Answer:
When I got this question I had to draw it out so if you have to do that, draw 3 stick figures holding hands, one representing the mother, father, and daughter. Then you write their weights on top of them and then draw an arrow pointing from the father to the mother.
Explanation:
use this formula :
=
then you fill it in :
=
= 
=

then you multiply that with the daughters weight :

and that's the answer :) : 37.89N
Answer:
The average induced emf in the coil is 0.0286 V
Explanation:
Given;
diameter of the wire, d = 11.2 cm = 0.112 m
initial magnetic field, B₁ = 0.53 T
final magnetic field, B₂ = 0.24 T
time of change in magnetic field, t = 0.1 s
The induced emf in the coil is calculated as;
E = A(dB)/dt
where;
A is area of the coil = πr²
r is the radius of the wire coil = 0.112m / 2 = 0.056 m
A = π(0.056)²
A = 0.00985 m²
E = -0.00985(B₂-B₁)/t
E = 0.00985(B₁-B₂)/t
E = 0.00985(0.53 - 0.24)/0.1
E = 0.00985 (0.29)/ 0.1
E = 0.0286 V
Therefore, the average induced emf in the coil is 0.0286 V
They should look for <span>a report from an independent scientific research firm,
even if they have to pay for it.
In preparing its report, the firm would have already surveyed many of the </span>
<span>citizens from several other towns that currently add fluoride to their water,
plus a lot of other relevant medical research on the subject.</span>
Answer: option d.
Explanation:1) The
direction of the
field lines inform about the
sign of the charges.
The field lines <span>
extend from the positive charges to the negative charges, so you can conclude that the charge C is positve and both charge A and charge B are negative:
</span><span>
</span><span>
</span><span>Charge C: positive
</span><span>
</span><span>Charge A: negative
</span><span>
</span><span>Charge B: netative
</span>
2) The
density of the lines (number of lines in a region) inform about the
magnitude of the electric field.
Since the charges are at the same distance, the magnitude of the electric field informs directly about the magnitude of the force and that about the magnitude of the charges.
Since, there are the
double of lines between C and B than between C and A, the magnitude of
charge B is the double than the magnitud of charge A.
From the five options given (a throug e) the only that is consistent with that charges A and B have the same sign, that charge C has different sign, and that charge B is the double of charge A is: