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Rzqust [24]
3 years ago
13

What is life like on planet mars?

Physics
1 answer:
Natali5045456 [20]3 years ago
4 0
It's boring.
You're forced to eat Mars Bars because it grows here.
Once in a while, a rogue Snickers Bar hurtles from asteroids.
That only happens about once a year.


(Am I funny yet? =_=)
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What is the best description of the chromosomes by the end of metaphase of mitosis?
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I did the test and the anwser is d- the chromosome pairs collect in a line across the middle of the cell
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3 years ago
A person holds a ladder horizontally at its center. Treating the ladder as a uniform rod of length 4.15 m and mass 7.98 kg, find
Ludmilka [50]

Answer:

4.535 N.m

Explanation:

To solve this question, we're going to use the formula for moment of inertia

I = mL²/12

Where

I = moment of inertia

m = mass of the ladder, 7.98 kg

L = length of the ladder, 4.15 m

On solving we have

I = 7.98 * (4.15)² / 12

I = (7.98 * 17.2225) / 12

I = 137.44 / 12

I = 11.45 kg·m²

That is the moment of inertia about the center.

Using this moment of inertia, we multiply it by the angular acceleration to get the needed torque. So that

τ = 11.453 kg·m² * 0.395 rad/s²

τ = 4.535 N·m

8 0
3 years ago
You stretch a spring by a distance of 0.3 m. The spring has a spring constant of 440 N/m. When you release the spring, it snaps
Sonja [21]

the answer is c 19.8

8 0
4 years ago
Compare and contrast two physical properties of apples and oranges
elena-14-01-66 [18.8K]

Color: Apples are red or green or yellow. Oranges are normally orange.

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5 0
3 years ago
The block slides on a horizontal frictionless surface. The block has a mass of 1.0kg and is pushed 5.0N at 45°. What is the magn
Firlakuza [10]

Answer:

3.5\:\mathrm{m/s^2}

Explanation:

Newton's 2nd law is given as \Sigma F = ma.

To find the acceleration in the horizontal direction, you need the horizontal component of the force being applied.

Using trigonometry to find the horizontal component of the force:

\cos 45^{\circ}=\frac{x}{5},\\\frac{\sqrt{2}}{{2}}=\frac{x}{5},\\x=\frac{5\sqrt{2}}{2}

Use this horizontal component of the force to solve for for the acceleration of the object:

\frac{5\sqrt{2}}{2}=1.0\cdot a,\\a=\frac{5\sqrt{2}}{2}\approx \boxed{3.5\:\mathrm{m/s^2}}

8 0
3 years ago
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