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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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What are possible formulas for impulse? Check all that apply. J = Fdeltat J = StartFraction force over change in time EndFractio
Alex

<u>The possible formulas for impulse are as follows:</u>

J = FΔt

J = mΔv

J = Δp

Answer: Option  A, E and F

<u>Explanation:</u>

The quantity which explains the consequences of a overall force acting on an object (moving force) is known as impulse. It is symbolised as J. When the average overall force acting on an object than such products are formed and in given duration than the start fraction force over change in time end fraction J = FΔt.

The impulse-momentum theorem explains that the variation in momentum of an object is same as the impulse applied to it: J = Δp J = mΔv if mass is constant J = m dv + v dm if mass changes. Logically, the impulse-momentum theorem is equivalent to Newton second laws of motion which is also called as force law.

6 0
3 years ago
What force is necessary to keep a mass of 0.8 kg revolving in a horizontal circle of radius 0.7 m with a period of 0.5 s? What i
inn [45]

Answer:

88.34 N directed towards the center of the circle

Explanation:

Applying,

F = mv²/r................... Equation 1

F = Force needed to keep the mass in a circle, m = mass of the mass, v = velocity of the mass, r = radius of the circle.

But,

v = 2πr/t................... Equation 2

Where t = time, π = pie

Substitute equation 2 into equation 1

F = m(2πr/t)²/r

F = 4π²r²m/t²r

F = 4π²rm/t²............. Equation 3

From the question,

Given: m = 0.8 kg, r = 0.7 m, t = 0.5 s

Constant: π = 3.14

Substitute these values into equation 3

F = 4(3.14²)(0.7)(0.8)/0.5²

F = 88.34 N directed towards the center of the circle

8 0
2 years ago
1. You get hit with a force of 45 N by a ball with a mass of 0.75 kg. How fast was the
dlinn [17]

Answer:

60 {ms}^{ - 2}

Explanation:

As we know,

=》Force = Mass × Acceleration

=》45 N = 0.75 × Acceleration

=》Acceleration = 45 ÷ 0.75

=》Acceleration = 60

hence, the Acceleration of the ball would be. 60 meters per second square

60m {s}^{ - 2}

6 0
3 years ago
Suppose a certain jet plane creates an intensity level of 124 dB at a distance of 5.01 m. What intensity level does it create on
Zolol [24]

Answer:71 dB

Explanation:

Given

sound Level \beta _1=124 dB

distance r_1=5.01 m

From sound Intensity

\beta =10dB\log (\frac{I_1}{I_0})

124=10dB\log (\frac{I_1}{I_0})

12.4=\log (\frac{I_1}{I_0})

I_1=(1\times 10^{-12})\times 10^{12.4}

I_1=2.51 W/m^2

we know Intensity I\propto ^\frac{1}{r^2}

I_1r_1^2=I_2r_2^2

I_2=I_1(\frac{r_1}{r_2})^2

I_2=2.51\cdot (\frac{5.01}{2.25\times 10^3})^2

I_2=1.24\times 10^{-5} W/m^2

Sound level corresponding to I_2

\beta _2=10\log (\frac{I_2}{I_0})

\beta _2=10\log (\frac{1.24\times 10^{-5}}{1\times 10^{-12}})

\beta _2=70.93\approx 71 dB

6 0
3 years ago
If 3,600 J of work is done in 3.0 s, what is the power?
Firlakuza [10]

Answer:

1,200 watts

Explanation:

1 watt = 1 Joule (J) of work / second

So, 3600 Joules of work / 3 seconds is:

3600 J / 3 seconds = 1,200 watts

8 0
3 years ago
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