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zmey [24]
3 years ago
15

Submarines use ______ of air to control their depths in water

Physics
1 answer:
jeka57 [31]3 years ago
8 0

Answer:

A supply of compressed air is maintained aboard the submarine in air flasks for life support and for use with the ballast tanks. In addition, the submarine has movable sets of short "wings" called hydroplanes on the stern (back) that help to control the angle of the dive.

Explanation:

Hope this helped u

Have a nice day

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The vector product of vectors A⃗ and B⃗ has magnitude 12.0 m2 and is in the +z-direction.Vector A⃗ has magnitude 4.0 m and is in
g100num [7]

Answer:

θ=180°

Explanation:

The problem says that the vector product of A and B is in the +z-direction, and that the vector A is in the -x-direction. Since vector B has no x-component, and is perpendicular to the z-axis (as A and B are both perpendicular to their vector product), vector B has to be in the y-axis.

Using the right hand rule for vector product, we can test the two possible cases:

  • If vector B is in the +y-axis, the product AxB should be in the -z-axis. Since it is in the +z-axis, this is not correct.

  • If vector B is in the -y-axis, the product AxB should be in the +z-axis. This is the correct option.

Now, the problem says that the angle θ is measured from the +y-direction to the +z-direction. This means that the -y-direction has an angle of 180° (half turn).

8 0
3 years ago
A stationary speed gun emits a microwave beam at 2.10*10^10Hz. It reflects off a car and returns 1030 Hz higher. What is the spe
lana [24]

Answer: V = 15 m/s

Explanation:

As  stationary speed gun emits a microwave beam at 2.10*10^10Hz. It reflects off a car and returns 1030 Hz higher. The observed frequency the car will be experiencing will be addition of the two frequency. That is,

F = 2.1 × 10^10 + 1030 = 2.100000103×10^10Hz

Using doppler effect formula

F = C/ ( C - V) × f

Where

F = observed frequency

f = source frequency

C = speed of light = 3×10^8

V = speed of the car

Substitute all the parameters into the formula

2.100000103×10^10 = 3×10^8/(3×10^8 -V) × 2.1×10^10

2.100000103×10^10/2.1×10^10 = 3×108/(3×10^8 - V)

1.000000049 = 3×10^8/(3×10^8 - V)

Cross multiply

300000014.7 - 1.000000049V = 3×10^8

Collect the like terms

1.000000049V = 14.71429

Make V the subject of formula

V = 14.71429/1.000000049

V = 14.7 m/s

The speed of the car is 15 m/s approximately

3 0
3 years ago
A)
Ksenya-84 [330]

Answer:

50N

Explanation:

is the opposing fotce

4 0
3 years ago
A child of mass 27 kg swings at the end of an elastic cord. At the bottom of the swing, the child's velocity is horizontal, and
snow_tiger [21]

Answer:

The magnitude of the rate of change of the child's momentum is 794.11 N.

Explanation:

Given that,

Mass of child = 27 kg

Speed of child in horizontal = 10 m/s

Length = 3.40 m

There is a rate of change of the perpendicular component of momentum.

Centripetal force acts always towards the center.

We need to calculate the magnitude of the rate of change of the child's momentum

Using formula of momentum

\dfrac{dp}{dt}=F

\dfrac{dP}{dt}=\dfrac{mv^2}{r}

Put the value into the formula

\dfrac{dP}{dt}=\dfrac{27\times10^2}{3.40}

\dfrac{dP}{dt}=794.11\ N

Hence, The magnitude of the rate of change of the child's momentum is 794.11 N.

7 0
4 years ago
A ladybug sits at the outer edge of a turntable, and a gentleman bug sits halfway between her and the axis of rotation. The turn
inn [45]

Answer:

The answer to the question is

The ladybug begins to slide

Explanation:

To solve the question we assume that the frictional force of the ladybug and the gentleman bug are the same

Where the  frictional force equals F_{Friction} = μ×N = m×g×μ

and the centripetal force is given by m·ω²·r

If we denote the properties of the ladybug as 1 and that of the gentleman bug as 2, we have

m₁×g×μ = m₁·ω²·r₁ ⇒ g×μ = ω²·r₁

and for the gentleman bug we have

m₂×g×μ = m₂·ω²·r₂ ⇒ g×μ = ω²·r₂

But r₁ = 2×r₂

Therefore substituting the values of r₁ =2×r₂ we have

g×μ = ω²·r₁ = g×μ = ω²·2·r₂

Therefore   ω²·r₂ = 0.5×g×μ for the ladybug. That is the ladybug has to overcome half the frictional force experienced by the gentleman bug before it start to slide

The ladybug begins to slide

6 0
4 years ago
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