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3 years ago

Explain what was ultimately responsible for London’s “killer fog” in 1952.

Physics

2 answers:

stellarik [79]3 years ago

6 0

Answer:

possible answer from edge

Explanation:

After World War II the only coal available to burn residentially, either for cooking or warmth, was of poor quality and contained large amounts of contaminants that made its fumes toxic. So when millions of homes in London burned this low quality coal and the local atmosphere over and around London stagnated, there was nowhere for the fumes to go. This stagnation allowed the toxic coal fumes to build up until they became physically harmful to humans and animals.

Fed [463]3 years ago

5 0

Answer:

Edge 2021 Practice

Explanation:

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A radar station detects an airplane approaching directly from the east. At first observation, the range to the plane is 375.0 m

Gekata [30.6K]

Answer:

819.78 m

Explanation:

<u>Given:</u>

OA = range of initial position of the airplane from the point of observation = 375 m
OB = range of the final position of the airplane from the point of observation = 797 m
$\theta$ = angle of the initial position vector from the observation point = $43^\circ$
$\alpha$ = angle of the final position vector from the observation point = $123^\circ$
$\vec{AB}$ = displacement vector from initial position to the final position

A diagram has been attached with the solution in order to clearly show the position of the plane.

$\vec{OA} = OA\cos \theta \hat{i}+OA \sin \theta \hat{j}\\\Rightarrow \vec{OA} = 375\ m\cos 43^\circ \hat{i}+375\ m\sin 43^\circ \hat{j}\\\Rightarrow \vec{OA} = (274.26\ \hat{i}+255.75\ \hat{j})\ m\\\vec{OB} = OB\cos \alpha \hat{i}+OB \sin \alpha \hat{j}\\\Rightarrow \vec{OB} = 797\ m\cos 123^\circ \hat{i}+797\ m\sin 123^\circ \hat{j}\\\Rightarrow \vec{OB} = (-434.08\ \hat{i}+668.42\ \hat{j})\ m$

Displacement vector of the airplane will be the shortest line joining the initial position of the airplane to the final position of the airplane which is given by:

$\vec{AB}=\vec{OB}-\vec{OA}\\\Rightarrow \vec{AB} = (-434.08\ \hat{i}+668.42\ \hat{j})\ m-(274.26\ \hat{i}+255.75\ \hat{j})\ m\\\Rightarrow \vec{AB} = (-708.34\ \hat{i}+412.67\ \hat{j})\ m$

The magnitude of the displacement vector = $\sqrt{(-708.34)^2+(412.67)^2}\ m = 819.78\ m$

Hence, the magnitude of the displacement of the plane is 819.67 m during the period of observation.

8 0

3 years ago

A 1569 kg car moves with a velocity of 15 m/s. What is its kinetic energy? Joules

Crank

Kinetic energy = 1/2 * mass * velocity^2

In this case,
KE = 1/2 * 1569 kg * (15 (m/s))^2 = 176,5 kN

8 0

3 years ago

8. An ant on a picnic table travels 30 cm eastward, then 25 cm northward, and finally 15 cm westward. What is the ant's displace

wel

Answer:

The displacement of the ant, R = 29.15 cm

The angle of the resultant displacement with its original position is, θ = 30° 57'

The direction of the displacement is towards the northeast.

Explanation:

Given data,

The displacement towards east, d₁ = 30 cm

The displacement towards north, d₂ = 25 cm

The displacement towards west, d₃ = 15 cm

The total displacement towards east,

d₄ = d₁ - d₃

= 30 - 15

= 15 cm

The total displacement of ant is given by the resultant displacement,

R = √(d₂² + d₄² + 2· d₂ d₄ CosФ)

Where Ф is the angle between the vectors, d₂ & d₄

Ф = 90°

Therefore,

R = √(d₂² + d₄²)

Substituting in the above equation,

R = √(25² + 15²)

= 29.15 cm

Hence, the displacement of the ant, R = 29.15 cm

The angle of the resultant displacement with its original position is,

θ = tan⁻¹ (d₄ / d₂)

= tan⁻¹ (15 / 25)

= tan⁻¹ 0.6

= 30° 57'

Hence, the angle of the resultant displacement the its original position is, θ = 30° 57'

The direction of the displacement is towards the northeast.

5 0

4 years ago

A player kicks a soccer ball in a high arc toward the opponent's goal. At the highest point in its trajectory

____ [38]

At the highest point in its trajectory, the ball's acceleration is zero but its velocity is not zero.

<h3>What's the velocity of the ball at the highest point of the trajectory?</h3>

At the highest point, the ball doesn't go more high. So its vertical velocity is zero.
However, the ball moves horizontal, so its horizontal component of velocity is non - zero i.e. u×cosθ.
u= initial velocity, θ= angle of projection

<h3>What's the acceleration of the ball at the highest point of projectile?</h3>

During the whole projectile motion, the earth exerts the gravitational force with a acceleration of gravity along vertical direction.
But as there's no acceleration along vertical direction, so the acceleration along vertical direction is zero.

Thus, we can conclude that the acceleration is zero and velocity is non-zero at the highest point projectile motion.

Disclaimer: The question was given incomplete on the portal. Here is the complete question.

Question: Player kicks a soccer ball in a high arc toward the opponent's goal. At the highest point in its trajectory

A- neither the ball's velocity nor its acceleration are zero.

B- the ball's acceleration points upward.

C- the ball's acceleration is zero but its velocity is not zero.

D- the ball's velocity points downward.

Learn more about the projectile motion here:

brainly.com/question/24216590

#SPJ1

7 0

2 years ago