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mariarad [96]
3 years ago
15

During a trampoline routine, a gymnast is tumbling in the air at 20 rad/s in a tuck position. He then extends into a layout posi

tion and doubles his radius of gyration just before landing on the trampoline bed. How fast is his angular velocity at this instant, just before landing on the trampoline bed? Group of answer choices 5 rad/s 10 rad/s 5 m/s 4 rad/s 80 rad/s
Physics
1 answer:
Yuki888 [10]3 years ago
5 0

Answer:

\omega = 5 rad/s

Explanation:

As we know that the gymnast has no external torque on it

so here we can say that angular momentum of the system will be conserved

so here we have

L_1 = L_f

I_1\omega_1 = I_2\omega_2

m_1r_1^2\omega_1 = m_1r_2^2\omega

r_1^2\omega_1 = r_2^2\omega

here we know that

r_2 = 2r_1

r_1^2\omega_1 = (2r_1)^2\omega

20 = 4\omega

\omega = 5 rad/s

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a 12.0L container is filled with a gas to a pressure of 2660 torr at 0°C. At what temp. will the pressure inside the container b
Alexxx [7]
\frac{p1}{t1} = \frac{p2}{t2}

Pressure Law: constant volume

Convert all Temperatures to Kelvin

0°C= 273K

\frac{2660}{273} = \frac{3040}{t2} \\ \\ 3040 \times 273 = t2 \times 2660 \\ \\ 829920 \div 2660 = t2 \\ \\ t2 = 312

answer= 312Kelvin
6 0
3 years 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
3 years ago
The pressure of a box pushes down on the floor is 50 Pa if the box weighs 400 N what is the area of the base of the box
Wewaii [24]

Answer:8m^2

Explanation:

Area=force÷pressure

Area=400÷50

A=8m^2

5 0
3 years ago
What is in the crysphere
butalik [34]

I think you mean the Cryosphere?

But the answer is D- Earths Ice

This word Cryosphere comes from the greek word "kryos" which means cold

Many people think of the cryosphere as being the north and south poles  but snow and ice can be found in a lot of places on the Earth

3 0
3 years ago
What is the ratio of escape speed from earth to circular orbital speed? ignore air resistance.
klio [65]
About 40 000 km/h
Here you go:

8 0
3 years ago
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