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3 years ago

The average speed of a snail is 0.020 miles/hour and that of a Leopard is 70 miles/hour. Convert these speeds in SI units.

1 answer:

8 0

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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?

$3.081x10^{16}m$

$9.46x10^{15}m$

(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 $1pc = 2.06x10^5 AU$

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

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

(b) How many meters are in a parsec?

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

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

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

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

$x = 9.46x10^{15}m$

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

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

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

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

5 0

3 years ago

A 33 kg gun is standing on a frictionless surface. The gun fires a 57.7g bullet with a muzzle velocity of 325m/s. The positive d

Stella [2.4K]

Kinetic Energy = (1/2) mv^2.

m = 57.7 g = 57.7/1000 = 0.00577 kg.
v = 325 m/s.

E = 0.5 * 0.00577 * 325^ 2. Use your calculator.

E = 304.728125 J.

That's the kinetic energy.

7 0

3 years ago

Uncle Harry weighs 750 N. What is his mass in kg?

Elenna [48]

Answer:

W = MG

750 = M * 10

M = 750/10

M = 75 kg

8 0

3 years ago

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PHYSICS HELP PLEASE!! MUST SHOW MATH WORK!!

SVETLANKA909090 [29]

Answer:

1) Distance traveled by bird = 403 meter

2)Average speed = 1.66 km /hour

3) Zcceleration = 2 $m/s^2$

Explanation:

1) Distance traveled = Speed * Time taken = 31 * 13 = 403 meter.

2) Average speed = Total distance covered / Time taken for that distance to cover.

Total distance covered = 2+0.5+2.5 = 5 km

Time taken = 3 hours

Average speed = 5/3 = 1.66 km /hour

3) Acceleration is defined as the rate of change of velocity, so acceleration a = change in velocity/time.

Change in velocity = 14 - 6 = 8 m/s

Time = 4 seconds

So acceleration = 8 / 4 = 2 $m/s^2$

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3 years ago

When you go on a hunting trip, you should leave a hunting plan with someone you trust. What information should the plan include?

bulgar [2K]

You should put when you will leave, where you will be, and what time you will get back.

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3 years ago

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The average speed of a snail is 0.020 miles/hour and that of a Leopard is 70 miles/hour. Convert these speeds in SI units. ​

The average speed of a snail is 0.020 miles/hour and that of a Leopard is 70 miles/hour. Convert these speeds in SI units.