from both of two opposite cube faces, with unifo
guajiro [1.7K]

Answer:

The charge inside the cube is null.

Explanation:

If we apply the gauss theorem with a cubical gaussian surface of the size of the cube:

\displaystyle\oint_{S} \vec{E}\,\vec{ds}=\frac{q_{in}}{\varepsilon_{0}}

If we consider than the direction of the electric field is \vec{E}=E_0\hat{x}, we can solve the problem differentiating  the integral for each face of the cube:

\displaystyle\oint_{S} \vec{E}\,\vec{ds}=\displaystyle\int_{S_1} \vec{E}\,\vec{ds_1}+\displaystyle\int_{S_2} \vec{E}\,\vec{ds_2}+\displaystyle\int_{S_3} \vec{E}\,\vec{ds_3}+\displaystyle\int_{S_4} \vec{E}\,\vec{ds_4}+\displaystyle\int_{S_5} \vec{E}\,\vec{ds_5}+\displaystyle\int_{S_6} \vec{E}\,\vec{ds_6}

\displaystyle\oint_{S} \vec{E}\,\vec{ds}=\displaystyle\int_{S_1} E_0\hat{x}\,\hat{x}ds_1+\displaystyle\int_{S_2} E_0\hat{x}\,\hat{-x}ds_2+\displaystyle\int_{S_3} E_0\hat{x}\,\hat{y}ds_3+\displaystyle\int_{S_4} E_0\hat{x}\,\hat{-y}ds_4+\displaystyle\int_{S_5} E_0\hat{x}\,\hat{z}ds_5+\displaystyle\int_{S_6} E_0\hat{x}\,\hat{-z}ds_6

E₀ is a constant and each surface is equal to each other, so: S_1=S_2=S_i=S

Therefore:

\displaystyle\oint_{S} \vec{E}\,\vec{ds}=E_0\displaystyle\int_{S_1} \,ds_1+E_0\displaystyle\int_{S_2} -1\,ds_2+0+0+0+0=E_0S-E_0S=0

\displaystyle\oint_{S} \vec{E}\,\vec{ds}=0=\frac{q_0}{\varepsilon_0} \longleftrightarrow q_0=0c

3 0
3 years ago
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