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3 years ago

Aristotle supported which of these views?

1 answer:

7 0

Aristotle believed that the Earth was orbited by the sun, moon, and the other planets which he referred to as Celestial Bodys

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Ryan applied a force of 10N and moved a book 30 cm in the direction of the force. How much was the workdone by Ryan?

Xelga [282]

<h2><u>Question</u><u>:</u><u>-</u></h2>

Ryan applied a force of 10N and moved a book 30 cm in the direction of the force. How much was the work done by Ryan?

<h2><u>Answer:</u><u>-</u></h2>

<h3>Given,</h3>

=> Force applied by Ryan = 10N

=> Distance covered by the book after applying force = 30 cm

30 cm = 0.3 m (distance)

=> Work done = Force × Distance

=> 10 × 0.3

=> 3 Joules

$\small \boxed{work \: done \: by \: Ryan \: = 3 \: Joules}$

4 0

2 years ago

I NEED HELP PLEASE, THANKS! :)

mrs_skeptik [129]

Answer:

1. Largest force: C; smallest force: B; 2. ratio = 9:1

Explanation:

The formula for the force exerted between two charges is

$F=K\dfrac{ q_{1}q_{2}}{r^{2}}$

where K is the Coulomb constant.

q₁ and q₂ are also identical and constant, so Kq₁q₂ is also constant.

For simplicity, let's combine Kq₁q₂ into a single constant, k.

Then, we can write

$F=\dfrac{k}{r^{2}}$

1. Net force on each particle

Let's

Call the distance between adjacent charges d.
Remember that like charges repel and unlike charges attract.

Define forces exerted to the right as positive and those to the left as negative.

(a) Force on A

$\begin{array}{rcl}F_{A} & = & F_{B} + F_{C} + F_{D}\\& = & -\dfrac{k}{d^{2}} - \dfrac{k}{(2d)^{2}} +\dfrac{k}{(3d)^{2}}\\& = & \dfrac{k}{d^{2}}\left(-1 - \dfrac{1}{4} + \dfrac{1}{9} \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{-36 - 9 + 4}{36} \right)\\\\& = & \mathbf{-\dfrac{41}{36} \dfrac{k}{d^{2}}}\\\\\end{array}$

(b) Force on B

$\begin{array}{rcl}F_{B} & = & F_{A} + F_{C} + F_{D}\\& = & \dfrac{k}{d^{2}} - \dfrac{k}{d^{2}} + \dfrac{k}{(2d)^{2}}\\& = & \dfrac{k}{d^{2}}\left(\dfrac{1}{4} \right)\\\\& = &\mathbf{\dfrac{1}{4} \dfrac{k}{d^{2}}}\\\\\end{array}$

(C) Force on C

$\begin{array}{rcl}F_{C} & = & F_{A} + F_{B} + F_{D}\\& = & \dfrac{k}{(2d)^{2}} + \dfrac{k}{d^{2}} + \dfrac{k}{d^{2}}\\& = & \dfrac{k}{d^{2}}\left( \dfrac{1}{4} +1 + 1 \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{1 + 4 + 4}{4} \right)\\\\& = & \mathbf{\dfrac{9}{4} \dfrac{k}{d^{2}}}\\\\\end{array}$

(d) Force on D

$\begin{array}{rcl}F_{D} & = & F_{A} + F_{B} + F_{C}\\& = & -\dfrac{k}{(3d)^{2}} - \dfrac{k}{(2d)^{2}} - \dfrac{k}{d^{2}}\\& = & \dfrac{k}{d^{2}}\left( -\dfrac{1}{9} - \dfrac{1}{4} -1 \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{-4 - 9 -36}{36} \right)\\\\& = & \mathbf{-\dfrac{49}{36} \dfrac{k}{d^{2}}}\\\\\end{array}$

(e) Relative net forces

In comparing net forces, we are interested in their magnitude, not their direction (sign), so we use their absolute values.

$F_{A} : F_{B} : F_{C} : F_{D} = \dfrac{41}{36} : \dfrac{1}{4} : \dfrac{9}{4} : \dfrac{49}{36}\ = 41 : 9 : 81 : 49\\\\\text{C experiences the largest net force.}\\\text{B experiences the smallest net force.}\\$

2. Ratio of largest force to smallest

$\dfrac{ F_{C}}{ F_{B}} = \dfrac{81}{9} = \mathbf{9:1}\\\\\text{The ratio of the largest force to the smallest is $\large \boxed{\mathbf{9:1}}$}$

7 0

2 years ago

Here is something to blow your mind. In one slice of "everything" pizza there are 650 Calories (1000 chemistry calories). That e

amid [387]

Answer:

Number of slices of pizza is 993.

Explanation:

It is given that, in one slice of "everything" pizza there are 650 Calories. The conversion factor from calories to joules is :

1 calorie = 4.184 joules

650 calories = 2719.6 joules

Total energy in the pizza, E = 2700000 J

Let there are n number of slices of pizza. It is given by :

$n=\dfrac{2700000}{2719.6}$

n = 992.79

n = 993

So, there are 993 slices of pizza. Hence, this is the required solution.

6 0

2 years ago

Question 1 of 15

Ierofanga [76]

Answer:

d 75 miles per hour

Explanation:

60 ÷ 20

3 × 25 = 75

hoped that helped!!

8 0

2 years ago

Twice each each lunar month, all year long, these tides occur. Whenever the Moon, Earth and Sun are aligned, the gravitational p

Zarrin [17]

That was a lucky pick.

Twice each each lunar month, all year long, whenever the Moon,
Earth and Sun are aligned, the gravitational pull of the sun adds
to that of the moon causing maximum tides.

This is the setup at both New Moon and Full Moon. It doesn't matter
whether the Sun and Moon are both on the same side of the Earth,
or one on each side. As long as all three bodies are lined up, we
get the biggest tides.

These are called "spring tides", when there is the greatest difference
between high and low tide.

At First Quarter and Third Quarter, when the sun, Earth, and Moon form a
right angle, there is the least difference between high and low tide. Then
they're called "neap tides".

4 0

2 years ago

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