Answer:
α = 13.7 rad / s²
Explanation:
Let's use Newton's second law for rotational motion
∑ τ = I α
we will assume that the counterclockwise turns are positive
F₁ 0 + F₂ R₂ - F₃ R₃ = I α
give us the cylinder moment of inertia
I = ½ M R₂²
α = (F₂ R₂ - F₃ R₃) 
let's calculate
α = (24 0.22 - 13 0.10)
2/12 0.22²
α = 13.7 rad / s²
Answer:
Rita and Katrina both followed similar paths into the Gulf.
Explanation:
Answer:
The first factor is the amount of charge on each object. The greater the charge, the greater the electric force. The second factor is the distance between the charges. The closer together the charges are, the greater the electric force is.
Explanation:
Temperature rise will be there in cylinder B more than in cylinder A because of internal energy.
what is internal energy?
The sum of the kinetic and chemical potential energies of all the particles in the system is the internal energy. Particles accelerate and pick up kinetic energy when energy is applied to increase the temperature.
Briefing:
Cylinder A uses the heat it absorbs to both work while expanding and to increase internal energy (or temperature).
While cylinder B solely uses the heat it absorbs to increase its internal energy
As a result, cylinder B's temperature rise is greater than cylinder A's.
To know more about internal energy visit:
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My calculator is about 1cm thick, 7cm wide, and 13cm long.
Its volume is (length) (width) (thick) = (13 x 7 x 1) = 91 cm³ .
The question wants me to assume that the density of my calculator
is about the same as the density of water. That doesn't seem right
to me. I could check it easily. All I have to do is put my calculator
into water, watch to see if sinks or floats, and how enthusiastically.
I won't do that. I'll accept the assumption.
If its density is actually 1 g/cm³, then its mass is about 91 grams.
The choices of answers confused me at first, until I realized that
the choices are actually 1g, 10² g, 10⁴ g, and 10⁶ g.
My result of 91 grams is about 100 grams ... about 10² grams.
Your results could be different.