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Semmy [17]
3 years ago
13

A small submarine has a volume of 30 m3 and has a mass of 40,000 kg. It needs external tanks that can be filled with air for buo

yancy. Assume the compressed air has a density of 37kg/m3 and the seawater has a density of 1030 kg/m3. If the external air tanks have a mass of 2000 kg, what is the minimum volume of the air tanks to provide enough buoyancy to lift the submarine in sea water?
Physics
1 answer:
iren [92.7K]3 years ago
6 0

Answer:

11.2m³

Explanation:

Applying Archimedes principle we have it that

Weight of submarine+ weight of tank+ weight of compound air =uptrust

So

(40000+2000+37V) x9.8= (30+V) 1030 x 9.8

38910= 993V

V= 11.2m³

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Andreyy89
What is the question?
4 0
3 years ago
Use this table of a school bus during morning pickups to calculate its average speed between 0 h and 2.340 h.
Gelneren [198K]

The average speed between 0 h and 2.340 h is 6.97 Km/h

Average speed is defined as the total distance travelled divided by the total time taken to cover the distance.

Average \: speed =  \frac{total \: distance}{total \: time}  \\  \\

With the above formula, we can obtain the average speed between 0 h and 2.340 h as illustrated below:

  • Total time = 2.340 – 0 = 2.340 h
  • Total distance = 16.3 – 0 = 16.3 Km
  • Average speed =?

Average \: speed =  \frac{total \: distance}{total \: time}  \\  \\Average \: speed =  \frac{16.3}{2.340}  \\  \\ Average \: speed = 6.97 \: Km/h \\  \\

Learn more about average speed: brainly.com/question/24884027

8 0
2 years ago
What is the separation distance, in meters, between masses m1 = 15 x 107 kg and m2 = 62 x 107 kg when the gravitational force be
nalin [4]

Answer:

94.1 m

Explanation:

From Coulombs law,

F  = Gm1m2/r²................... Equation 1

where F = force, m1 = first mass, m2 = second mass, G = universal constant, r = distance of separation.

Make r the subject of the equation,

r = √(Gm1m2/F)................. Equation 2

Given: F = 7×10² N, m1 = 15×10⁷ kg, m2 = 62×10⁷ kg,

Constant: G = 6.67×10⁻¹¹ Nm²/kg²

Substitute into equation 2

r = √( 6.67×10⁻¹¹×15×10⁷×62×10⁷/7×10²)

r √(886.16×10)

r √(88.616×10²)

r = 9.41×10

r = 94.1 m.

Hence the distance of separation = 94.1 m

3 0
4 years ago
Singly charged gas ions are accelerated from rest through a voltage of 10.3 V. At what temperature (in K) will the average kinet
Natasha_Volkova [10]

Answer:

Temperature of the gas molecules is 7.96 x 10⁴ K

Explanation:

Given :

Ions accelerated through voltage, V = 10.3 volts

The work done to change the position of singly charged gas ions is given by the relation :

W = q x V

Here q is charge of the ions and its value is 1.6 x 10⁻¹⁹ C.

Average kinetic energy of gas molecules is given by the relation:

K.E. = \frac{3}{2}kT

Here T is temperature and k is Boltzmann constant and its value is 1.38 x 10⁻²³ J/K.

According to the problem, the average kinetic energy of gas is equal to the work done to move the singly charged ions, i.e. ,

K.E. = W

\frac{3}{2}kT = qV

Rearrange the above equation in terms of T :

T= \frac{2qV}{3k}

Substitute the suitable values in the above equation.

T=\frac{2\times1.6\times10^{-19}\times10.3 }{3\times1.38\times10^{-23} }

T = 7.96 x 10⁴ K

5 0
3 years ago
A beam of microwaves with λ = 0.9 mm is incident upon a 9 cm slit. At a distance of 1.5 m from the slit, what is the approximate
liq [111]

Answer:

3 cm

Explanation:

According to the question,

D=1.5 m.

d=9 cm.

\lambda =0.9 mm.

Now the approximate slit's image width is equal to width of central maxima.

And width of central maxima is twice the width from center to first maxima

So,

y=2\frac{\lambda (D)}{d}.

Substitute all the variable in above equation.

y=\frac{(2)0.9\times 10^{-2} m(1.5 m) }{0.09 m}.

y=3 cm.

5 0
3 years ago
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