Cause he left out the noble gases out of the periodic table for one good reason, 1: He did not know them
Can you input a picture??
Answer:
= 4.86 s
= 1.98 s
Explanation:
<u><em>Given:</em></u>
Length = l = 1 m
Acceleration due to gravity of moon = 
= 1.67 m/s²
Acceleration due to gravity of Earth = 
= 10 m/s²
<u><em>Required:</em></u>
Time period = T = ?
<u><em>Formula:</em></u>
T = 2π 
<u><em>Solution:</em></u>
<u>For moon</u>
<em>Putting the givens,</em>
T = 2(3.14) 
T = 6.3 
T = 6.3 × 0.77
T = 4.86 sec
<u>For Earth,</u>
<em>Putting the givens</em>
T = 2π 
T = 2(3.14) 
T = 6.3 × 0.32
T = 1.98 sec
Answer:
C. Technician B
Explanation:
Excessive Galvanic activity:
To check for excessive galvanic activity, voltmeter is used to check the coolant. If the voltmeter is giving a reading greater than 0.5 V, there is excessive galvanic activity. Excessive galvanic activity is solved by flushing the coolant fluid from engine and refiling it.
Electrolysis problem:
When the system is not properly ground, the cooling system accepts stray current and the coolant becomes an electrolyte which might eat up the radiator. To test for excessive electrolysis, start the engine and turn on all electrical accessories, if the reading is more than 0.5 V, there is electrolysis problem. Ground wires and connections should be checked at this point to stop stray current.
In our case, the first reading is 0.2 V(engine turned off) which is normal and there is no excessive galvanic activity. This means that Technician A is not correct. The second reading is 0.8 V when the engine and all electrical accessories are turned on. This reading is greater than 0.5 V which means there is an electrolysis problem. This means that Technician B is correct and ground wires and connections should be inspected and repaired.
I don't think that 4m has anything to do with the problem.
anyway. here.
A___________________B_______C
where A is the point that the train was released.
B is where the wheel started to stick
C is where it stopped
From A to B, v=2.5m/s, it takes 2s to go A to B so t=2
AB= v*t = 2.5 * 2 = 5m
The train comes to a stop 7.7 m from the point at which it was released so AC=7.7m
then BC= AC-AB = 7.7-5 = 2.7m
now consider BC
v^2=u^2+2as
where u is initial speed, in this case is 2.5m/s
v is final speed, train stop at C so final speed=0, so v=0
a is acceleration
s is displacement, which is BC=2.7m
substitute all the number into equation, we have
0^2 = 2.5^2 + 2*a*2.7
0 = 6.25 + 5.4a
a = -6.25/5.4 = -1.157
so acceleration is -1.157m/(s^2)
