(a) Let be the maximum linear speed with which the ball can move in a circle without breaking the cord. Its centripetal/radial acceleration has magnitude
where is the radius of the circle.
The tension in the cord is what makes the ball move in its plane. By Newton's second law, the maximum net force on it is
so that
Solve for :
(b) The net force equation in part (a) leads us to the relation
so that is directly proportional to the square root of . As the radius increases, the maximum linear speed will also increase, so the cord is less likely to break if we keep up the same speed.
Answer:
which pic...? there is no picture attached to your question
Answer:
PE = 44.1 J
Explanation:
Ok, to have the specific data, the first thing we must do is convert from grams to kilograms. Since mass must always be in kilograms (kg)
We have:
- 1 kilograms = 1000 grams.
We convert it using a rule of 3, replacing, simplifying units and solving:
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Earth's gravity is known to be 9.8 m/s², so we have:
Data:
- m = 0.3 kg
- g = 9.8 m/s²
- h = 15 m
- PE = ?
Use formula of potencial energy:
Replace and solve:
Since the decimal number, that is, the number after the comma is less than 5, it cannot be rounded, then we have this result.
The potential energy of the volleyball is <u>44.1 Joules.</u>
Greetings.
Answer:
The appropriate solution is:
(a)
(b)
(c)
Explanation:
According to the question, the value is:
Power of bulb,
= 60 W
Distance,
= 1.0 mm
Now,
(a)
⇒
On applying cross-multiplication, we get
⇒
⇒
⇒
(b)
As we know,
⇒
By putting the values, we get
⇒
(c)
⇒
⇒
⇒
The main formula to be used here is
Force = (mass) x (acceleration).
We'll get to work in just a second. But first, I must confess to you that I see
two things happening here, and I only know how to handle one of them. So
my answer will be incomplete, but I believe it will be more reliable than the
first answer that was previously offered here.
On the <u>right</u> side ... where the 2 kg and the 3 kg are hanging over the same
pulley, those weights are not balanced, so the 3 kg will pull the 2kg down, with
some acceleration. I don't know what to do with that, because . . .
At the <em>same time</em>, both of those will be pulled <u>up</u> by the 10 kg on the other side
of the upper pulley.
I think I can handle the 10 kg, and work out the acceleration that IT has.
Let's look at only the forces on the 10 kg:
-- The force of gravity is pulling it down, with the whatever the weight of 10 kg is.
-- At the same time, the rope is pulling it UP, with whatever the weight of 5 kg is ...
that's the weight of the two smaller blocks on the other end of the rope.
So, the net force on the 10 kg is the weight of (10 - 5) = 5 kg, downward.
The weight of 5 kg is (mass) x (gravity) = (5 x 9.8) = 49 newtons.
The acceleration of 10 kg, with 49 newtons of force on it, is
Acceleration = (force) / (mass) = 49/10 = <em>4.9 meters per second²</em>