Answer:
repel each other
Explanation:
The magnitude of the charge of an electron is called... ... If a positively-charged glass rod is suspended so that it turns easily and another positively-charged glass rod is brought close to it, the two rods will... Repel each other.
We're u can never put it back together
Answer:Technician A
Explanation:
Technician A statement is correct as
The battery is required to start the vehicle which, in effect, rotates the alternator at sufficient speed to keep the battery charged. This means if the battery is low it is not possible to start the vehicle and thus we are unable to test the alternator.
That is the battery is pre-requisite to test the alternator. So the battery must be at least a 75 % charge to test the alternator.
Answer:
largest lead = 3 m
Explanation:
Basically, this problem is about what is the largest possible distance anchorman for team B can have over the anchorman for team A when the final leg started that anchorman for team A won the race. This show that anchorman for team A must have higher velocity than anchorman for team B to won the race as at the starting of final leg team B runner leads the team A runner.
So, first we need to calculate the velocities of both the anchorman
given data:
Distance = d = 100 m
Time arrival for A = 9.8 s
Time arrival for B = 10.1 s
Velocity of anchorman A = D / Time arrival for A
=100/ 9.8 = 10.2 m/s
Velocity of anchorman B = D / Time arrival for B
=100/10.1 = 9.9 m/s
As speed of anchorman A is greater than anchorman B. So, anchorman A complete the race first than anchorman B. So, anchorman B covered lower distance than anchorman A. So to calculate the covered distance during time 9.8 s for B runner, we use
d = vt
= 9.9 x 9.8 = 97 m
So, during the same time interval, anchorman A covered 100 m distance which is greater than anchorman B distance which is 97 m.
largest lead = 100 - 97 = 3 m
So if his lead no more than 3 m anchorman A win the race.
Answer:
The correct answer is "0.32 mL".
Explanation:
The given values are:
Density of gold bar,
d = 19.3 g/mL
Mass of gold bar,
m = 6.3 grams
Now,
The volume will be:
⇒ 
or,
⇒ 
On substituting the values, we get
⇒ 
⇒ 
