Answer:
a) Since the height of the baseball at 99 m was 8.93 m and the fence at that distance is 3m tall, the hit was a home run.
b) The total distance traveled by the baseball was 108.7 m.
Explanation:
a) To know if the hit was a home run we need to calculate the height of the ball at 99 m:
Where:
: is the final height =?
: is the initial height = 1 m
: is the initial vertical velocity = v₀sin(45)
v₀: is the initial velocity = 32.5 m/s
g: is the gravity = 9.81 m/s²
t: is the time
First, we need to find the time by using the following equation:
Now, the height is:
Since the height of the baseball at 99 m was 8.93 m and the fence at that distance is 3m tall, the hit was a home run.
b) To find the distance traveled by the baseball first we need to find the time of flight:
By solving the above quadratic equation we have:
t = 4.73 s
Finally, with that time we can find the distance traveled by the baseball:
Hence, the total distance traveled by the baseball was 108.7 m.
I hope it helps you!
Answer:
If all these three charges are positive with a magnitude of each, the electric potential at the midpoint of segment would be approximately .
Explanation:
Convert the unit of the length of each side of this triangle to meters: .
Distance between the midpoint of and each of the three charges:
Let denote Coulomb's constant (.)
Electric potential due to the charge at : .
Electric potential due to the charge at : .
Electric potential due to the charge at : .
While forces are vectors, electric potentials are scalars. When more than one electric fields are superposed over one another, the resultant electric potential at some point would be the scalar sum of the electric potential at that position due to each of these fields.
Hence, the electric field at the midpoint of due to all these three charges would be:
.
A - the objects are too small
GRAVITATIONAL FORCE IS EXPERIENCED BY ALL OBJECTS IN THE UNIVERSE ALL THE TIME. BUT THE ORDINARY OBJECTS YOU SEE EVERY DAY HAVE MASSES SO SMALL THAT THEIR ATTRACTION TOWARD EACH OTHER IS HARD TO DETECT. -https://www.ftsd.org/cms/lib6/MT01001165/Centricity/ModuleInstance/630/CHAPTER_2_NOTES_FOR_EIGHTH_GRADE_PHYSICAL_SCIENCE.pdf
Answer:
The average velocity of the car is, V = 74.04 m/s
Explanation:
Given data,
The initial velocity of the car, u = 0 m/s
The displacement of the ca, S = 1100 m
The time period of travel, t = 14 s
The velocity of the car at point k, v = 65 m/s
Using the II equation of motion,
S = ut + ½ at²
Substituting the given values,
1100 = 0 + ½ x a x 14²
a = 11.22 m/s²
Using the III equation of motion
v² = u² + 2 as
v = √(2as) (∵ u = 0)
Substituting,
v = √(2 x 11.22 x 1100)
= 157.11 m/s
The average speed of the car,
V = 74.04 m/s
Hence, the average velocity of the car is, V = 74.04 m/s
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