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2 years ago

How do you do the bisects?

Physics

1 answer:

Evgen [1.6K]2 years ago

4 0

Answer:

Answer Below!!

Explanation:

STEP 1: Draw a straight line with a ruler. STEP 2:Put the pin of a compass at the end of the line you want to bisect.Set the compass to more than half the length of the line and draw an arc crossing the line. STEP 3: Keep the width of the compass the same, and from the opposite end of the line draw another arc.

Hope I Helped

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Water waves in a small tank are .06 m long. They pass a given point at a rate of 14.8 waves every three seconds. What is the spe

snow_lady [41]

Answer:

Speed = 0.296m/2

Period = 0.203 s

Explanation:

If by 'long' you mean the wavelength of the waves, then the wavelength $\lambda=0.06m$ .

The frequency $f$ of the waves is 14.8 waves every 3 seconds or

$f=14.8/3 =4.33Hz$ .

Now the relationship between wavelength $\lambda$ , frequency $f$ and speed $v$ of the waves is:

$v=\lambda f$

We put in the values $\lambda=0.06m$ and $f=4.933Hz$ and get:

$\boxed{v=0.06*4.922=0.296m/s}$

Now the period $T$ is just the inverse of the frequency, or

$T=\frac{1}{f}$

$\boxed{T=\frac{1}{4.933}=0.203\:seconds }$

4 0

3 years ago

A 0.35-kgkg cord is stretched between two supports, 7.4 mm apart. When one support is struck by a hammer, a transverse wave trav

Agata [3.3K]

Answer:

The tension in the string is $T = 1.49*10^{-6}N$ .

Explanation:

For a string with tension $T$ and linear density $\mu_d$ carrying a transverse wave at speed $v$ it is true that

$v = \sqrt{\dfrac{T}{\mu_d} }$

solving for $T$ we get:

$T = \dfrac{v^2}{\mu_d}.$

Now, the transverse wave covers the distance of 7.4mm in 0.88s, which means it's speed is

$v =\dfrac{7.4*10^{-3}m}{0.88s} \\\\v = 8.4*10^{-3}s$

And it's linear density (mass per unit length) is

$\mu_d = \dfrac{0.35kg}{7.4*10^{-3}m} \\\\\mu_d = 47.3kg/m$

Therefore, the tension in the cord is

$T = \dfrac{(8.4*10^{-3}m/s^2)^2}{47.3kg/m}.$

$\boxed{T = 1.5*10^{-6}N}$

or in micro newtons

$T =1.5\mu N$

4 0

3 years ago

A slab of glass 8.0 cm thick is placed upon a printed page. If the refractive index of the glass is 1.5, how far from the surfac

7nadin3 [17]

Answer:

5.3 cm

Explanation:

This question is an illustration of real and apparent distance.

From the question, we have the following given parameters

Real Distance, R = 8.0cm

Refractive Index, μ = 1.5

Required

Determine the apparent distance (A)

The relationship between R, A and μ is:

μ = R/A

i.e.

Refractive Index = Real Distance ÷ Apparent Distance

Substitute values in the above formula

1.5 = 8/A

Multiply both sides by A

1.5 * A = A * 8/A

1.5A = 8

Divide both side by 1.5

1.5A/1.5 = 8/1.5

A = 8/1.5

A = 5.3cm

Hence, the letters would appear at a distance of 5.3cm

8 0

3 years ago

Why does a geostationary satellite must orbit around Earth's equator, rather than in some other orbit (such as around the poles)

LiRa [457]

Answer:

A satellite on non-equatorial orbit would show daily motion even if its period is exactly 1 sidereal day.

Explanation:

5 0

2 years ago

What environmental factors affect sound?

Mariana [72]

Answer:

Here are 5:

Distance from source to receiver

Wind speed and direction

Wind gradients

Temperature gradients

Atmospheric attenuation

and there are many more...

Hope that was helpful.Thank you!!!

7 0

3 years ago