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postnew [5]
3 years ago
7

SONAR was developed during World War II, to find submarines. Submarines were sneaking up on ships underwater and releasing torpe

does to blow up the ship.
Is that an example of science affects technology
Physics
2 answers:
masha68 [24]3 years ago
8 0

Answer:

As long as you describe the chemical reactions within the torpedo and explain how torpedoes advanced modern warfare.

Explanation:

the modern torpedo is a self-propelled weapon with an explosive warhead, launched above or below the water surface, propelled underwater towards a target, and designed to detonate either on contact with its target or in proximity to it.  and even in the early 2000s we still use them .All American submarines carry the MK 48 ADCAP torpedo. The Los Angeles and Virginia class submarines are designed to carry around 25 torpedoes and have 4 torpedo tubes to shoot them, while the Seawolf class, a Cold War behemoth with 8 torpedo tubes, can carry up to 50 torpedoes.

Reika [66]3 years ago
3 0
As long as you describe the chemical reactions within the torpedo and explain how torpedoes advanced modern warfare.
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A 4.0-kg object is supported by an aluminum wire of length 2.0 m and diameter 2.0 mm. How much will the wire stretch?
forsale [732]

Answer:

The extension of the wire is 0.362 mm.

Explanation:

Given;

mass of the object, m = 4.0 kg

length of the aluminum wire, L = 2.0 m

diameter of the wire, d = 2.0 mm

radius of the wire, r = d/2 = 1.0 mm = 0.001 m

The area of the wire is given by;

A = πr²

A = π(0.001)² = 3.142 x 10⁻⁶ m²

The downward force of the object on the wire is given by;

F = mg

F = 4 x 9.8 = 39.2 N

The Young's modulus of aluminum is given by;

Y = \frac{stress}{strain}\\\\Y = \frac{F/A}{e/L}\\\\Y = \frac{FL}{Ae} \\\\e = \frac{FL}{AY}

Where;

Young's modulus of elasticity of aluminum = 69 x 10⁹ N/m²

e = \frac{FL}{AY} \\\\e = \frac{(39.2)(2)}{(3.142*10^{-6})(69*10^9)} \\\\e = 0.000362 \ m\\\\e = 0.362 \ mm

Therefore, the extension of the wire is 0.362 mm.

8 0
2 years ago
The melting point of potassium thiocyanate determined by a student in the laboratory turned out to be 174.5 oC. The accepted val
yaroslaw [1]

Answer:

0.75%

Explanation:

Measured value of melting point of potassium thiocyanate = 174.5 °C

Actual value of melting point of potassium thiocyanate = 173.2 °C

<em>Error in the reading = |Experimental value - Theoretical value|</em>

<em>= |174.5 - 173.2|</em>

<em>= |1.3|</em>

<em>Percentage error = (Error / Theoretical value) × 100</em>

<em>= (1.3 / 173.2)×100</em>

<em>= 0.75 %</em>

∴ Percentage error in the reading is 0.75%

4 0
2 years ago
Why is the top of a wave called a crest
Molodets [167]

Answer: A crest is the highest point the medium rises to and a trough is the lowest point the medium sinks to. It is also a point on the wave where the displacement of the medium is at a maximum.

4 0
3 years ago
What happened to the balloon when it was placed on the bottle with the baking soda and vinegar and why.
user100 [1]

Answer:

The ballon would be inflated. The reason is that the sodium bicarbonate in baking soda reacts with acetic acid in vinegar to produce gas.

Explanation:

The main component of baking soda is sodium bicarbonate, {\rm Na_{2}CO_{3}}.

Vinegar is mostly a solution of acetic acid {\rm CH_{3}COOH} in water.

Acids such as acetic acid react with carbonate salts. One of the products of such reactions is carbon dioxide {\rm CO_{2}}, a gas.

In this question, when the acetic acid in vinegar reacts with sodium bicarbonate in the baking soda, the following reaction would occur:

\begin{aligned}& {\rm Na_{2}CO_{3}}\, (aq) + 2\, {\rm CH_{3}COOH}\, (aq) \\ &\to 2\, {\rm CH_{3}COONa}\, (aq) + {\rm CO_{2}}\, (g)\end{aligned}.

The {\rm CO_{2}} produced would then inflate the ballon placed on the opening of the bottle.

5 0
2 years ago
The de Broglie wave is produced only by sub atomic particle and photon. O True O False
Alina [70]

Answer:

True

Explanation:

Matter can be in the form of a particle or a wave. This is known as the dual nature of matter. This concept was proposed by Louis de Broglie and was named after him. This phenomenon has been observed for all the elementary particles.

The de Broglie wavelength is given by

\lambda=\frac{h}{p}=\frac{h}{mv}

Where

h = Planck's constant

p = Particles momentum

m = Mass of particle

v = Velocity of particle

8 0
3 years ago
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