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3 years ago

8

What is an example of transparent

2 answers:

nataly862011 [7]3 years ago

7 0

Answer:

When something is transparent, it means that it allows light to pass through or is see-through . For example:

1) glass

2) air

3) some plastics

poizon [28]3 years ago

4 0

Something that is see through

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One of the key disadvantages of buying fitness equipment is __________. A. privacy B. flexibility C. maintenance D. choice Pleas

Alex Ar [27]

Answer:

<h2>C. maintenance </h2>

Explanation:

I personally believe one key disadvantages is the cost of maintaining the equipment unlike the gym where you have to subscribe for the month or the year and forget about anything, owning the gym equipment comes with the extra cost and responsibilities of maintenance to the owners.

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3 years ago

Which statement describes a question that can be answered by a scientific

goblinko [34]

It’s d! hope I can help

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4 years ago

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A physical science test book has a mass of 2.2kg. . What is the weight on the earth

dem82 [27]

Weight = mg, g ≈ 9.8 m/s²

Weight = 2.2 * 9.8 ≈ 21.56 N

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4 years ago

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Suppose the rocket in the Example was initially on a circular orbit around Earth with a period of 1.6 days. Hint (a) What is its

ruslelena [56]

Answer:

a

The orbital speed is $v= 2.6*10^{3} m/s$

b

The escape velocity of the rocket is $v_e= 3.72 *10^3 m/s$

Explanation:

Generally angular velocity is mathematically represented as

$w = \frac{2 \pi}{T}$

Where T is the period which is given as 1.6 days = $1.6 *24 *60*60 = 138240 sec$

Substituting the value

$w = \frac{2 \pi}{138240}$

$= 4.54*10^ {-5} rad /sec$

At the point when the rocket is on a circular orbit

The gravitational force = centripetal force and this can be mathematically represented as

$\frac{GMm}{r^2} = mr w^2$

Where G is the universal gravitational constant with a value $G = 6.67*10^{-11}$

M is the mass of the earth with a constant value of $M = 5.98*10^{24}kg$

r is the distance between earth and circular orbit where the rocke is found

Making r the subject

$r = \sqrt[3]{\frac{GM}{w^2} }$

$= \sqrt[3]{\frac{6.67*10^{-11} * 5.98*10^{24}}{(4.45*10^{-5})^2} }$

$= 5.78 *10^7 m$

The orbital speed is represented mathematically as

$v=wr$

Substituting value

$v= (5.78*10^7)(4.54*10^{-5})$

$v= 2.6*10^{3} m/s$

The escape velocity is mathematically represented as

$v_e = \sqrt{\frac{2GM}{r} }$

Substituting values

$= \sqrt{\frac{2(6.67*10^{-11})(5.98*10^{24})}{5.78*10^7} }$

$v_e= 3.72 *10^3 m/s$

7 0

4 years ago

A small bolt with a mass of 41.0 g sits on top of a piston. The piston is undergoing simple harmonic motion in the vertical dire

EleoNora [17]

Answer:

Amplitude will be equal to 0.091 m

Explanation:

Given mass of the slits = 41 gram = 0.041 kg

Frequency f = 1.65 Hz

So angular frequency $\omega =2\pi f=2\times 3.14\times 1.65=10.362rad/sec$

Angular frequency is equal to $\omega =\sqrt{\frac{k}{m}}$

$10.362 =\sqrt{\frac{k}{0.041}}$

Squaring both side

$107.371 ={\frac{k}{0.041}}$

k = 4.40 N/m

For vertical osculation

$mg=kA$

$0.041\times 9.8=4.40\times A$

A = 0.091 m

So amplitude will be equal to 0.0391 m

8 0

3 years ago