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elena-s [515]
3 years ago
14

#1 “A persons body continuing to move forward even though the car comes to a sudden stop” Which newtons Law Do they pertain? A:

First Law B: Second Law C: Third Law
#2 “A fighter Pilot Feels Massive Amounts of forcé when his plane turns sharply” Which Newton Law?

#3 “ A Paddle wheel boat pushed on water and the water pushes back causing the boat to move” Which Netwons Law?
Physics
1 answer:
Aloiza [94]3 years ago
3 0
<span>#1 “A persons body continuing to move forward even though the car comes to a sudden stop” Which newtons Law Do they pertain?
A: First Law </span>

<span>#2 “A fighter Pilot Feels Massive Amounts of forcé when his plane turns sharply” Which Newton Law?
A: First Law.
</span><span><span>You were following the Newton's first law and kept your velocity straight until you departed from linear motion when you turn sharply; you are forced to follow the curve. <span>The force that the jet exerts on you is called centripetal force and is suitable for the center of curvature of the forced traveling path.</span></span>

</span><span>#3 “ A Paddle wheel boat pushed on water and the water pushes back causing the boat to move” Which Netwons Law?
C: Third Law</span>


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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
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Select the correct answer.
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A. Alternating current
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How are musical instruments grouped?
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A and B are equivalent.  That's one way instruments are often grouped. (the "sopranos", the "altos", the "bass")

C is another way instruments are often grouped; (the "woods", the "brass")

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A simple and common technique for accelerating electrons is shown in the figure, where there is a uniform electric field between
Sunny_sXe [5.5K]

Answer : 4.483\times 10^{15}\ m/s^2.

Explanation:

It is given that,

Electric field strength, E=2.55\times 10^{4}\ N/C

We know that,

Charge of electron, q=1.6\times 10^{-19}\ C

Mass of electron, m=9.1\times 10^{-31}\ kg

From the definition of electric field, F=qE...............(1)

According to Newton's second law, F = ma..........(2)

From equation (1) and (2)

ma=qE

a=\dfrac{qE}{m}

a=\dfrac{1.6\times 10^{-19}\ C\times 2.55\times 10^{4}\ N/C }{9.1\times 10^{-31}\ kg}

a=0.4483\times 10^{16}\ m/s^2

or

a=4.483\times 10^{15}\ m/s^2

So, the horizontal component of acceleration of an electron is 4.483\times 10^{15}\ m/s^2.

Hence, it is the required solution.

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3 years ago
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