Answer:
The acceleration experienced by the occupants of the spaceship during launch is 282652.782 meters per square second.
Explanation:
Let suppose that spaceship is accelerated uniformly. A yard equals 0.914 meters. A feet equals 0.304 meters. If air viscosity and friction can be neglected, then acceleration (
), measured in meters per square second, is estimated by this kinematic formula:
(1)
Where:
- Travelled distance, measured in meters.
, 
- Initial and final speeds of the spaceship, measured in meters.
If we know that 
, 
and 
, then the acceleration experimented by the spaceship is:
The acceleration experienced by the occupants of the spaceship during launch is 282652.782 meters per square second.
Answer:
The average velocity is 50 km/h south
Explanation:
The average velocity of an object is its total displacement divided by
the total time taken.
That means it is the rate at which an object changes its position from
one place to another.
Average velocity is a vector quantity.
The SI unit is meters per second.
A bicycle that starts 100 km south and is 120 km south of town after
0.4 hour.
The displacement = 120 - 100 = 20 km south
The time = 0.4 hour
The average velocity = 
, where D is the displacement
and t is the time
The average velocity of the bicycle = 
km/h
<em>The average velocity is 50 km/h south</em>
If you want it in meter per second, change the kilometer to meter
and change the hour to seconds
1 km = 1000 m
1 hour = 60 × 60 = 3600 seconds
The average velocity of the bicycle = 
m/s south
The answer is radio waves
Answer:
A scalar is a quantity that is fully described by a magnitude only. It is described by just a single number. Some examples of scalar quantities include speed, volume, mass, temperature, power, energy, and time.
Examples of scalar quantity are:
Distance.
Speed.
Mass.
Temperature.
Energy.
Work.
Volume.
Area.
Explanation:
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Answer:
The only incorrect statement is from student B
Explanation:
The planet mercury has a period of revolution of 58.7 Earth days and a rotation period around the sun of 87 days 23 ha, approximately 88 Earth days.
Let's examine student claims using these rotation periods
Student A. The time for 4 turns around the sun is
t = 4 88
t = 352 / 58.7 Earth days
In this time I make as many rotations on itself each one with a time to = 58.7 Earth days
#_rotaciones = t / to
#_rotations = 352 / 58.7
#_rotations = 6
therefore this statement is TRUE
student B. the planet rotates 6 times around the Sun
t = 6 88
t = 528 s
The number of rotations on itself is
#_rotaciones = t / to
#_rotations = 528 / 58.7
#_rotations = 9
False, turn 9 times
Student C. 8 turns around the sun
t = 8 88
t = 704 days
the number of turns on itself is
#_rotaciones = t / to
#_rotations = 704 / 58.7
#_rotations = 12
True
The only incorrect statement is from student B
