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3 years ago

Why would a flat sheet of paper and a wad of paper with the same mass not fall through the air at the same rate?

Physics

2 answers:

Gelneren [198K]3 years ago

7 0

The flat sheet of paper has more surface area than the crumpled ball

tatyana61 [14]3 years ago

7 0

The flat sheet has a larger surface area, so there will be more air resistance upon it. You have two forces on your paper, gravity (down) and normal (up.) Although gravity is stronger than the normal of air, the difference between the down force and the up is smaller. For example:

If gravity is 10 newtons and the air resistance is 5, then the force is:
10N-5N= 5N

If gravity is in a vacuum, then air resistance is 0:
10N-0N= 10N

In the first example, you get half as much force bringing your paper down, so it will fall at a slower rate.

When the paper is crumpled, the air resistance is lower because their is less surface area for the normal to work on.

If gravity is 10 newtons and the air resistance is 2, then the force is:
10N-2N= 8N

More force, falling faster.

Ignore this if you don't want to be confused:
Technically, they do fall at the same rate, because all objects fall at the same speed regardless of mass. The illusion of time difference comes from friction.

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1. Two blocks travel along a level frictionless surface. Block A is initially moving to the right at 5.0 m/s, while block B is i

ad-work [718]

Answer: 2.67 m/s

Explanation:

Given

Mass of block A is $m_a=2\ kg$

mass of block B is $m_b=3\ kg$

The initial velocity of block A $u_a=5\ m/s$

the initial velocity of block B is $u_b=0$

After collision velocity of block A is $v_a=1\ m/s$

Conserving momentum

$m_au_a+m_bu_b=m_av_a+m_bv_b\\\\2\times 5+3\times0=2\times 1+3\times v_b\\\\v_b=\dfrac{8}{3}=2.67\ m/s$

The momentum of block A after the collision is $P_a=2\times 1=2\ kg.m/s$

Therefore, there is no change in sign.

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2 years ago

Which of the following is the best explanation of work

Hunter-Best [27]

Answer:

Explanation:

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3 years ago

Read 2 more answers

Astronomers can now report that active star formation was going on at a time when the universe was only 20% as old as it is toda

disa [49]

Answer:

First, as you may know, the light travels at a given velocity.

In vaccum, this velocity is c = 3x10^8 m/s.

And we know that:

distance = velocity*time

Now, if some object (like a star ) is really far away, the light that comes from that star may take years to reach the Earth.

This means that the images that the astronomers see today, actually happened years and years ago (So the night sky is like a picture of the "past" of the universe)

Also, for example, if an astronomer sees some particular thing, he can apply a model (a "simplification" of some phenomena that is used to simplify it an explain it) and with the model, the scientist can infer the information of the given thing some time before it was seen.

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3 years ago

Benny has 20 jellybeans and wants to share with his friends how many will each friend get? there are 5 friends.

trasher [3.6K]

Answer each friend will get 3.33333 repeating if he is included. if only his friends are getting them then each one gets 4

Explanation:

devide 20/6 and 20/5 respectively.

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3 years ago

You have been hired as a technical consultant for an early-morning cartoon series for children to make sure that the science is

katen-ka-za [31]

The initial potential energy of the wagon containing gold boxes will enable

it roll down the hill when cut loose.

The Lone Ranger and Tonto have approximately <u>5.1 seconds</u>.

Reasons:

Mass wagon and gold = 166 kg

Location of the wagon = 77 meters up the hill

Slope of the hill = 8°

Location of the rangers = 41 meters from the canyon

Mass of Lone Ranger, m₁ = 65 kg

Mass of Tonto m₂ = 66 kg

Solution;

Height of the wagon above the level ground, h = 77 m × sin(8°) ≈ 10.72 m

Potential energy = m·g·h

Where;

g = Acceleration due to gravity ≈ 9.81 m/s²

Potential energy of wagon, P.E. ≈ 166 × 9.81 × 10.72 = 17457.0912

Potential energy of wagon, P.E. ≈ 17457.0912 J

By energy conservation, P.E. = K.E.

$K.E. = \mathbf{\dfrac{1}{2} \cdot m \cdot v^2}$

Where;

v = The velocity of the wagon a the bottom of the cliff

Therefore;

$\dfrac{1}{2} \times 166 \times v^2 = 17457.0912$

$v = \sqrt{\dfrac{17457.0912}{\dfrac{1}{2} \times 166} } \approx 14.5$

Velocity of the wagon, v ≈ 14.5 m/s

Momentum = Mass, m × Velocity, v

Initial momentum of wagon = m·v

Final momentum of wagon and ranger = (m + m₁ + m₂)·v'

By conservation of momentum, we have;

m·v = (m + m₁ + m₂)·v'

$\therefore v' = \mathbf{ \dfrac{m \cdot v}{(m + m_1 + m_2) }}$

Which gives;

$\therefore v' = \dfrac{166 \times 14.5}{(166 + 65 + 66) } \approx 8.1$

The velocity of the wagon after the Ranger and Tonto drop in, v' ≈ 8.1 m/s

$Time = \dfrac{Distance}{Velocity}$

$\mathrm{The \ time \ the\ Lone \ Ranger \ and \ Tonto \ have, \ t} = \dfrac{41 \, m}{8.1 \, m/s} \approx 5.1 \, s$

The Lone Range and Tonto have approximately <u>5.1 seconds</u> to grab the

gold and jump out of the wagon before the wagon heads over the cliff.

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2 years ago