A ball is tossed with enough speed straight up so that it is in the air several seconds. (a) What is the velocity of the ball wh
1 answer:
Answer:
The velocity of the ball at it's highest point is zero.
Explanation:
When we toss a ball upwards as we know that it's motion is an uniformly accelerated motion with acceleration = in the downward direction.
Hence as we throw the ball upwards it's speed will go on decreasing as it gains height. The ball will stop gaining height when it's velocity becomes zero and the ball starts to descend again towards earth.
If the ball is tossed at an angle with the horizontal in that case it's velocity at the top most point will not be zero but in our case since the ball is tossed vertically it's velocity at the topmost point will become zero.
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Explanation:
Formula to determine the critical crack is as follows.
= 1,
= 24.1
[/tex]\sigma_{y}[/tex] = 570
and,
= 427.5
Hence, we will calculate the critical crack length as follows.
a =
=
=
Therefore, largest size is as follows.
Largest size = 2a
=
=
Thus, we can conclude that the critical crack length for a through crack contained within the given plate is .
Answer:
G = 6,786 10⁻¹¹ m³ / s² kg
Explanation:
The law of universal gravitation is
F = G m M/ r²
Where G is the gravitational constant, m and M are the masses of the bodies and r is the distance from their centers
Let's use Newton's second law
F = m a
The acceleration is centripetal
a =
We replace
G m M / r² = m
G = r² / M
Let's replace and calculate
G = 2.7 10⁻³ (3.88 10⁸)² / 5.99 10²⁴
G = 6,786 10⁻¹¹ m³ / s² kg
Let's perform a dimensional analysis
[N m²/kg²] = [kg m/s² m² / kg²] = [m³ / s² kg]