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3 years ago

When was World War 3 Started?

Physics

2 answers:

trasher [3.6K]3 years ago

5 0

Answer:

Korean War: 25 June 1950 – 27 July 1953

Many then believed that the conflict was likely to soon escalate into a full-scale war between the three countries, the US, the USSR, and China. CBS war correspondent Bill Downs wrote in 1951 that, "To my mind, the answer is: Yes, Korea is the beginning of World War III.

Explanation:

IRINA_888 [86]3 years ago

4 0

Answer:

1950-1953

Explanation:

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3 years ago

Two thin concentric spherical shells of radii r1 and r2 (r1 < r2) contain uniform surface charge densities V1 and V2, respect

Lyrx [107]

Answer:

Answer is explained in the explanation section below.

Explanation:

Solution:

We know that the Electric field inside the thin hollow shell is zero, if there is no charge inside it.

So,

a) 0 < r < r1 :

We know that the Electric field inside the thin hollow shell is zero, if there is no charge inside it.

Hence, E = 0 for r < r1

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E x 4 $\pi$ $r^{2}$ = $\frac{Q1}{E_{0} }$

So,

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E = $\frac{Q1}{E_{0} . 4 \pi. r^{2} }$ x $\frac{r1^{2} }{r1^{2} }$

Rearranging the above equation, we will get Electric Field for r1 < r < r2:

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c) r > r2 :

Electric Field = ?

E x 4 $\pi$ $r^{2}$ = $\frac{Q1 + Q2}{E_{0} }$

Rearranging the above equation for E:

E = $\frac{Q1+Q2}{E_{0} . 4 \pi. r^{2} }$

E = $\frac{Q1}{E_{0} . 4 \pi. r^{2} }$ + $\frac{Q2}{E_{0} . 4 \pi. r^{2} }$

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$\frac{Q1}{E_{0} . 4 \pi. r^{2} }$ = (σ1 x $r1^{2}$ ) /( $E_{0}$ x $r^{2}$ )

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$\frac{Q2}{E_{0} . 4 \pi. r^{2} }$ = (σ2 x $r2^{2}$ ) /( $E_{0}$ x $r^{2}$ )

So,

E = $\frac{Q1}{E_{0} . 4 \pi. r^{2} }$ + $\frac{Q2}{E_{0} . 4 \pi. r^{2} }$

Replacing the above equations to get E:

E = (σ1 x $r1^{2}$ ) /( $E_{0}$ x $r^{2}$ ) + (σ2 x $r2^{2}$ ) /( $E_{0}$ x $r^{2}$ )

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d) Under what conditions, E = 0, for r > r2?

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3 years ago

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3 years ago

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harkovskaia [24]

Answer:

m = 3 kg

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Explanation:

From the equations of motion;

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t = 2s/(u+v)

t = time taken

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u = initial speed = 25 m/s

v = final velocity = 0

Substituting the given values;

t = (2×62.5)/(25+0)

t = 5

Since, t = 5 the acceleration during this period is;

acceleration a = ∆v/t = (v-u)/t

a = (25)/5

a = 5 m/s^2

Force F = mass × acceleration

F = ma

Making m the subject of formula;

m = F/a

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Substituting the values

m = 15/5

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3 years ago

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Answer:

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3 years ago