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ddd [48]
3 years ago
7

In the simulation, there are three balls on the floor. Drag each of them up off the floor, and then let go. See what happens to

the balls. In what way do the balls behave the same? In what way do they behave differently?
Physics
1 answer:
Vlad1618 [11]3 years ago
7 0

Answer:

I hope this helps and I'm not to late

A way the balls behave the same way is by bouncing about 1 time after throwing the balls up. A way the balls act differently is the blue ball is bouncier than all the balls, the red ball bounces about 2 times before stopping, and the green ball doesn’t really bounce except for one time.

Explanation:

you also can use paraphrase to help you reword bye bye!!

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Planet geos in orbit a distance of 1
nadya68 [22]

Planet Geos in orbit a distance of 1 A.U. (astronomical unit) from the star Astra has an orbital period of 1 "year." If planet Logos is 4 A.U. from Astra, how long does Logos require for a complete orbit?

TB = <span>8</span> years

4 0
3 years ago
Read 2 more answers
What is the polarity of each of the earths magnetic poles ? Explain you answer
RideAnS [48]

Answer:

When you put un-like poles together (South facing North) you can feel magnetic attraction. In the Northern Hemisphere, your compass needle points North, but if you think about it for a moment, you will discover that the magnetic pole in the Earth's Northern Hemisphere has to be a South polarity.

3 0
3 years ago
Volume of a cylinder with a diameter of 1.55 cm and a height of 1.34 cm
Bogdan [553]

Answer:

2.53 cm³

Explanation:

Volume of cylinder = πr²h; where h is the height and r is the radius and

π = 3.14 approx.

Volume = 3.14 * (1.55/2)² * 1.34 = 2.53 cm³

I divided 1.55 by 2 because we were given the diameter and not radius. Diameter = radius / 2

4 0
3 years ago
How to calculate? I got 15000N for (a)
MAXImum [283]
Your answer for part a is correct.  Using Newton's 3rd Law, the force on the rocket by the exhaust leaving the rocket is the same magnitude (opposite direction) as the downward force applied to the exhaust.
In part b the net force is the upward force from the exhaust thrust minus the downward force of gravity:

F_{net}=F_{thrust}-mg=15,000N-1500kg \times 1.6 \frac{m}{s^2}=12,600N

Then using the Second Law, we get for part c:

F=ma \\ a= \frac{F}{m} = \frac{12600N}{1500kg} =8.4 \frac{m}{s^2}

The force and acceleration are in the upward direction
3 0
3 years ago
Consider the two-body system at the right. A 22.7-N block is placed upon an inclined plane which is inclined at a 17.2 degree an
artcher [175]

Answer:

Given that

For A weight Wt= 22.7 N

m₁ = 22.7/10 = 2.27 kg

Force alone inclined plane

Wt₁ = m₁ g sin θ

Wt₁  =22.7 sin 17.2°

Wt₁ = 6.7 N

For B weight Wt₂= 34.5-N

m₂ = 3.45 kg

coefficient of friction ,μ= 0.219

θ = 17.2 degree

The friction force on the block A

Fr= μ m₁ g cos θ

Fr= 0.216 x 22.7 x cos 17.2°

Fr= 4.68 N

Lets take acceleration of system is a m/s²

Tension = T

From Newtons law

Wt₂ - Wt₁ - Fr = (m₁ +m₂) a

34.5 -  6.7 - 4.68 = (2.27 + 3.45 ) a

a= 4.05 m/s²

Block B

Wt₂ - T = m₂ a

T = 34.5 - 4.05 x 3.45

T= 20.52 N

3 0
4 years ago
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