All categories

3 years ago

based on your reading "take a closer look" what can you say about the image that forms on the retina of your eye?

Physics

2 answers:

shepuryov [24]3 years ago

8 0

Answer:

it is an upside-down image but the brain realizes it and corrects it in its orientation so we see it in right direction.

Nuetrik [128]3 years ago

3 0

Answer:

The lens of the eye is a convex lens. The image that it forms on the retina is upside down.

Explanation:

The eye makes an image, but it is upside down, the brain corrects the image and turns it right side up

You might be interested in

__is the distance traveled during a specific unit of time.<br> 21 pts

denpristay [2]

Answer:

The speed of an object

7 0

3 years ago

Read 2 more answers

When light passes through a convex lens, it does this.....

Tpy6a [65]

Explanation:

6. Converge or come together

7. convex

3 0

3 years ago

Read 2 more answers

You can feel the heat waves from a small electric heater on your hands and face even though the heater is on the other side of t

NeX [460]

That's an example of heat transfer by radiation.

8 0

3 years ago

What two things are a included when describing force? Question 3 options: Direction and mass Strength and size Size and directio

Vinvika [58]

Force is a vector quantity and denotes any interaction that change the motion of an object in a certain direction. When influence by force an object can change its velocity or

The two things that are a included when describing force are: magnitude and direction. To fully describe the force acting upon an object, you must describe both the magnitude (size or numerical value) and the direction.

6 0

3 years ago

A rock is thrown upward from a bridge into a river below. The function f(t)=−16t2+44t+88 determines the height of the rock above

babymother [125]

1) 88 ft

2) 4.09 s

3) 1.38 s

4) 118.2 m

Explanation:

For an object thrown upward and subjected to free fall, the height of the object at any time t is given by the suvat equation:

$h(t) = h_0 + ut - \frac{1}{2}gt^2$ (1)

where

$h_0$ is the height at time t = 0

$u$ is the initial vertical velocity

$g=32 ft/s^2$ is the acceleration due to gravity

The function that describes the height of the rock above the surface at a time t in this problem is

$f(t)=-16t^2+44t+88$ (2)

By comparing the terms with same degree of eq(1) and eq(2), we observe that

$h_0 = 88 ft$

which means that the rock is at height h = 88 ft when t = 0: therefore, this means that the height of the bridge above the water is 88 feet.

The rock will hit the water when its height becomes zero, so when

$f(t)=0$

which means when

$0=-16t^2+44t+88$

First of all, we can simplify the equation by dividing each term by 4:

$0=-4t^2+11t+22$

This is a second-order equation, so we solve it using the usual formula and we find:

$t_{1,2}=\frac{-b \pm \sqrt{b^2-4ac}}{2a}=\frac{-11\pm \sqrt{(11)^2-4(-4)(22)}}{2(-4)}=\frac{-11\pm \sqrt{121+352}}{-8}=\frac{-11\pm 21.75}{-8}$

Which gives only one positive solution (we neglect the negative solution since it has no physical meaning):

t = 4.09 s

So, the rock hits the water after 4.09 seconds.

Here we want to find how many seconds after being thrown does the rock reach its maximum height above the water.

For an object in free fall motion, the vertical velocity is given by the expression

$v=u-gt$

where

u is the initial velocity

g is the acceleration due to gravity

t is the time

The object reaches its maximum height when its velocity changes direction, so when the vertical velocity is zero:

$v=0$

which means

$0=u-gt$

Here we have

$u=+44 ft/s$ (initial velocity)

$g=32 ft/s^2$ (acceleration due to gravity)

Solving for t, we find the time at which this occurs:

$t=\frac{u}{g}=\frac{44}{32}=1.38 s$

The maximum height of the rock can be calculated by evaluating f(t) at the time the rock reaches the maximum height, so when

t = 1.38 s

The expression that gives the height of the rock at time t is

$f(t)=-16t^2+44t+88$

Substituting t = 1.38 s, we find:

$f(1.38)=-16(1.38)^2 + 44(1.38)+88=118.2 m$

So, the maximum height reached by the rock during its motion is

$h_{max}=118.2 m$

Which means 118.2 m above the water.

3 0

4 years ago