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mote1985 [20]
4 years ago
12

If a person can't accompdate and the glasses is +2.50D. At which distance will the person see cleary

Physics
1 answer:
snow_tiger [21]4 years ago
5 0

I believe the correct answer is 40 cm.

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In which phase of mitosis are chromosome first seen as a result of chromatin coiling?
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<span>Mitosis is a a means for cells to split and produce exact copies of themselves. The process produces two identical copies of the original cell and occurs throughout the human body. Mitosis is divided up into four main phases known as prophase, metaphase, anaphase and telophase. The chromosomes first become visible in early prophase.</span>
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3 years ago
8. A driver starts his parked car and within 12 hours reaches a speed of 75km/h, as he travels
erastovalidia [21]

Answer:

56mph

Explanation:

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3 years ago
A package is dropped from a helicopter moving upward at 15 m/s
daser333 [38]

The distance the package above the ground when it was released, s ≈ 530 meters

<h3 /><h3>What are kinematic equations?</h3>

The kinematic equation of motion gives the interrelationships of the variables of motion.The correct option for the distance the package above the ground when it was released, is the third option;

It is given that:

The velocity of the helicopter from which the package was dropped = 15 m/s

The time it takes the package to strike the ground = 12 seconds

The required parameter:

The height of the package from the ground when it was dropped

The kinematic equation of motion relating distance, s, time, t, acceleration due to gravity, g, initial velocity, u, and final velocity, v, is applied as follows;

The package continues the upward motion for some time, t₁, given as follows;

Upward motion of the package

v = u - g·t₁

v = 0 at highest point reached by the package;

Therefore;

0 = 15 m/s - 9.81 m/s²  × t₁

t₁ = 15 m/s/(9.81 m/s²) ≈ 1.5295022 seconds

The time the package takes to return to the initial starting point, t₂ = t₁

The time the package falls after returning to the point it was dropped, t₃, is given as follows;

t₃ = t - (t₂ + t₁) = t - 2 × t₁

∴ t₃ = 12 s - 2 × 1.5295022 s ≈ 8.940996 s

From the symmetry of the motion of a projectile, the velocity of the package when returns to its staring point where it was dropped = u (Downwards) = 15 m/s

The distance the package falls, s, which is the distance the package above the ground when it was released, is given as follows;

s = u·t + (1/2)·g·t²

s = 15× 8.940996  + (1/2) × 9.81 × 8.940996² = 526.22755346 ≈ 530

The distance the package falls, s ≈ 530 m = The height of the

The distance the package above the ground when it was released, s ≈ 530 meters

Learn more about the kinematic equations of motion here:

brainly.com/question/16995301

#SPJ4

7 0
2 years ago
The most interpersonal constructive passion response to relational conflict is..
Ilya [14]

Answer:

loyalty

Explanation:

7 0
4 years ago
A very light ping-pong ball moving east at a speed of 4 m/s collides with a very heavy stationary bowling ball. The Ping-Pong ba
KengaRu [80]

Answer:

They experience the same magnitude impulse

Explanation:

We have a ping-pong ball colliding with a stationary bowling ball. According to the law of conservation of momentum, we have that the total momentum before and after the collision must be conserved:

where is the initial momentum of the ping-poll ball

is the initial momentum of the bowling ball (which is zero, since the ball is stationary)

is the final momentum of the ping-poll ball

is the final momentum of the bowling ball

We can re-arrange the equation as follows or

which means (1) so the magnitude of the change in momentum of the ping-pong ball is equal to the magnitude of the change in momentum of the bowling ball.

However, we also know that the magnitude of the impulse on an object is equal to the change of momentum of the object:

(2) therefore, (1)+(2) tells us that the ping-pong ball and the bowling ball experiences the same magnitude impulse:

3 0
3 years ago
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