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2 years ago

What was Operation Overlord?

Physics

2 answers:

docker41 [41]2 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]2 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 certain alarm clock ticks four times each second, with each tick representing half a period. The balance wheel consists of a t

Semenov [28]

Answer:

a. $I=2.77x10^{-8} kg*m^2$

b. $K=4.37 x10^{-6} N*m$

Explanation:

The inertia can be find using

$I = m*r^2$

$m = 0.95 g * \frac{1 kg}{1000g}=9.5x10^{-4} kg$

$r=0.54 cm * \frac{1m}{100cm} =5.4x10^{-3}m$

$I = 9.5x10^{-4}kg*(5.4x10^{-3}m)^2$

$I=2.77x10^{-8} kg*m^2$

now to find the torsion constant can use knowing the period of the balance

T=0.5 s

$T=2\pi *\sqrt{\frac{I}{K}}$

Solve to K'

$K = \frac{4\pi^2* I}{T^2}=\frac{4\pi^2*2.7702 kg*m^2}{(0.5s)^2}$

$K=4.37 x10^{-6} N*m$

3 0

2 years ago

Which formula is used to find fluctuation of the shape of body

Sladkaya [172]

Answer:

varn=n1+1ehvkT–1

Explanation:

This is Einstein's equation.

5 0

2 years ago

Calculate the de broglie wavelength (in picometers) of a hydrogen atom traveling at 440 m/s.

Aleonysh [2.5K]

De broglie wavelength, $\lambda = \frac{h}{mv}$ , where h is the Planck's constant, m is the mass and v is the velocity.

$h = 6.63*10^{-34}$

Mass of hydrogen atom, $m = 1.67*10^{-27}kg$

v = 440 m/s

Substituting

Wavelength $\lambda = \frac{h}{mv} = \frac{6.63*10^{-34}}{1.67*10^{-27}*440} = 0.902 *10^{-9}m = 902 *10^{-12}m$

$1 pm = 10^{-12}m\\ \\ So , \lambda =902 pm$

So the de broglie wavelength (in picometers) of a hydrogen atom traveling at 440 m/s is 902 pm

7 0

2 years ago

A lens of focal length 15.0 cm is held 10.0 cm from a page (the object ). Find the magnification .

nevsk [136]

Answer:

Magnification, m = 3

Explanation:

It is given that,

Focal length of the lens, f = 15 cm

Object distance, u = -10 cm

Lens formula :

$\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$

v is image distance

$\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{15}+\dfrac{1}{(-10)}\\\\v=-30\ cm$

Magnification,

$m=\dfrac{v}{u}\\\\m=\dfrac{-30}{10}\\\\m=3$

So, the magnification of the lens is 3.

3 0

2 years ago

A circular loop of wire with a radius of 12.0 cm and oriented in the horizontal xy-plane is located in a region of uniform magne

Ulleksa [173]

(a) 34 V

The average emf induced in the loop is given by Faraday-Newmann-Lenz law:

$\epsilon = -\frac{\Delta \Phi_B}{\Delta t}$ (1)

where

$\Delta \Phi_B$ is the variation of magnetic flux through the coil

$\Delta t = 2.0 ms = 0.002 s$ is the time interval

We need to find the magnetic flux before and after. The magnetic flux is given by:

$\Phi_B = BA$

where

B is the magnetic field intensity

A is the area of the coil

The radius of the coil is r = 12.0 cm = 0.12 m, so its area is

$A=\pi r^2 = \pi (0.12 m)^2 = 0.045 m^2$

At the beginning, the magnetic field is

$B_i = 1.5 T$

so the flux is

$\Phi_i = B_i A = (1.5 T)(0.045 m^2)=0.068 Wb$

while after the removal of the coil, the magnetic field is zero, so the flux is also zero:

$\Phi_f = 0$

so the variation of magnetic flux is

$\Delta \Phi = 0-0.068 Wb=-0.068 Wb$

And substituting into (1) we find the average emf in the coil

$\epsilon=-\frac{-0.068 Wb}{0.002 s}=34 V$

(b) Counterclockwise

In order to understand the direction of the induced current, we have to keep in mind the negative sign in Lenz's law (1), which tells that the direction of the induced current must be such that the magnetic field produced by this current opposes the variation of magnetic flux in the coil.

In this situation, the magnetic flux through the coil is decreasing, since the coil is removed from the field. So, the induced current must be such that it produces a magnetic field whose direction is the same as the direction of the external magnetic field, which is upward along the positive z-direction.

Looking down from above and using the right-hand rule on the loop (thumb: direction of the current, other fingers wrapped: direction of magnetic field), we see that in order to produce at the center of the coil a magnetic field which is along positive z-direction, the induced current must be counterclockwise.

4 0

2 years ago