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3 years ago

What was Operation Overlord?

Physics

2 answers:

docker41 [41]3 years ago

8 0

Answer: a. the code name for the Allied invasion of Normandy

Explanation: The operation Overlord is the code name for the invasion of Normandy, an Allied operation that launched the successful invasion of German-occupied Western Europe during WW II. This operation took place in 1944, and started with the Nornamdy landings, that are knowed as the D-day

The correct option is a: the code name for the Allied invasion of Normandy

ratelena [41]3 years ago

8 0

Answer: The code name for the Allied invasion of Normandy. The correct option is A.

Explanation:

Operation overload is the code name for the allied invasion of Normandy. The overall Commander of operation overload was General Flight Eisenhower. It was launched on 6 June, 1944.

In 1940, the German leader Adolf Hitler defeated France and took charge of Eastern Europe. The operation overload required the type of logistical issues that no army had ever had to cope with before and the plan was for the Allies to have landed a vast amount of both men and equipment by the end of the D- day itself. They came up with the decision to land in Normandy in France to take the German by surprise after making them believe that Pays dear Calais was the main target..

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a length of wire is cut into five equal pieces. the five pieces are then connected in parallel, with the resulting resistance be

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The equivalent resistance of n resistors connected in parallel is given by
$\frac{1}{R_{eq}} = \frac{1}{R_1}+ \frac{1}{R_2}+...+ \frac{1}{R_n}$ (1)

In our problem, the resulting resistance of the 5 pieces connected in parallel is $R_{Eq}=2.00 \Omega$ , and since the 5 pieces are identical, their resistance R is identical, so we can rewrite (1) as
$\frac{1}{R_{Eq} }= \frac{1}{2 \Omega}= \frac{1}{R}+ \frac{1}{R} + \frac{1}{R} + \frac{1}{R} + \frac{1}{R} = \frac{5}{R}$
From which we find $R= 5 \cdot 2 \Omega = 10 \Omega$ .

So, each piece of wire has a resistance of $10 \Omega$ . Before the wire was cut, the five pieces were connected as they were in series. The equivalent resistance of a series of n resistors is given by
$R_{Eq}=R_1 + R_2 + ...+R_n$
So if we apply it at our case, we have
$R_{eq}=R+R+R+R+R=5 R= 5\cdot 10 \Omega= 50 \Omega$

therefore, the resistance of the original wire was $50 \Omega$ .

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Answer:

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