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3 years ago

A submerged submarine alters its _____ to rise or fall in the water.

Physics

2 answers:

son4ous [18]3 years ago

8 0

Answer:

density

Explanation:

Dima020 [189]3 years ago

3 0

Are there any answer choices?
Submarines use buoyancy by filling ballast tanks up with water. When they are filled with water, they are more dense than the surrounding water, so they are able to sink. If they want to rise, they fill these tanks up with air so that the density is less than the water it surrounds.
Hope this helps! :)

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A bar magnet is doplar. This means is has ___ poles. A. Two B. Zero C. Three D. One

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This means it has TWO poles

The 2 poles are NORTH and SOUTH

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2 years ago

If the mass of a material is 45 grams and the volume of the material is 11 cm^3, what would the density of the material be?

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The density of the material would be 4.1 g/cm³.

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A person travels by car from one city to another with differem constan1 speeds between pairs of cities. She drives for 30.0 min

Montano1993 [528]

Answer:

a.52.9 km/h

b.90 km

Explanation:

We are given that

$v_1=89km/h$

$t_1=30min$

$v_2=100km/h$

$t_2=12min$

$v_3=40km/h$

$t_3=45 min$

Time spend on eating lunch and buying ga=15 min.

a.Total time=30+12+45+15=102 minute= $\frac{102}{60}=1.7 hour$

1 hour=60 minutes

Distance= $speed\times time$

$d_1=v_1\times t_1=80\times\frac{30}{60}=40km$

$d_2=100\times \frac{12}{60}=20 km$

$d_3=40\times \frac{45}{60}=30 km$

Total distance= $d_1+d_2+d_3=40+20+30=90km$

Average speed= $\frac{total\;speed}{total\;time}$

Using the formula

Average speed= $\frac{90}{1.7}=52.9Km/h$

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2 years ago

A mass is attached to a spring with an unknown spring constant. The spring gains 10 J of elastic potential energy if stretched b

Firlakuza [10]

From the given information in the question, the correct option is Option 1: 14 cm.

A non-stretched elastic spring has a conserved potential energy which gives it the ability to perform work. The elastic potential energy can be expressed as:

PE = $\frac{1}{2}$ k $x^{2}$

Where PE is the energy, k is the spring constant and x is extension.

i. Given that: PE = 10 J and x = 10 cm, then;

PE = $\frac{1}{2}$ k $x^{2}$

10 = $\frac{1}{2}$ k $10^{2}$

20 = 100k

k = 0.2 J/cm

ii. To determine how far the spring is needed to be stretched, given that PE = 20 J.

PE = $\frac{1}{2}$ k $x^{2}$

20 = $\frac{1}{2}$ (0.2) $x^{2}$

40 = 0.2 $x^{2}$

$x^{2}$ = 200

x = $\sqrt{200}$

= 14.1421

x = 14.14 cm

So that;

x is approximately 14.00 cm.

Thus, the spring need to be stretched to 14.00 cm to give the spring 20 J of elastic potential energy.

For more information, check at: brainly.com/question/1352053.

8 0

2 years ago