Right when it’s about to go down and stopped.
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:
1.03 m/s
Explanation:
I'm too lazy to write the explanation down but my teacher graded this and it was right
Time required : 3 s
<h3>Further explanation
</h3>
Power is the work done/second.
To do 33 J of work with 11 W of power
P = 11 W
W = 33 J
Answer:
24 Coulumbs
Explanation:
Given data
time= 1 minute= 6 seconds
P=2 W
R= 12 ohm
We know that
P= I^2R
P/R= I^2
2/12= I^2
I^2= 0.166
I= √0.166
I= 0.4 amps
We know also that
Q= It
substitute
Q= 0.4*60
Q= 24 Columbs
Hence the charge is 24 Coulumbs
