Which bar graph could represent the reaction rates of a reversible reaction<br> that has just begun?

As a laudably skeptical physics student, you want to test Coulomb's law. For this purpose, you set up a measurement in which a p

monitta

Answer:

force F = 1.66 × $10^{-13}$ N

Explanation:

given data

proton and an electron = 865 nm

solution

we get here force that is express as

force F = k q1 q2 ÷ r² ......................1

put here value and we get

force F = 9 × $10^{9}$ × $\frac{1.6\times (10^{-19})^{2}}{865 \times (10^{-9})^{2}}$

force F = 1.66 × $10^{-13}$ N

4 0

3 years ago

One normal afternoon in undisclosed City X, the chocolate factory workers overload the Easter egg machine. A group of angsty tee

musickatia [10]

Answer:

The correct option is;

C. 1,715 m

Explanation:

We are given the information from the group of teen at the City edge

Time of arrival of explosion sound = 5 s after sighting

Time of sighting explosion = 5 s before hearing the boom

Speed of sound in air ≈ 343 m/s

Speed of light = 299,792 km/s

Therefore, distance covered by sound in 5 seconds is given by the following equation;

$Speed = \frac{Distance}{Time}$

$\therefore 343 \ m/s= \frac{Distance}{5 \, s}$

Hence Distance = 343 m/s × 5 s = 1715 m

To check, we compare the time it would take for the light to cover 1715 m

That is $Time = \frac{Distance}{Speed} = \frac{1715}{299,792,000} = 0.00000572 \, s$ which is instantaneous hence the distance can be approximated by the time duration for the speed of sound.

Therefore, the distance of the students from the factory is approximately 1,715 m

8 0

3 years ago

What is an astronomer? Answer in the simpliest way please.

Sphinxa [80]

They would be an expert in astronomy
astronomy:the branch of science that deals with celestial objects

8 0

3 years ago

Use the diagram below to answer the following question:

d1i1m1o1n [39]

Answer:

3.0 cm

Explanation:

We can solve this problem by using the mirror equation:

$\frac{1}{f}=\frac{1}{p}+\frac{1}{q}$

where

f is the focal length of the mirror

p is the distance of the object from the mirror

q is the distance of the image from the mirror

In this problem we have:

f = 1.5 cm is the focal length of the mirror (positive for a concave mirror)

p = 3.0 cm is the distance of the object from the mirror

Therefore, the distance of the image is:

$\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{1.5}-\frac{1}{3.0}=\frac{1}{3.0}\\\rightarrow q=3.0 cm$

And the positive sign means that the image is real.

(The second part of the exercise is just the description of the image of the first exercise).

5 0

2 years ago

Are cells made of tisse?

zepelin [54]

Answer:

Yes

Explanation:

Cells make up tissues. Hope this helped

5 0

3 years ago

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Which bar graph could represent the reaction rates of a reversible reaction that has just begun?