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2 years ago

Why do you think that some asteroids tumble end over end through space while other asteroids rotate around their axis?

1 answer:

7 0

Imagine a lonely asteroid there in space, untouched, not rotating, just still. Then another asteroid passes by and comes into contact with that first asteroid. Upon collision, based on the conservation of elastic momentum, the asteroid that was once still moves; it may even spin if the incoming asteroid hit it at its side. Now asteroids have three angles or rotation (three dimensions): θ (theta - to the x-axis), φ (phi - to the y-axis), and let's say ψ (psi - to the z-axis). So these asteroids wobble through space, spinning like crazy.

I may have gone too in depth. Sorry, lol.
I hope this helped!

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A rod of length 0.82 m, rotating with an angular speed, 4.2 rad/s, about axes that pass perpendicularly through one end, has a m

mariarad [96]

Answer:

Explanation:

KE = ½Iω²

ΚΕ = ½(mL²/3)ω²

ΚΕ = ½(0.63(0.82²)/3)4.2²

ΚΕ = 1.24541928

KE = 1.2 J

5 0

2 years ago

a 1-kg discus is thrown with a velocity of 19 m/s at an angle of 35 degrees from the vertical direction. calculate the vertical

nlexa [21]

Answer:

Vx = 10.9 m/s , Vy = 15.6 m/s

Explanation:

Given velocity V= 19 m/s

the angle 35 ° is taken from Y-axis so the angle with x-axis will be 90°-35° = 55°

θ = 55°

to Find Vx = ? and Vy= ?

Vx = V cos θ

Vx = 19 m/s × cos 55°

Vx = 10.9 m/s

Vx = V sin θ

Vy = 19 m/s × sin 55°

Vy = 15.6 m/s

6 0

2 years ago

Read 2 more answers

A 50.0 N box sliding on a rough horizontal floor, and the only horizontal force acting on it is friction. You observe that at on

Vikki [24]

Answer:

-4.0 N

Explanation:

Since the force of friction is the only force acting on the box, according to Newton's second law its magnitude must be equal to the product between mass (m) and acceleration (a):

$F_f = ma$ (1)

We can find the mass of the box from its weight: in fact, since the weight is W = 50.0 N, its mass will be

$m=\frac{W}{g}=\frac{50.0 N}{9.8 m/s^2}=5.1 kg$

And we can fidn the acceleration by using the formula:

$a=\frac{v-u}{t}$

where

v = 0 is the final velocity

u = 1.75 m/s is the initial velocity

t = 2.25 s is the time the box needs to stop

Substituting, we find

$a=\frac{0-1.75 m/s}{2.25 s}=-0.78 m/s^2$

(the acceleration is negative since it is opposite to the motion, so it is a deceleration)

Therefore, substituting into eq.(1) we find the force of friction:

$F_f = (5.1 kg)(-0.78 m/s^2)=-4.0 N$

Where the negative sign means the direction of the force is opposite to the motion of the box.

6 0

2 years ago

What is a good hypothesis when putting pop rocks into soda?

8_murik_8 [283]

That it will erupt upon contact. Hope it helps!

6 0

3 years ago

12.

vazorg [7]

Answer: A

Explanation: How large a parachute is (in other words, the parachute's surface area) affects its air resistance, or drag force. ... In the case of these parachutes, the drag force is opposite to the force of gravity, so the drag force slows the parachutes down as they fall.

7 0

3 years ago

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