Ask question

Ask question

All categories

2 years ago

14

Aaron created the chart to list the benefits and limitations of using synthetic polymers . In which row has Aaron made an error

in his chart?

2 answers:

Sever21 [200]2 years ago

8 0

Answer:

Its B trust me

djyliett [7]2 years ago

8 0

Answer:the answer is row2

Explanation:

You might be interested in

?/1 Jorge traveled 5 miles north to school. He then traveled 3 miles west to the store. Then he left the store and traveled 5 mi

aleksklad [387]

Answer:

He's 3 miles west of school.

Explanation:

He went 5 miles up and 5 miles down which means that he really didn't go up or down. In between that, he went 3 miles west so if the 5 milers don't count, this puts him at 3 miles west of school.

4 0

2 years ago

What gases can CFC and HCFC refrigerants decompose into at high temperatures

nekit [7.7K]

Answer:

Hydrochloric and Hydrofluoric Acids.

3 0

2 years ago

An 7.5 × binocular has 3.7-cm-focal-length eyepieces. What is the focal length of the objective lenses? Express your answer to t

elixir [45]

To solve this problem we will apply the concept of magnification, which is given as the relationship between the focal length of the eyepieces and the focal length of the objective. This relationship can be expressed mathematically as,

$\mu = \frac{f_0}{f_e}$

Here,

$\mu$ = Magnification

$f_e$ = Focal length eyepieces

$f_0$ = Focal length of the Objective

Rearranging to find the focal length of the objective

$f_0 = \mu f_e$

Replacing with our values

$f_0 = 7.5* 3.7cm$

$f_0 = 27.75cm$

Therefore the focal length of th eobjective lenses is 27.75cm

5 0

2 years ago

A point charge of 5.0 x 10^-7 C moves to the right at 2.6 x 10^5 m/s in a magnetic field that is directed into the screen and ha

-BARSIC- [3]

The magnitude of the magnetic force acting on the charge is 2.34×10⁻³ N.

<h3>What is magnetic force?</h3>

A magnetic force is the force that act in a magnetic field.

To calculate the magnetic force, we use the formula below.

Formula:

F = qvB.........Equation 1

Where:

F = magnetic force
q = point charge
v = Velocity of the the charge
B = Field strength

From the question,

Given:

q = 5.0×10⁻⁷ C
v = 2.6×10⁵ m/s
B = 1.8×10⁻² T

Substitute these values into equation 2

F = (5.0×10⁻⁷)(2.6×10⁵)(1.8×10⁻²)
F = 23.4×10⁻⁴
F = 2.34×10⁻³ N

Hence, the magnitude of the magnetic force acting on the charge is 2.34×10⁻³ N.

Learn more about magnetic force here: brainly.com/question/2279150

#SPJ12

5 0

2 years ago

To practice problem-solving strategy 22.1: gauss's law. an infinite cylindrical rod has a uniform volume charge density ρ (where

BabaBlast [244]

Let say the point is inside the cylinder

then as per Gauss' law we have

$\int E.dA = \frac{q}{\epcilon_0}$

here q = charge inside the gaussian surface.

Now if our point is inside the cylinder then we can say that gaussian surface has charge less than total charge.

we will calculate the charge first which is given as

$q = \int \rho dV$

$q = \rho * \pi r^2 *L$

now using the equation of Gauss law we will have

$\int E.dA = \frac{\rho * \pi r^2* L}{\epcilon_0}$

$E. 2\pi r L = \frac{\rho * \pi r^2* L}{\epcilon_0}$

now we will have

$E = \frac{\rho r}{2 \epcilon_0}$

Now if we have a situation that the point lies outside the cylinder

we will calculate the charge first which is given as it is now the total charge of the cylinder

$q = \int \rho dV$

$q = \rho * \pi r_0^2 *L$

now using the equation of Gauss law we will have

$\int E.dA = \frac{\rho * \pi r_0^2* L}{\epcilon_0}$

$E. 2\pi r L = \frac{\rho * \pi r_0^2* L}{\epcilon_0}$

now we will have

$E = \frac{\rho r_0^2}{2 \epcilon_0 r}$

7 0

3 years ago