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Ipatiy [6.2K]
3 years ago
6

An angler experiences a glare of light coming from the water's surface. Which type of glasses should she wear to minimize the gl

are?
Physics
2 answers:
vodka [1.7K]3 years ago
7 0

Explanation :

An angler experiences a glare of light coming from the water's surface. Polarized glasses are used to minimize the glare coming from water.

When light is reflected from flat surfaces, it has the ability to move in a multi-direction. The reflected light has a very high intensity which can cause glare and can decreases visibility.

By using polarized sunglasses, the glare can reflect off to the ground, water bodies etc.

Vaselesa [24]3 years ago
6 0
The glasses that she should wear are the glasses that block horizontally polarized light coming from glares, like polaroid sunglasses, because these type of glasses block out the light being reflected from the water's surface by the vertically oriented polarizers in the lenses. 
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Calculate the displacement in m and velocity in m/s at the following times for a rock thrown straight down with an initial veloc
svet-max [94.6K]

Incomplete question as time is missing.I have assumed some times here.The complete question is here

Calculate the displacement and velocity at times of (a) 0.500 s, (b) 1.00 s, (c) 1.50 s, (d) 2.00 s, and (e) 2.50 s for a rock thrown straight down with an initial velocity of 10 m/s from the Verrazano Narrows Bridge in New York City. The roadway of this bridge is 70.0 m above the water.

Explanation:

Given data

Vi=10 m/s

S=70 m

(a) t₁=0.5 s

(b) t₂=1 s

(c) t₃=1.5 s

(d) t₄=2 s

(e) t₅=2.5 s

To find

Displacement S from t₁ to t₅

Velocity V from t₁ to t₅

Solution

According to kinematic equation of motion and given information conclude that v is given by

v=v_{i}+gt\\

Also get the equation of displacement

S=v_{i}t+(1/2)gt^{2}

These two formula are used to find velocity as well as displacement for time t₁ to t₅

For t₁=0.5 s

v_{1}=v_{i}+gt\\v_{1}=(10m/s)+(9.8m/s^{2} ) (0.5s)\\v_{1}=14.9m/s\\  And\\S_{1} =v_{i}t+(1/2)gt^{2}\\ S_{1}=(10m/s)(0.5s)+(1/2)(9.8m/s^{2} )(0.5s)^{2} \\S_{1}=6.225m

For t₂

v_{2}=v_{i}+gt\\v_{2}=(10m/s)+(9.8m/s^{2} ) (1s)\\v_{2}=19.8m/s\\  And\\S_{2} =v_{i}t+(1/2)gt^{2}\\ S_{2}=(10m/s)(1s)+(1/2)(9.8m/s^{2} )(1s)^{2} \\S_{2}=14.9m

For t₃

v_{3}=v_{i}+gt\\v_{3}=(10m/s)+(9.8m/s^{2} ) (1.5s)\\v_{3}=24.7m/s\\  And\\S_{3} =v_{i}t+(1/2)gt^{2}\\ S_{3}=(10m/s)(1.5s)+(1/2)(9.8m/s^{2} )(1.5s)^{2} \\S_{3}=26.025m

For t₄

v_{4}=v_{i}+gt\\v_{4}=(10m/s)+(9.8m/s^{2} ) (2s)\\v_{4}=29.6m/s\\  And\\S_{4} =v_{i}t+(1/2)gt^{2}\\ S_{4}=(10m/s)(2s)+(1/2)(9.8m/s^{2} )(2s)^{2} \\S_{4}=39.6m

For t₅

v_{5}=v_{i}+gt\\v_{5}=(10m/s)+(9.8m/s^{2} ) (2.5s)\\v_{5}=34.5m/s\\  And\\S_{5} =v_{i}t+(1/2)gt^{2}\\ S_{5}=(10m/s)(2.5s)+(1/2)(9.8m/s^{2} )(2.5s)^{2} \\S_{5}=55.625m

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4 years ago
The process of burning fuel is called?
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The process of burning fuel is Combustion
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4 years ago
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an object tied to the end of the string moves in a circle . the force exerted by the string depends on the mass of the object it
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Answer:Fc=mv^2/r

Explanation:

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4 years ago
A typical ten-pound car wheel has a moment of inertia of about 0.35kg⋅m2. The wheel rotates about the axle at a constant angular
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Answer:

The rotational kinetic energy of the rotating wheel is 529.09 J  

Explanation:

Given;

moment of inertia I = 0.35kg⋅m²

number of revolutions = 35.0

time of revolution, t = 4.00 s

Angular speed (in revolution per second), ω = 35/4 = 8.75 rev/s

Angular speed (in radian per second), ω = 8.75 rev/s x 2π = 54.985 rad/s

Rotational kinetic energy, K = ¹/₂Iω²

Rotational kinetic energy, K = ¹/₂ x 0.35 x (54.985)²

Rotational kinetic energy, K = 529.09 J  

Therefore, the rotational kinetic energy of the rotating wheel is 529.09 J  

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Until a train is a safe distance from the station it must travel at 5 m/s. Once the train is on open track it can speed up to
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Answer: 45x8=360 :)

Explanation:

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