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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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Three long wires are connected to a meter stick and hang down freely. Wire 1 hangs from the 50-cm mark at the center of the mete
disa [49]

Answer:

Three long wires are connected to a meter stick and hang down freely. Wire 1 hangs from the 50-cm mark at the center of the meter stick and carries 1.50 A of current upward. Wire 2 hangs from the 70-cm mark and carries 4.00 A of current downward. Wire 3 is to be attached to the meterstick and to carry a specific current, and we want to attach it at a location that results ineach wire experiencing no net force.

(a) Determine the position of wire 3.

b) Determine the magnitude and direction of current in wire 3

Explanation:

a) F_{net} \text {on wire }3=0

\frac{\mu_0 I_1 I_3}{2 \pi x} = \frac{\mu I_2 I_3}{2 \pi (0.2+x)} \\\\\frac{1.5}{x} =\frac{4}{0.2+x} \\\\0.03+1.5x=4x\\\\x=0.012m\\\\=1.2cm

position of wire = 50 - 1.2

= 48.8cm

b)  F_{net} \text {on wire }1=0

\frac{\mu _0 I_1 I_3}{2 \pi (1.2)} = \frac{\mu _0 I_1 I_2}{2 \pi (20)} \\\\\frac{I_3}{1.2} =\frac{4}{20} \\\\I_3=0.24A

Direction ⇒ downward

5 0
3 years ago
bumper car A (281 kg) moving +2.82 m/s makes an elastic collision with bumper car B (209 kg) moving +1.72 m/s what is the veloci
spin [16.1K]

Answer: V1 = 3.559 - 0.744V2

Explanation: Given that the

bumper car A has

Mass M1 = (281 kg) moving with

Velocity U1 = 2.82 m/s

bumper car B has

Mass M2 = (209 kg) moving with

Velocity U2 = 1.72 m/s

Where U1, U2 are the initial velocity of the two cars

Since the collision is elastic, we will use the formula below,

M1U1 + M2U2 = M1V1 + M2V2

Substitute the values into the formula

281×2.28 + 209×1.72 = 281V1 + 209V2

640.68 + 359.48 = 281V1 + 209V2

1000.16 = 281V1 + 209V2

Make V1 the subject of formula

281V1 = 1000.16 - 209V2

V1 = 1000.16/281 - 209V2/281

V1 = 3.559 - 0.744V2

Therefore, the velocity of car A which

is V1 after the collision will be expressed as V1 = 3.559 - 0.744V2

6 0
3 years ago
The cluster of decisions that managers make to assist the organization to achieve its goals is known as:
vfiekz [6]
The correct answer is letter C. SWOT analysis. The cluster of decisions that managers make to assist the organization to achieve its goals is known as SWOT Analysis.
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8 0
3 years ago
Use the work-energy theorem to determine the force required to stop a 1000 kg car moving at a speed of 20.0 m/s if there is a di
Vlad1618 [11]

Answer:

4.44 kN in the opposite direction of acceleration.

Explanation:

Given that, the initial speed of the car is, u=20m/s

And the mass of the car is, m=1000 kg

The total distance covered by the car before stop, s=45m

And the final speed of the car is, u=0m/s

Now initial kinetic energy is,

KE_{i}=\frac{1}{2}mu^{2}

Substitute the value of u and m in the above equation, we get

KE_{i}=\frac{1}{2}(1000kg)\times (20)^{2}\\KE_{i}=20000J

Now final kinetic energy is,

KE_{f}=\frac{1}{2}mv^{2}

Substitute the value of v and m in the above equation, we get

KE_{f}=\frac{1}{2}(1000kg)\times (0)^{2}\\KE_{i}=0J

Now applying work energy theorem.

Work done= change in kinetic energy

Therefore,

F.S=KE_{f}-KE_{i}\\F\times 45=(0-200000)J\\F=\frac{-200000J}{45}\\ F=-4444.44N\\F=-4.44kN

Here, the force is negative because the force and acceleration in the opposite direction.

6 0
2 years ago
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MaRussiya [10]
Storm!! Hope I could help
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