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3 years ago

When you whisper you produce a 10-dB sound.

1 answer:

4 0

All you would do is for a, 10 times 2 is 20 so it would be 20-dB
For b, 10 times 4 is 40 so it would be 40-dB
<span>For c, 10 times 8 is 80 so it would be 80-dB</span>

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• The average length is ________ cm. This is the mean or average. b)• Subtract the highest value from the lowest value: ________

coldgirl [10]

16, 5 , 3 = 16+5+3= 24 + 3

So at the end put 24 + 3 cm
And put 16 for the lengths
For the value 5 and for the diving thingy 3

7 0

3 years ago

)) What do these two changes have in common?

zloy xaker [14]

Answer:

Both are only physical changes

Explanation:

A physical change is a change that does not involve or alter the chemical composition of the substances involved. Physical changes form no new substance and can be easily separated into individual constituents. Example of physical changes are change in state, boiling, melting etc.

According to this question, two processes were given as follows:

1. mixing chocolate syrup into milk

2. rain forming in a cloud

These two processes are similar in the sense that they are both examples of physical changes.

7 0

3 years ago

How does radiation help treat cancer?

sergejj [24]

Answer:

it cools down the cancer cells

Explanation:

The cancer cells are rapidly reproducing and when you cool it down it slows down the mutation/ reproduction process.

3 0

3 years ago

Read 2 more answers

Calculate the orbital period of a dwarf planet found to have a semimajor axis of a = 4.0x 10^12 meters in seconds and years.

padilas [110]

Explanation:

We have,

Semimajor axis is $4\times 10^{12}\ m$

It is required to find the orbital period of a dwarf planet. Let T is time period. The relation between the time period and the semi major axis is given by Kepler's third law. Its mathematical form is given by :

$T^2=\dfrac{4\pi ^2}{GM}a^3$

G is universal gravitational constant

M is solar mass

Plugging all the values,

$T^2=\dfrac{4\pi ^2}{6.67\times 10^{-11}\times 1.98\times 10^{30}}\times (4\times 10^{12})^3\\\\T=\sqrt{\dfrac{4\pi^{2}}{6.67\times10^{-11}\times1.98\times10^{30}}\times(4\times10^{12})^{3}}\\\\T=4.37\times 10^9\ s$

Since,

$1\ s=3.17\times 10^{-8}\ \text{years}\\\\4.37\times 10^9\ s=4.37\cdot10^{9}\cdot3.17\cdot10^{-8}\\\\4.37\times 10^9\ s=138.52\ \text{years}$

So, the orbital period of a dwarf planet is 138.52 years.

3 0

3 years ago

A gymnast is swinging on a high bar. The distance between his waist and the bar is 0.905 m, as the drawing shows. At the top of

Citrus2011 [14]

Answer:

5.959 m/s

Explanation:

m = Mass of gymnast

u = Initial velocity

v = Final velocity

$h_i$ = Initial height

$h_f$ = Final height

From conservation of Energy

$\frac{1}{2}mv^2+mgh_f=\frac{1}{2}mu^2+mgh_i\\\Rightarrow\frac{1}{2}mv^2+mg0=\frac{1}{2}m0^2+mgh_i\\\Rightarrow \frac{1}{2}mv^2=mgh_i\\\Rightarrow v=\sqrt{2gh_i}$

$h_i=2r$

$v=\sqrt{4gr}\\\Rightarrow v=\sqrt{4\times 9.81\times 0.905}\\\Rightarrow v=5.959\ m/s$

Velocity of gymnast at bottom of swing is 5.959 m/s

5 0

3 years ago