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WINSTONCH [101]
3 years ago
13

Which statement explains why looking at a dimly lit object with peripheral vision at night results in the clearest image of the

object? When one looks to the side of an object:
A.two images of the object are sent to the brain, which results in a clearer image.
B.its image falls in the periphery of the retina, which is denser in rods.
C.its image falls on the fovea, which results in the clearest image.
D.a single image of the object is sent to the brain, which results in a clearer image.
Physics
1 answer:
FrozenT [24]3 years ago
8 0

Answer:

Option C. Its image falls on the periphery of the retina, denser in rods

Explanation:

This can be explained as the photo receptors present in the outer periphery are more sensitive to light and enable us to see clearly dimly lit objects.

Also the peripheral vision is the the one on the retinal periphery which is highly rich in concentration of rods than cones and thus a dimly lit object at night with peripheral vision gives a clearer image of the object.

The central vision makes use of the central part, i.e., fovea which is dense with the concentration of cones, photo-receptors which enable us to see different colors and functions well in bright light.

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14. a ball is thrown horizontally from the roof of a building 75 m tall with a speed of 4.6 m/s. a. how much later does the ball
Burka [1]

The time taken to hit the ground is 3.9 s, the range is 18m and the final velocity is 42.82 m/s

<h3>Motion Under Gravity</h3>

The motion of an object under gravity is the vertical motion of the object under the influence of acceleration due to gravity.

Given that a ball is thrown horizontally from the roof of a building 75 m tall with a speed of 4.6 m/s.

a. how much later does the ball hit the ground?

The time can be calculated by considering the vertical component of the motion with the use of formula below.

h = ut + 1/2gt²

Where

  • Height h = 75 m
  • Initial velocity u = 0 ( vertical velocity )
  • Acceleration due to gravity g = 9.8 m/s²
  • Time t = ?

Substitute all the parameters into the formula

75 = 0 + 1/2 × 9.8 × t²

75 = 4.9t²

t² = 75/4.9

t² = 15.30

t = √15.3

t = 3.9 s

b. how far from the building will it land?

The range can be found by using the formula

R = ut

Where u = 4.6 m/s ( horizontal velocity )

R = 4.6 × 3.9

R = 18 m

c. what is the velocity of the ball just before it hits the ground?

The final velocity will be

v = u + gt

v = 4.6 + 9.8 × 3.9

v = 4.6 + 38.22

v = 42.82 m/s

Therefore, the answers are 3.9 s, 18 m and 42.82 m/s

Learn more about Vertical motion here: brainly.com/question/24230984

#SPJ1

7 0
1 year ago
The capacitance of A conductor is affected by the presence of A second conductor that is uncharged and isolated electrically. Wh
Komok [63]
What happens is the potential value of the conductor decreases due to the presence of second conductor
as the capacitance is given by C = q/v
the value of v deceases as v-v1
thus the new capacitance is = C' = q/v-v1 thus the lowering of v increases the capacitance
5 0
4 years ago
A student throws a small rock straight upwards. The rock rises to its highest point and then falls back down. How does the speed
Vladimir [108]

Answer:

we assume that it starts with a velocity of 10m/s. At 2m height above ground level, its velocity decreases at 3m above ground level

for its way down the velocity at 3m on its way down is more than its velocity at 2m on its way down.

Explanation:

A student throws a small rock straight upwards. The rock rises to its highest point and then falls back down. How does the speed of the rock at 2m on the way down compare with its speed at 2m on the way up?

It decreases in speed on its way down and increases in speed on its way down.

it decreases in speed on its way up because the the vertical motion is against the earths gravitational pull on an object to the earth's center

.It increases in speed on his way down because its under the influence of gravity

from newton's equation of motion we can check by

using V^2=u^2+2as

we assume that it starts with a velocity of 10m/s. At 2m height above ground level, its velocity decreases at 3m above ground level

for its way down the velocity at 3m on its way down is more than its velocity at 2m on its way down.

5 0
3 years ago
What is (3.25 x 10^5) + (7.5 x 10^4)?
ivanzaharov [21]

Answer:

400000

Explanation:

So first solve one part:

(3.25 * 10^5)

(3.25 * 100,000)

= 325000

Then solve the next part:

(7.5 * 10^4)

(7.5 * 10000)

= 75000

Now lastly, add the two answers:

325000 + 75000 = 400000

Therefore,

(3.25 x 10^5) + (7.5 x 10^4) = 400000

5 0
3 years ago
Read 2 more answers
A ship maneuvers to within 2.46×10³ m of an island’s 1.80 × 10³ m high mountain peak and fires a projectile at an enemy ship 6.1
Nesterboy [21]

Answer:

The distance close to the peak is 597.4 m.

Explanation:

Given that,

Distance of the first ship from the mountain d=2.46\times10^{3}\ m

Height of islandh=1.80\times10^{3}\ m

Distance of the enemy ship from the mountain d'=6.10\times10^{2}\ m

Initial velocity v=2.55\times10^{2}\ m/s

Angle = 74.9°

We need to calculate the horizontal component of initial velocity

Using formula of horizontal component

v_{x}=v\cos\theta

Put the value into the formula

v_{x}=2.55\times10^{2}\cos74.9

v_{x}=66.42\ m/s

We need to calculate the vertical component of initial velocity

Using formula of vertical component

v_{y}=v\sin\theta

Put the value into the formula

v_{y}=2.55\times10^{2}\sin74.9

v_{y}=246.19\ m/s

We need to calculate the time

Using formula of time

t=\dfrac{d}{v_{x}}

t=\dfrac{2.46\times10^{3}}{66.42}

t=37.03\ sec

We need to calculate the height of the shell on reaching the mountain

Using equation of motion

H= v_{y}t-\dfrac{1}{2}gt^2

Put the value in the equation

H=246.19\times37.03-\dfrac{1}{2}\times9.8\times(37.03)^2

H=2397.4\ m

We need to calculate the distance close to the peak

Using formula of distance

H'=H-h

Put the value into the formula

H'=2397.4-1800

H'=597.4\ m

Hence, The distance close to the peak is 597.4 m.

6 0
3 years ago
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