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serious [3.7K]
2 years ago
15

Why are the Soviets ahead in the race to capture the V-2 rocket and Werner von Braun?

Physics
1 answer:
Dahasolnce [82]2 years ago
5 0

The second world war, and its war weapons, such as the v-2 rockets, had a great impact on the world until today, to answer this question we need that...

<h3>V-2 factory </h3>

On April 11, 1945, US troops took the town of Bleicherode, in the Kohnstein region, where the V-2 factory was located. From there about 100 complete V-2s and thousands of parts and equipment were "captured" as war loot and transferred to the United States, where they formed the basis for practical studies of the missile defense program.

With this information, we can say that because it was a base discovered by US troops, none of the alternatives is correct, as it was not the Soviets who discovered it, and that the base was also located in the central part of Germany.

<u>The </u><u>v-2 factory</u><u> was located in </u><u>Kohnstein, central germany,</u><u> by this claim, </u><u>none of the alternatives is correct.</u>

Learn more about second world war brainly.com/question/7013432

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How much work must be done to bring three electrons from a great distance apart to 3.0×10−10 m from one another (at the corners
Natasha_Volkova [10]

Answer:

Potential\ Energy=Work \ Done=2.301*10^{-18} J

Work  done to bring three electrons from a great distance apart to 3.0×10−10 m from one another (at the corners of an equilateral triangle) is 2.301*10^{-18} Joules

Explanation:

The potential energy is given by:

U=Q*V

where:

Q is the charge

V is the potential difference

Potential Difference=V=\frac{kq}{r}

So,

Potential\ Energy=\frac{Qkq}{r} \\Q=q\\Potential\ Energy=\frac{kq^2}{r}

Where:

k is Coulomb Constant=8.99*10^9 Nm^2/C^2

q is the charge on electron=-1.6*10^-19 C

r is the distance=3.0*10^{-10}m

For 3 Electrons Potential Energy or work Done is:

Potential\ Energy=3*\frac{kq^2}{r}

Potential\ Energy=3*\frac{(8.99*10^9)(-1.6*10^{-19})^2}{3*10^{-10}}\\Potential\ Energy=2.301*10^{-18} J

Work  done to bring three electrons from a great distance apart to 3.0×10−10 m from one another (at the corners of an equilateral triangle) is 2.301*10^{-18} Joules

7 0
3 years ago
A particle executes simple harmonic motion with an amplitude of 1.69 cm. At what positive displacement from the midpoint of its
m_a_m_a [10]

Answer: 0.0146m

Explanation: The formula that defines the velocity of a simple harmonic motion is given as

v = ω√A² - x²

Where v = linear velocity, A = amplitude = 1.69cm = 0.0169m, x = displacement.

The maximum speed of a simple harmonic motion is derived when x = A, hence v = ωA

One half of maximum speed = speed of motion

3ωA/2 = ω√A² - x²

ω cancels out on both sides of the equation, hence we have that

A/2 = √A² - x²

(0.0169)/2 = √(0.0169² - x²)

0.00845 = √(0.0169² - x²)

By squaring both sides, we have that

0.00845² = 0.0169² - x²

x² = 0.0169² - 0.00845²

x² = 0.0002142

x = √0.0002142

x = 0.0146m

5 0
3 years ago
After a long night of cramming for a test, your college room-mate hits his head forcefully against the wall of your room in desp
zhannawk [14.2K]

Answer: the wall contracts the force exerted by his head. The wall produces the opposite force which is equal to the force his head bangs the wall with.

Explanation: if his head exerts a much greater force than the wall can counteract the wall will be destroyed, if the wall exerts a much greater force than his head exerts he will be pushed far back and might even suffer a broken head.

The wall in this case provides the opposite reactive force.

6 0
3 years ago
Help me please, need more assistance
Dmitrij [34]

Explanation:

12) q = mCΔT

125,600 J = (500 g) (4.184 J/g/K) (T − 22°C)

T = 82.0°C

13) Solving for ΔT:

ΔT = q / (mC)

a) ΔT = 1 kJ / (0.4 kg × 0.45 kJ/kg/K) = 5.56°C

b) ΔT = 2 kJ / (0.4 kg × 0.45 kJ/kg/K) = 11.1°C

c) ΔT = 2 kJ / (0.8 kg × 0.45 kJ/kg/K) = 5.56°C

d) ΔT = 1 kJ / (0.4 kg × 0.90 kJ/kg/K) = 2.78°C

e) ΔT = 2 kJ / (0.4 kg × 0.90 kJ/kg/K) = 5.56°C

f) ΔT = 2 kJ / (0.8 kg × 0.90 kJ/kg/K) = 2.78°C

14) q = mCΔT

q = (2000 mL × 1 g/mL) (4.184 J/g/K) (80°C − 20°C)

q = 502,000 J

20) q = mCΔT

q = (2000 g) (4.184 J/g/K) (100°C − 15°C) + (400 g) (0.9 J/g/K) (100°C − 15°C)

q = 742,000 J

24) q = mCΔT

q = (0.10 g) (0.14 J/g/K) (8.5°C − 15°C)

q = -0.091 J

6 0
3 years ago
Explain that a universe is a large collection of billions of galaxies and is expanding continuously?
pychu [463]
It is the effect of the Big Bang and dark energy.
4 0
3 years ago
Read 2 more answers
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