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2 years ago

What is so unusual about plutos orbit

1 answer:

3 0

The answer is the orbit is tilted by 17° relative to the other eight planets. Everything else is false lol Its the most eccentricity, its orbital period is 248 years and neptunes orbital period is 165 years and twice of that is 330z

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Two astronauts (each with mass 100 kg) are drifting together through space. They are connected to each other by a rope 5 m in le

Nana76 [90]

Answer:

1000 kgm²/s, 400 J

1000 kgm²/s, 1000 J

600 J

Explanation:

m = Mass of astronauts = 100 kg

d = Diameter

r = Radius = $\frac{d}{2}$

v = Velocity of astronauts = 2 m/s

Angular momentum of the system is given by

$L=mvr+mvr\\\Rightarrow L=2mvr\\\Rightarrow L=2\times 100\times 2\times 2.5\\\Rightarrow L=1000\ kgm^2/s$

The angular momentum of the system is 1000 kgm²/s

Rotational energy is given by

$K=I\omega^2\\\Rightarrow K=\frac{1}{2}(mr^2)\left(\frac{v}{r}\right)^2\\\Rightarrow K=mv^2\\\Rightarrow K=100\times 2^2\\\Rightarrow K=400\ J$

The rotational energy of the system is 400 J

There no external toque present so the initial and final angular momentum will be equal to the initial angular momentum 1000 kgm²/s

$L_i=L_f\\\Rightarrow 2mv_ir_i=2mv_fr_f\\\Rightarrow v_f=\frac{v_ir_i}{r_f}\\\Rightarrow v_f=\frac{2\times 2.5}{0.5}\\\Rightarrow v_f=10\ m/s$

Energy

$E_2=mv_f^2\\\Rightarrow E_2=100\times 10\\\Rightarrow E_2=1000\ J$

The new energy will be 1000 J

Work done will be the change in the kinetic energy

$W=E_2-E\\\Rightarrow W=1000-400\\\Rightarrow W=600\ J$

The work done is 600 J

5 0

3 years ago

Which event is an example of a contact force?

Scorpion4ik [409]

Answer:

D. a person pulling a sled.

Explanation:

contact force only occurs when something directly comes in contact with another object.

a. is wrong because that is called magnetic force.

b. gravitational force

c. is an electrical force

4 0

3 years ago

What layer is above the troposphere

goldenfox [79]

The stratosphere is the layer above the troposphere

6 0

3 years ago

Read 2 more answers

A car travels at uniform acceleration over a tine interval of 20.0s. its initial velocity is 11.0m/s and its final velocity is 3

SIZIF [17.4K]

Since the acceleration is uniform, we can calculate it from the data we are given:

a = (vf - vi)/2

where vf=33 m/s and vi=11 m/s

Then use Suvat's equation:

x(t) = vi*t + 0.5 * a * t

where t=20s

3 0

3 years ago

In the vertical jump, an athlete starts from a crouch and jumps upward as high as possible. Even the best athletes spend little

horrorfan [7]

Answer:

t_up / t_down = 6.83

Explanation:

Find:

Calculate the ratio of the time he is above y_max/2 to the time it takes him to go from the floor to that height.

Solution:

- Compute the velocity v_o at y_max:

v_i^2 = v_f^2 - 2*g*y_max

0 = v_o^2 - 2*g*y_max

v_o = sqrt (2*g*y_max)

- The total time spend by athlete above height y_max / 2 is:

y - y_o = v_o*t_up - 0.5*g*t^2_up

v_o = 0.5*g*t_up

- Equate two equations:

sqrt (2*g*y_max) = 0.5*g*t_up

t_up = 2*sqrt(2*g*y_max) / g

- The total time taken by athlete to reach height y_max / 2 from ground is:

y - y_o = v_o*t_down - 0.5*g*t^2_down

g*t_^2down - 2*v_o*t_down + y_max = 0

- Solve the quadratic and evaluate t_down:

t_down = (v_o +/- sqrt (v^2_o - g*y_max)) / g

Substitute for v_o = sqrt (2*g*y_max)

t_down = (sqrt(2g*y_max) +/- sqrt(g*y_max)) / g

- We will use the minus quantity, because we need the first part of the journey from ground from the two times he passes the height of y_max/2.

Hence,

t_down = (sqrt(2g*y_max) - sqrt(g*y_max)) / g

t_down = (sqrt(g*y_max) / g) * (sqrt(2) - 1)

- Compute the ratio t_up to t_down:

t_up / t_down = 2*sqrt(2*g*y_max) / g * g / (sqrt(g*y_max)*(sqrt(2) - 1)

= 2*sqrt(2) / (sqrt(2) - 1)

= 6.83

8 0

3 years ago