Answer:
0.015m^3
Explanation:
1 m^3 = 1000 liters
x m^3 = 15 liters
Cross multiply
xm^3 x 1000 l = 15 l
Divide both sides by 1000
xm^3 x1000/1000 = 15/1000
xm^3 = 0.015m^3
Therefore 15 liter = 0.015m^3
Answer:
After a male ejaculates, many sperms move to the upper vagina (via contractions from the vagina) through the cervix and across the length of the uterus, reaching the fallopian tubes. Here they will meet the egg cell ready to be fertilized
Some examples of projective tests are the Rorschach Inkblot Test, the Thematic Apperception Test (TAT), the Contemporized-Themes Concerning Blacks test, the TEMAS (Tell-Me-A-Story), and the Rotter Incomplete Sentence Blank (RISB).
Some examples of projective tests are the Rorschach Inkblot Test, the Thematic Apperception Test (TAT), the Contemporized-Themes Concerning Blacks test, the TEMAS (Tell-Me-A-Story), and the Rotter Incomplete Sentence Blank (RISB).
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
Answer:
Final speed of the train is 7.5 m/s
Explanation:
It is given that,
Uniform acceleration of the train is, a = 1.5 m/s²
It starts from rest and travels for 5.0 s. We have to find the final velocity of the train. By using first equation of motion as :
Here, train starts from rest so, u = 0
v = 7.5 m/s
So, the final velocity of the train is 7.5 m/s. Hence, this is the required solution.
