The ball is against the vector of gravity. Then, the gravity will be negative.
The ball will stop in the air after approx. 4.72 seconds. And will take the same time to hit the ground.
It will stay approx. 9.44 seconds in the air.
A ball is dropped from the top of a building.After 2 seconds, it’s velocity is measured to be 19.6 m/s. Calculate the accelerati
2 answers:
Explanation:
The given data is as follows.
Initial velocity; u = 0, Final velocity; v = 19.6 m/s
time; t = 2 seconds
As the relation between initial velocity, final velocity and acceleration is as follows.
v = u + at
Hence, putting the given values into the above formula as follows.
v = u + at
19.6 m/s = 0 +
a = 9.8
Thus, we can conclude that acceleration of the dropped ball is 9.8 .
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The ball is against the vector of gravity. Then, the gravity will be negative.
The ball will stop in the air after approx. 4.72 seconds. And will take the same time to hit the ground.
It will stay approx. 9.44 seconds in the air.
The ball was dropped from a height 20 meters
Explanation:
The given is
1. A ball is dropped from the top of a cliff
2. By the time it reaches the ground, all the energy in its gravitational
potential energy store has been transferred into its kinetic energy
store, that mean K.E = P.E
3. The ball is travelling at 20 m/s when it hits the ground
4. The gravitational field strength is 10 N/kg
We need to find the height that the ball dropped from it
The ball dropped from the top of a cliff means the initial speed is 0
→ K.E =
where v is the final speed, in the initial speed and m
is the mass
→ v = 20 m/s and = 0 m/s
→ K.E =
→ K.E =
→ K.E = 200 m joules ⇒ when the ball hits the ground
→ P.E = m g h
where g is the gravitational field strength, m is the mass and h is
the height
→ g = 10 N/kg
→ P.E = m(10)(h)
→ P.E = 10 m h joules
→ P.E = K.E
→ 10 m h = 200 m
Divide both sides by 10 m
→ h = 20 meters
The ball was dropped from a height 20 meters
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Answer:
E) are almost circular, with low eccentricities.
Explanation:
Kepler's laws establish that:
All the planets revolve around the Sun in an elliptic orbit, with the Sun in one of the focus (Kepler's first law).
A planet describes equal areas in equal times (Kepler's second law).
The square of the period of a planet will be proportional to the cube of the semi-major axis of its orbit (Kepler's third law).
Where T is the period of revolution and a is the semi-major axis.
Planets orbit around the Sun in an ellipse with the Sun in one of the focus. Because of that, it is not possible to the Sun to be at the center of the orbit, as the statement on option "C" says.
However, those orbits have low eccentricities (remember that an eccentricity = 0 corresponds to a circle)
In some moments of their orbit, planets will be closer to the Sun (known as perihelion). According with Kepler's second law to complete the same area in the same time, they have to speed up at their perihelion and slow down at their aphelion (point farther from the Sun in their orbit).
Therefore, option A and B can not be true.
In the celestial sphere, the path that the Sun moves in a period of a year is called ecliptic, and planets pass very closely to that path.