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.
<h3><u>Answer;</u></h3>
volume = 6.3 × 10^-2 L
<h3><u>Explanation</u>;</h3>
Volume = mass/density
Mass = 0.0565 Kg,
Density = 900 kg/m³
= 0.0565 kg/ 900 kg /m³
= 6.3 × 10^-5 M³
but; 1000 L = 1 m³
Hence, <u>volume = 6.3 × 10^-2 L</u>
Answer
Explanation:
As the three resistors are connected in series, the expression to be used for the
calculation of RT equivalent resistance
is:
RT = R1 + R2 + R3
We replace the data of the statement in the previous expression and it remains:
5 10 15 RT + R1 + R2 + R3 + +
We perform the mathematical operations that lead us to the result we are looking for:
RT - 30Ω
