All categories

3 years ago

Can you label the parts of the eye?

Physics

1 answer:

photoshop1234 [79]3 years ago

7 0

Answer:

Iris
Pupil
Corona
Anterior chamber
lens
Vitreous humor
Blood vessels
Optic nerve
Hyaloid canal
Retina

Hope it helps :)

You might be interested in

When figure skating how do you quickly transition from backwards to forwards without falling or just going in a circle? I'm fair

olga2289 [7]

I am a competitive figure skater. There are certain turns you can use such as a mowhawk, where you set one foot down that is facing the opposite direction from which you are gliding. There is a two foot turn, where you either go on or toes and turn backwards, or lean somewhat on your heals and turn forwards. Use your hips to help turn. And a 3 turn, which is basically a 2 foot turn on 1 foot. But remember, it takes practice, and you may fall a couple times.

8 0

4 years ago

Is a table made from minerals?

Serga [27]

A table can be made from minerals but other kinds of solids as well

3 0

3 years ago

Read 2 more answers

a ball kicked with a velocity of 8m/s at an angle of 30 degree to horizontal. calculate the time of flight of the ball. (g=10ms^

posledela

Answer:

Approximately $0.8\; \rm s$ (assuming that air resistance is negligible.)

Explanation:

Let $v_0$ denote the initial velocity of this ball. Let $\theta$ denote the angle of elevation of that velocity.

The initial velocity of this ball could be decomposed into two parts:

Initial vertical velocity: $v_0(\text{vertical}) = v_0 \cdot \sin(\theta)$ .
Initial horizontal velocity: $v_0(\text{vertical}) = v_0 \cdot \cos(\theta)$ .

If air resistance on this ball is negligible, $v_0(\text{vertical})$ alone would be sufficient for finding the time of flight of this ball.

Calculate $v_0(\text{vertical})$ given that $v_0 = 8 \; \rm m \cdot s^{-1}$ and $\theta = 30^\circ$ :

$\begin{aligned}& v_0(\text{vertical}) \\ &= v_0 \cdot \sin(\theta) \\ &= \left(8 \; \rm m \cdot s^{-1} \right) \cdot \sin\left(30^{\circ}\right) \\ &= 4\;\rm m \cdot s^{-1} \end{aligned}$ .

Assume that air resistance on this ball is zero. Right before the ball hits the ground, the vertical velocity of this ball would be exactly the opposite of the value when the ball was launched.

Since $v_0(\text{vertical}) = 4\; \rm m \cdot s^{-1}$ , the vertical velocity of this ball right before landing would be $v_1(\text{vertical}) = -4\; \rm m \cdot s^{-1}$ .

Calculate the change to the vertical velocity of this ball:

$\begin{aligned}& \Delta v(\text{vertical}) \\ & = v_1(\text{vertical}) - v_0(\text{vertical}) \\ &= -8\; \rm m \cdot s^{-1}\end{aligned}$ .

In other words, the vertical velocity of this ball should have change by $8\; \rm m \cdot s^{-1}$ during the entire flight (from the launch to the landing.)

The question states that the gravitational field strength on this ball is $g = 10\; \rm m \cdot s^{-2}$ . In other words, the (vertical) downward gravitational pull on this ball could change the vertical velocity of the ball by $10\; \rm m\cdot s^{-1}$ each second. What fraction of a second would it take to change the vertical velocity of this ball by $8\; \rm m \cdot s^{-1}$ ?

$\begin{aligned}t &= \frac{\Delta v(\text{initial})}{g} \\ &= \frac{8\; \rm m \cdot s^{-1}}{10\; \rm m \cdot s^{-2}} = 0.8\; \rm s\end{aligned}$ .

In other words, it would take $0.8\; \rm s$ to change the velocity of this ball from the initial velocity at launch to the final velocity at landing. Therefore, the time of the flight of this ball would be $0.8\; \rm s\!$ .

5 0

3 years ago

Which is an example of thigmotropism?

sertanlavr [38]

Answer A, climbing vines. Thigmotropism is growth in response to mechanical contact.

5 0

3 years ago

Read 2 more answers

Please answer my quastions

Neko [114]

Answer:

a) 120 N

b) 5 N

c) 0.2 N

d) Mass remains the same, and weight decreases.

Explanation:

Use the formula W = mg, where mass is in kg, and gravitational field strength in N/kg.

W = mg

= 12 × 10

= 120 N

500 g = 0.5 kg

W = mg

= 0.5 × 10

= 5 N

20 g = 0.02 kg

W = mg

= 0.02 × 10

= 0.2 N

Mass remains the same, and weight decreases.

6 0

3 years ago