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3 years ago

In addition to 1 m = 39.37 in, the following exact conversion equivalents are

Physics

1 answer:

Norma-Jean [14]3 years ago

5 0

Answer:

3.76 m/s.

Explanation:

Velocity in mile per hour (mph) = 8.4 miles per hour

Velocity in metre per second (m/s) =?

Next, we shall convert 8.4 mph to ft/h. This can be obtained as follow :

1 mph = 5280 ft/h

Therefore,

8.4 mph = 8.4 mph / 1 mph × 5280 ft/h

8.4 mph = 44352 ft/h

Next, we shall convert 44352 ft/h to in/h. This is illustrated below:

1 ft/h = 12 in/h

Therefore,

44352 ft/h = 44352 ft/h / 1 ft/h × 12 in/h

44352 ft/h = 532224 in/h

Next, we shall convert 532224 in/h to m/h. This can be obtained as follow:

39.37 in/h = 1 m/h

Therefore,

532224 in/h = 532224 in/h / 39.37 in/h × 1 m/h

532224 in/h = 13518.51664 m/h

Next, we shall convert 13518.51664 m/h to m/min. This is illustrated below:

1 m/h = 1/60 m/min

Therefore,

13518.51664 m/h = 13518.51664 m/h / 1 m/h × 1/60 m/min

13518.51664 m/h = 225.30861 m/min

Finally, we shall convert 225.30861 m/min to m/s. This is illustrated below:

1 m/min = 1/60 m/s

Therefore,

225.30861 m/min = 225.30861 m/min / 1 m/min × 1/60 m/s

225.30861 m/min = 3.76 m/s

Therefore,

8.4 miles per hour is equivalent to 3.76 m/s.

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irina1246 [14]

Answer:

Small, icy bodies that have highly eccentric orbits and can be found in the Oort cloud or the Kuiper belt are called COMETS.

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3 years ago

When a high-energy proton or pion traveling near the speed of light collides with a nucleus, it may travel 2.5 x 10-15 m before

Tasya [4]

Explanation:

It is given that,

When a high-energy proton or pion traveling near the speed of light collides with a nucleus, $d=2.5\times 10^{-15}\ m$

Speed of light, $c=3\times 10^{8}\ m\s$

Let t is the time interval required for the strong interaction to occur. The speed is given by :

$c=\dfrac{d}{t}$

$t=\dfrac{d}{c}$

$t=\dfrac{2.5\times 10^{-15}\ m}{3\times 10^{8}\ m/s}$

$t=8.33\times 10^{-24}\ s$

So, the time interval required for the strong interaction to occur is $8.33\times 10^{-24}\ s$ . Hence, this is the required solution.

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3 years ago

Where will the spacecraft be when the gravitational forces acting on it are equal?

klio [65]

It would be not be able to move yet it would be in the air

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3 years ago

One mole of magnesium (6 × 1023 atoms) has a mass of 24 grams, as shown in the periodic table on the inside front cover of the t

natka813 [3]

This question involves the concepts of density, volume, and mass.

The approximate diameter of a magnesium atom is "3.55 x 10⁻¹⁰ m".

<h3>STEP 1 (FINDING MASS OF INDIVIDUAL ATOM)</h3>

It is given that:

Mass of one mole = 24 grams

Mass of 6 x 10²³ atoms = 24 grams

Mass of 1 atom = $\frac{24\ grams}{6\ x\ 10^{23}\ atoms}$ = 4 x 10⁻²³ grams

<h3>STEP 2 (FINDING VOLUME OF A SINGLE ATOM)</h3>

$\rho = \frac{m}{V}\\\\V=\frac{m}{\rho}$

where,

$\rho$ = density = 1.7 grams/cm³
m = mass of single atom = 4 x 10⁻²³ grams
V = volume of single atom = ?

Therefore,

$V=\frac{4\ x\ 10^{-23}\ grams}{1.7\ grams/cm^3}$

V = 2.35 x 10⁻²³ cm³

<h3>STEP 3 (FINDING DIAMETER OF ATOM)</h3>

The atom is in a spherical shape. Hence, its Volume can be given as follows:

$V =\frac{\pi d^3}{6}\\\\d=\sqrt[3]{ \frac{6V}{\pi}}\\\\d=\sqrt[3]{ \frac{6(2.35\ x\ 10^{-23}\ cm^3)}{\pi}}$

d = 0.355 x 10⁻⁷ cm = 3.55 x 10⁻¹⁰ m

Learn more about density here:

brainly.com/question/952755

7 0

2 years ago

A gold bar has a density of 19.3 g/mL. If the gold bar has a mass of 6.3 grams, what is the volume?

Natali [406]

Answer:

The correct answer is "0.32 mL".

Explanation:

The given values are:

Density of gold bar,

d = 19.3 g/mL

Mass of gold bar,

m = 6.3 grams

Now,

The volume will be:

⇒ $Density = \frac{Mass}{Volume}$

or,

⇒ $Volume=\frac{Mass}{Density}$

On substituting the values, we get

⇒ $=\frac{6.3 \ g}{19.3 \ g/mL}$

⇒ $=0.32 \ mL$

7 0

2 years ago