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3 years ago

What does the pupil of the eye controls

Physics

1 answer:

Alja [10]3 years ago

5 0

The correct answer would be D

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11 kg is a familiar weight for a bag of flour. You are baking cookies for a Save The Rain Forest fund drive. It takes 500 g of f

arlik [135]

Answer: 22 batches.

Explanation:

Given that 11 kg is a familiar weight for a bag of flour. Also, it is given that It takes 500 g of flour to make one batch of cookies.

How many batches of cookies can you make with one bag of flour

Let's first convert 11 kg into grams (g) by multiplying it by 1000

11 × 1000 = 11000 g

Divide 11000 by 500

11000/500 = 22

Therefore, 22 batches of cookies can be made with one bag of flour.

8 0

3 years ago

Is 1 milliliter/second a small or large flow? Why?

ozzi

Answer:

1 ml/second is a small flow

Explanation:

7 0

2 years ago

Read 2 more answers

What does the pupil of the eye controls

Alja [10]

The correct answer would be D

5 0

3 years ago

A 6 kg cart starting from rest rolls down a hill and at the bottom has a speed of 10 m/s. What is the height of the hill?

Arisa [49]

Answer:

h = 5.09 m

Explanation:

Applying the Law of conservation of energy to this situation, we can write:

$Kinetic\ Energy\ Gained\ by\ the\ Cart = Potential\ Energy\ Lost\ by\ the\ Cart\\\frac{1}{2}mv^2 = mgh\\\\h = \frac{v^2}{2g}$

where,

h = height of the hill = ?

v = speed of cart at the end = 10 m/s

g = acceleration due to gravity = 9.81 m/s²

Therefore,

$h = \frac{(10\ m/s)^2}{(2)(9.81\ m/s^2)}\\\\$

4 0

3 years ago

Suppose the ring rotates once every 4.30 ss . If a rider's mass is 58.0 kgkg , with how much force does the ring push on her at

Stells [14]

Answer:

422.36 N

Explanation:

given,

time of rotation = 4.30 s

T = 4.30 s

Assuming the diameter of the ring equal to 16 m

radius, R = 8 m

$v = \dfrac{2\pi R}{T}$

$v = \dfrac{2\pi\times 8}{4.30}$

v = 11.69 m/s

now, Force does the ring push on her at the top

$- N - m g = \dfrac{-mv^2}{R}$

$N + m g = \dfrac{mv^2}{R}$

$N = \dfrac{mv^2}{R}- m g$

$N = m(\dfrac{v^2}{R}- g)$

$N = 58\times (\dfrac{11.69^2}{8}- 9.8)$

N = 422.36 N

The force exerted by the ring to push her is equal to 422.36 N.

6 0

2 years ago