All categories

2 years ago

A sine bar is used to determine the angle of a part feature. The length of the sine

Physics

1 answer:

Bond [772]2 years ago

3 0

A sine bar is used to get the angular measurement of a part feature. The angle of the part feature is 18.24°

From the question

Given that,

Length of the sine bar, L = 8.000 in

Diameter of the rolls = 1.000 in

Height under the roll, H = 2.0000 + 0.5000 + 0.0050

= 2.505 in

From sine bar formula,

we know that,

H = sin A x L

sin A = H ÷ L

where,

A ⇒ angle of part feature

H ⇒ height under the roll

L ⇒ length of the sine bar

Substituting values in the above equation,

sin A = H / L

A = sin⁻¹ ( 2.505 ÷ 8 )

A = sin⁻¹ (0.3131)

A = 18.24⁰

Hence the angle of the part feature = 18.24°

To learn more about sine bar : brainly.com/question/19391560

#SPJ1

You might be interested in

Two thin concentric spherical shells of radii r1 and r2 (r1 < r2) contain uniform surface charge densities V1 and V2, respect

Lyrx [107]

Answer:

Answer is explained in the explanation section below.

Explanation:

Solution:

We know that the Electric field inside the thin hollow shell is zero, if there is no charge inside it.

So,

a) 0 < r < r1 :

We know that the Electric field inside the thin hollow shell is zero, if there is no charge inside it.

Hence, E = 0 for r < r1

b) r1 < r < r2:

Electric field =?

Let, us consider the Gaussian Surface,

E x 4 $\pi$ $r^{2}$ = $\frac{Q1}{E_{0} }$

So,

Rearranging the above equation to get Electric field, we will get:

E = $\frac{Q1}{E_{0} . 4 \pi. r^{2} }$

Multiply and divide by $r1^{2}$

E = $\frac{Q1}{E_{0} . 4 \pi. r^{2} }$ x $\frac{r1^{2} }{r1^{2} }$

Rearranging the above equation, we will get Electric Field for r1 < r < r2:

E= (σ1 x $r1^{2}$ ) /( $E_{0}$ x $r^{2}$ )

c) r > r2 :

Electric Field = ?

E x 4 $\pi$ $r^{2}$ = $\frac{Q1 + Q2}{E_{0} }$

Rearranging the above equation for E:

E = $\frac{Q1+Q2}{E_{0} . 4 \pi. r^{2} }$

E = $\frac{Q1}{E_{0} . 4 \pi. r^{2} }$ + $\frac{Q2}{E_{0} . 4 \pi. r^{2} }$

As we know from above, that:

$\frac{Q1}{E_{0} . 4 \pi. r^{2} }$ = (σ1 x $r1^{2}$ ) /( $E_{0}$ x $r^{2}$ )

Then, Similarly,

$\frac{Q2}{E_{0} . 4 \pi. r^{2} }$ = (σ2 x $r2^{2}$ ) /( $E_{0}$ x $r^{2}$ )

So,

E = $\frac{Q1}{E_{0} . 4 \pi. r^{2} }$ + $\frac{Q2}{E_{0} . 4 \pi. r^{2} }$

Replacing the above equations to get E:

E = (σ1 x $r1^{2}$ ) /( $E_{0}$ x $r^{2}$ ) + (σ2 x $r2^{2}$ ) /( $E_{0}$ x $r^{2}$ )

Now, for

d) Under what conditions, E = 0, for r > r2?

For r > r2, E =0 if

σ1 x $r1^{2}$ = - σ2 x $r2^{2}$

4 0

3 years ago

Energy from solar radiation may be ________ or taken in by a surface or an object.

tester [92]

Answer:

Absorbed

Explanation:

I hope this helps you

7 0

3 years ago

When you must give something up in order to get something else, it is called...

natali 33 [55]

Opportunity cost refers to what you have to give up to buy what you want in terms of other goods or services. When economists use the word “cost,” we usually mean opportunity cost.

6 0

3 years ago

Why does a small particle size generally cause a reaction to proceed faster?

Ugo [173]

I think its a. i am not sure though.

7 0

3 years ago

Where would you find the lowest-density seawater?

PolarNik [594]

In rivers that are close to sea

4 0

3 years ago