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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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Not too long ago houses were protected from excessive currents by fuses rather than circuit breakers. sometimes a fuse blew out

olganol [36]

Answer: Resistance = $8.21 \times 10^{-8} \Omega$

The approximate diameter of a penny is, d = 20 mm

thickness of penny is, L = 1.5× $10^{-3}$ mm

The area of penny along circular face is, $A = \frac{\pi d^2}{4} =\frac{\pi (20 mm\frac{1 m}{1000 m})^2}{4}$

= 3.14× $10^{-4}$ m²

The resistivity of copper is ρ = 1.72 x 10-8 Ωm.

Resistance,

$R = \rho \frac{L}{A} = (1.72 \times 10^{-8} \Omega m)\frac{1.5 \times 10^{-3} m}{3.14 \times 10^{-4} m^2} = 8.21 \times 10^{-8} \Omega$

5 0

2 years ago

A flywheel is a mechanical device used to store rotational kinetic energy for later use. Consider a flywheel in the form of a un

Kamila [148]

Answer:

a) 6738.27 J

b) 61.908 J

c) $\frac{4492.18}{v_{car} ^{2} }$

Explanation:

The complete question is

A flywheel is a mechanical device used to store rotational kinetic energy for later use. Consider a flywheel in the form of a uniform solid cylinder rotating around its axis, with moment of inertia I = 1/2 mr2.

Part (a) If such a flywheel of radius r1 = 1.1 m and mass m1 = 11 kg can spin at a maximum speed of v = 35 m/s at its rim, calculate the maximum amount of energy, in joules, that this flywheel can store?

Part (b) Consider a scenario in which the flywheel described in part (a) (r1 = 1.1 m, mass m1 = 11 kg, v = 35 m/s at the rim) is spinning freely at its maximum speed, when a second flywheel of radius r2 = 2.8 m and mass m2 = 16 kg is coaxially dropped from rest onto it and sticks to it, so that they then rotate together as a single body. Calculate the energy, in joules, that is now stored in the wheel?

Part (c) Return now to the flywheel of part (a), with mass m1, radius r1, and speed v at its rim. Imagine the flywheel delivers one third of its stored kinetic energy to car, initially at rest, leaving it with a speed vcar. Enter an expression for the mass of the car, in terms of the quantities defined here.

moment of inertia is given as

$I$ = $\frac{1}{2}$ $mr^{2}$

where m is the mass of the flywheel,

and r is the radius of the flywheel

for the flywheel with radius 1.1 m

and mass 11 kg

moment of inertia will be

$I$ = $\frac{1}{2}$ $*11*1.1^{2}$ = 6.655 kg-m^2

The maximum speed of the flywheel = 35 m/s

we know that v = ωr

where v is the linear speed = 35 m/s

ω = angular speed

r = radius

therefore,

ω = v/r = 35/1.1 = 31.82 rad/s

maximum rotational energy of the flywheel will be

E = $Iw^{2}$ = 6.655 x $31.82^{2}$ = 6738.27 J

b) second flywheel has

radius = 2.8 m

mass = 16 kg

moment of inertia is

$I$ = $\frac{1}{2}$ $mr^{2}$ = $\frac{1}{2}$ $*16*2.8^{2}$ = 62.72 kg-m^2

According to conservation of angular momentum, the total initial angular momentum of the first flywheel, must be equal to the total final angular momentum of the combination two flywheels

for the first flywheel, rotational momentum = $Iw$ = 6.655 x 31.82 = 211.76 kg-m^2-rad/s

for their combination, the rotational momentum is

$(I_{1} +I_{2} )w$

where the subscripts 1 and 2 indicates the values first and second flywheels

$(I_{1} +I_{2} )w$ = (6.655 + 62.72)ω

where ω here is their final angular momentum together

==> 69.375ω

Equating the two rotational momenta, we have

211.76 = 69.375ω

ω = 211.76/69.375 = 3.05 rad/s

Therefore, the energy stored in the first flywheel in this situation is

E = $Iw^{2}$ = 6.655 x $3.05^{2}$ = 61.908 J

c) one third of the initial energy of the flywheel is

6738.27/3 = 2246.09 J

For the car, the kinetic energy = $\frac{1}{2}mv_{car} ^{2}$

where m is the mass of the car

$v_{car}$ is the velocity of the car

Equating the energy

2246.09 = $\frac{1}{2}mv_{car} ^{2}$

making m the subject of the formula

mass of the car m = $\frac{4492.18}{v_{car} ^{2} }$

3 0

2 years ago

A 1150 kg car is on a 8.70° hill. using x-y axis tilted down the plane, what is the x-component of the weight?

Fed [463]

I assume the x-y axis are tilted such that the x-axis is parallel to the surface of the hill while the y-axis is perpendicular to it.

In this case, the x-component of the weight is given by:
$W_x =mg \sin \theta$
where
m is the mass of the car
g is the acceleration of gravity
$\theta$ is the angle of the hill

Substituting numbers into the formula, we find
$W_x=(1150 kg)(9.81 m/s^2)(\sin 8.70^{\circ})=1706 N$

6 0

2 years ago

A car initially traveling at 24 m/s slams on the brakes and moves forward 196 m before coming to a complete halt. What was the m

Marrrta [24]

Answer:

$-1.47 m/s^2$

Explanation:

We can use the following SUVAT equation to solve the problem:

$v^2 - u^2 = 2ad$

where

v = 0 is the final velocity of the car

u = 24 m/s is the initial velocity

a is the acceleration

d = 196 m is the displacement of the car before coming to a stop

Solving the equation for a, we find the acceleration:

$a=\frac{v^2-u^2}{2d}=\frac{0-(24)^2}{2(196)}=-1.47 m/s^2$

4 0

2 years ago

A sister spins her brother in a circle of radius R at angular speed wi. Then the sister decreases her angular speed

Igoryamba

Answer:

The change in the centripetal acceleration of the brother,

Δa = V₂²/R - V₁²/R

Explanation:

Given data,

A sister spins her brother in a circle of radius, R

The angular velocity of the brother, ω₁ = V₁/R

The angular velocity of the brother, ω₂ = V₂/R

The centripetal acceleration is given by the relation

a = V²/R

Therefore change in the centripetal acceleration of the brother,

Δa = V₂²/R - V₁²/R

6 0

2 years ago