What terms are needed to completely describe velocity?

Compare and contrast the CONFLICT (choose one) in the short story you read with the elements appearing in The Watsons Go to Birm

forsale [732]

Answer:

they were in two places in flint and Birmingham and in Birmingham it is hot but flint of cold the Simi is they both have Sunday school for Joetta

Explanation:

use in your own words teachers know when your not trust me.

4 0

2 years ago

you decide to work part time at a local supermarket. The job pays eight dollars and 60 per hour and you work 20 hours per week.

aleksley [76]

Answer:

I guess that we want to find how much money you get each week.

We know that the job pays $8.60 per hour.

We know that you work 20 hours per week.

Then the gross pay (the total money that you earn) in a week is 20 times $8.60, or:

20*$8.60 = $172.

Now we know that your employer witholds:

10% + 7.65% + 5% = 22.65%

Then your employer withholds 22.65% of your gross pay.

if the 100% of your gross pay is $172

Then the 22.65% will be:

(22.65%/100%)*$172 = 0.2265*$172 = $38.96

This means that your employer withholds $38.96 of your weekly gross pay.

Then each week you get:

$172 - $38.96 = $133.04

4 0

2 years ago

How do I calculate the tension in the horizontal string?

matrenka [14]

ANSWER

T₂ = 10.19N

EXPLANATION

Given:

• The mass of the ball, m = 1.8kg

First, we draw the forces acting on the ball, adding the vertical and horizontal components of each one,

In this position, the ball is at rest, so, by Newton's second law of motion, for each direction we have,

$\begin{gathered} T_{1y}-F_g=0_{}_{}_{} \\ T_2-T_{1x}=0 \end{gathered}$

The components of the tension of the first string can be found considering that they form a right triangle, where the vector of the tension is the hypotenuse,

$\begin{gathered} T_{1y}=T_1\cdot\cos 30\degree \\ T_{1x}=T_1\cdot\sin 30\degree \end{gathered}$

We have to find the tension in the horizontal string, T₂, but first, we have to find the tension 1 using the first equation,

$T_1\cos 30\degree-m\cdot g=0$

Solve for T₁,

$T_1=\frac{m\cdot g}{\cos30\degree}=\frac{1.8kg\cdot9.8m/s^2}{\cos 30\degree}\approx20.37N$

Now, we use the second equation to find the tension in the horizontal string,

$T_2-T_1\sin 30\degree=0$

Solve for T₂,

$T_2=T_1\sin 30\degree=20.37N\cdot\sin 30\degree\approx10.19N$

Hence, the tension in the horizontal string is 10.19N, rounded to the nearest hundredth.

8 0

11 months ago

Fiber optics are an important part of our modern internet. In these fibers, two different glasses are used to confine the light

professor190 [17]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

$\theta_{max} =18.38^o$

b

New $n_{cladding} =1.491$

Explanation:

From the question we are told that

The refractive index of the core is $n_{core} = 1.497$

The refractive index of the cladding is $n_{cladding} = 1.421$

Generally according to Snell's law

$n_{core} * sin(90- \theta) = n_{cladding} * sin (90)$

Where $\theta_{max}$ is the largest angle a largest angle a ray will make with respect to the interface of the fiber and experience total internal reflection