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3 years ago

10

The stopping distance d of a car after the brakes are applied varies directly as the square of the speed r. If a car travelling

70 mph can stop in 270 ft, how many feet will it take the same car to stop whenit is travelling 60 mph?

1 answer:

Crank3 years ago

4 0

270/70^2 = x/80^2

Cross multiply

270 (6400) = 4900x

x = 270(6400)/4900

352 and 32/49 feet

Hope this helps

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What is the algebraic expression for the component of the normal force in the vertical direction?

inna [77]

The algebraic expression for the component of the normal force in the vertical direction is Force= product of mass and area.

<h3>What is Force?</h3>

A force is an effect that can alter an object's motion according to physics. An object with mass can change its velocity, or accelerate, as a result of a force. An obvious way to describe force is as a push or a pull. A force is a vector quantity since it has both magnitude and direction. Or, there is a specific meaning to the word "force." At this level, calling a force a push or a pull is entirely appropriate. A force is not something an object "has in it" or that it "contains." One thing experiences a force from another. There are both living things and non-living objects in the concept of a force.

To learn more about Force, visit

brainly.com/question/13014979

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5 0

1 year ago

The astronomical unit (AU) is defined as the mean center-to-center distance from Earth to the Sun, namely 1.496x10^(11) m. The p

Rudiy27

Answer:

a) How many parsecs are there in one astronomical unit?

$4.85x10^{-6}pc$

(b) How many meters are in a parsec?

$3.081x10^{16}m$

(c) How many meters in a light-year?

$9.46x10^{15}m$

(d) How many astronomical units in a light-year?

$63325AU$

(e) How many light-years in a parsec?

3.26ly

Explanation:

The parallax angle can be used to find out the distance using triangulation. Making a triangle between the nearby star, the Sun and the Earth, knowing that the distance between the Earth and the Sun ( $1.496x10^{11} m$ ) is defined as 1 astronomical unit:

$\tan{p} = \frac{1AU}{d}$

Where d is the distance to the star.

Since p is small it can be represent as:

$p(rad) = \frac{1AU}{d}$ (1)

Where p(rad) is the value of in radians

However, it is better to express small angles in arcseconds

$p('') = p(rad)\frac{180^\circ}{\pi rad}.\frac{60'}{1^\circ}.\frac{60''}{1'}$

$p('') = 2.06x10^5 p(rad)$

$p(rad) = \frac{p('')}{2.06x10^5}$ (2)

Then, equation 2 can be replace in equation 1:

$\frac{p('')}{2.06x10^5} = \frac{1AU}{d}$

$\frac{d}{1AU} = \frac{2.06x10^5}{p('')}$ (3)

From equation 3 it can be see that $1pc = 2.06x10^5 AU$

<em>a) How many parsecs are there in one astronomical unit? </em>

$1AU . \frac{1pc}{2.06x10^5AU}$ ⇒ $4.85x10^{-6}pc$

<em>(b) How many meters are in a parsec? </em>

$2.06x10^{5}AU . \frac{1.496x10^{11}m}{1AU}$ ⇒ $3.081x10^{16}m$

<em>(c) How many meters in a light-year? </em>

To determine the number of meters in a light-year it is necessary to use the next equation:

$x = c.t$

Where c is the speed of light ( $c = 3x10^{8}m/s$ ) and x is the distance that light travels in 1 year.

In 1 year they are 31536000 seconds

$x = (3x10^{8}m/s)(31536000s)$

$x = 9.46x10^{15}m$

<em>(d) How many astronomical units in a light-year?</em>

$9.46x10^{15}m . \frac{1AU}{1.496x10^{11}m}$ ⇒ $63325AU$

<em>(e) How many light-years in a parsec?</em>

$2.06x10^{5}AU . \frac{1ly}{63235AU}$ ⇒ $3.26ly$

5 0

4 years ago

Why is the overall charge of the atom neutral or zero?

MrRissso [65]

Answer:

B

Explanation:

this is because the neutrons do not have a charge, the things that have charge in an atom are electrons and protons.

and in the nucleus of an atom, there are protons and neutrons so you can see that A is not the answer

if you see the periodic table, you will know that the number of electrons and protons are equal, so the charges cancel each other out, hence the charge of an atom will be neutral

let me give you a tip which I got from my teacher, never write there is no charge in the atom, this suggests that there is no protons or electrons.

instead, write, the it is neutral

hope it helps if not please report it so that someone else gets to try it out

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3 years ago

What is an non-example of a switch

loris [4]

Dont really understand what youre asking but switches can be used to turn things on and off.

3 0

3 years ago

Use the data provided to calculate the gravitational potential energy of each cylinder mass. Round your answers to the nearest t

Goryan [66]

Answer: 14.7kJ, 29.4kJ, 44.1kJ

Explanation:

<em>The gravitational potential energy is the energy that a body or object possesses, due to its position in a gravitational field. </em>

<em />

In the case of the Earth, in which the gravitational field is considered constant, the value of the gravitational potential energy $U_{p}$ will be:

$U_{p}=mgh$

Where $m$ is the mass of the object, $g=9.8m/s^{2}$ the acceleration due gravity and $h=500m$ the height of the object.

Knowing this, let's begin with the calculaations:

For m=3kg

$U_{p}=(3kg)(9.8m/s^{2})(500m)$

$U_{p}=14700J=14.7kJ$

For m=6kg

$U_{p}=(6kg)(9.8m/s^{2})(500m)$

$U_{p}=29400J=29.4kJ$

For m=9kg

$U_{p}=(9kg)(9.8m/s^{2})(500m)$

$U_{p}=44100J=44.1kJ$

6 0

3 years ago