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3 years ago

The persistence of vision for normal eye is

Physics

2 answers:

attashe74 [19]3 years ago

6 0

Answer:

The answer is 1/16

Explanation:

1. Persistence of vision refers to the optical illusion that occurs when visual perception of an object does not cease for some time after the rays of light proceeding from it have ceased to enter the eye. 2. The persistence of vision for normal eye is 1/16 if a second.

pashok25 [27]3 years ago

4 0

Answer:

idk

Explanation:

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The crest of one wave with an amplitude of 4 m meets up with the crest of a second

trapecia [35]

Answer:

Constructive

Explanation:

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2 years ago

Which of these is not true about the proton?

bulgar [2K]

Answer:

Option C is the untrue statement.

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3 years ago

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A 1000 kg truck moving at 2.0 m/s runs into a concrete wall. It takes 0.5 s for the truck to completely stop. What is the magnit

guajiro [1.7K]

Answer:

Momentum is given by

. Impulse is the change of momentum,

and is also equal to force times time:

. Rearranging,

−

000

−

4000

Explanation:

Momentum before the collision is

2000

⋅

000

−

Assuming the truck comes to a complete halt, the momentum after the collision is

−

The change in momentum,

, is initial minus final

→

−

000

−

000

This is called the impulse:

. Impulse is also equal (check the units) to force times time:

We can rearrange this expression to make

the subject:

−

000

−

4000

The negative sign just means the force acting is in the opposite direction to the initial momentum.

(This will be the average force acting during the collision: collisions are chaotic so the force is unlikely to be constant.)

3 0

3 years ago

Mercury has a radial acceleration of 3.96 × 10−2 m/s2 and its orbital period is T = 88 days. What is the radius of Mercury’s orb

Maslowich

Answer: 58,045,522,878.8 meters

Explanation:

Ok, the data we have is

Period = T = 88 days

Radial acceleration = ar = 3.96x10^-2 m/s^2

And we know that the equation for the radial acceleration is:

ar = v^2/r = r*w^2

Where v is the velocity. r is the radius and w is the angular velocity.

And we know that:

w = 2*pi*f

where f is the frequency, and:

T = 1/f.

Then we can write:

w = 2*pi/T

and our equation becomes:

ar = r*(2*pi/T)^2

Now we solve this for r.

First we need to use the same units in both equations, so we want to write T in seconds.

T = 88 days,

A day has 24 hours, and one hour has 3600 seconds:

T = 88*24*3600 s =7,603,200s

Then:

3.96x10^-2 m/s^2 = r*(2*3.14/7,603,200s)^2

r = (3.96x10^-2 m/s^2) /(2*3.14/7,603,200s)^2 = 58,045,522,878.8 meters

5 0

3 years ago

Approximate the Sun as a uniform sphere of radius 6.96 X 108 m, rotating about its central axis with a period of 25.4 days. Supp

In-s [12.5K]

Answer:

T = 184 seconds

Explanation:

First in order to solve this, we need to know which is the expression to calculate the period. This is an exercise of angular velocity, so:

T = 2π/w

Where w: angular speed (in rad/s)

So, let's calculate first the innitial angular speed:

w = 2π/T

Converting days to seconds:

25.4 days * 24 h/day * 3600 s/h = 2,194,560 s

Then the angular speed:

w = 2π / 2,194,560 = 2.863x10^-6 rad/s

Now, the innitial angular momentum is:

I = (2/5)Mr² replacing data:

I = 2/5* (6.96x10^8)² * M = 1.94x10^17m² * M

so the initial angular momentum would be:

L = Iω = 2.863x10^-6 * 1.94x10^17 M

L = 5.55x10^11 m²/s * M = final angular momentum

Now the final I = 2/5Mr²

Final I = 2/5 * (6.37x10^6)² * M = 1.62x10^13m² * M

Then 5.55x10^11m²/s * M = 1.62x10^13m² * M * ω → M cancels

ω = 3.42x10^-2 rad/s

Then the new period

T = 2π/ω = 2*3.14 / 3.42x10^-2

T = 184 seconds

8 0

3 years ago