Answer:
a) t = 0.74s
b) D = 4.76m
c) Vf = 5.35m/s
Explanation:
The ball starts rolling when Vf = ωf*R.
We know that:
Vf = Vo - a*t
ωf = ωo + α*t
With a sum of forces on the ball:




With a sum of torque on the ball:



Replacing both accelerations:


t=0.74s
The distance will be:


Final velocity:

Vf=5.35m/s
Answer:
A jar of mixed nuts is the correct answer.
Explanation:
A heterogeneous mixture is a type of mixture that is not uniform in the appearance and composition varies throughout.
The heterogeneous mixture consists of multiple phases.
A jar of mixed nuts is an example of the heterogeneous mixture because they do not have a uniform composition component differ in proportion and they do not mix, instead, they form two different layers and they can be easily separated.
Answer:
23376 days
Explanation:
The problem can be solved using Kepler's third law of planetary motion which states that the square of the period T of a planet round the sun is directly proportional to the cube of its mean distance R from the sun.

where k is a constant.
From equation (1) we can deduce that the ratio of the square of the period of a planet to the cube of its mean distance from the sun is a constant.

Let the orbital period of the earth be
and its mean distance of from the sun be
.
Also let the orbital period of the planet be
and its mean distance from the sun be
.
Equation (2) therefore implies the following;

We make the period of the planet
the subject of formula as follows;

But recall that from the problem stated, the mean distance of the planet from the sun is 16 times that of the earth, so therefore

Substituting equation (5) into (4), we obtain the following;

cancels out and we are left with the following;

Recall that the orbital period of the earth is about 365.25 days, hence;

Answer:
Kinda? Depends what the question is fully asking
Explanation:
Acceleration is a change in velocity. So I guess if the velocity of something is -2 m/s and its positively accelerating at a value of +1 m/s, then that means every second its velocity changes by +1m/s.
So that -2 m/s thing after one second will be going -1 m/s.
After another second it'll be going 0 m/s.
After another itll be going +1 m/s and so on.
So at one point for a brief moment, it can have an acceleration but be at 0 m/s velocity.
Answer:
<em>The Antarctic blue whale </em>
Explanation:
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