Solve for P if P varies as the square of Q, and P=4 when Q=16. Find R if R varies inversely as the square of its radius, and R=0.80 when r=0.4.

P varies as the square of Q, so we can write this relationship as:

$$P = kQ^2$$

where $k$ is the constant of proportionality.

We are given $P = 4$ when $Q = 16$. Substitute these values into the equation:

$$4 = k(16)^2$$ $$4 = k(256)$$

Divide by 256 to find $k$:

$$k = \frac{4}{256}$$ $$k = \frac{1}{64}$$

Substitute the value of $k$ back into the general equation:

$$P = \frac{1}{64}Q^2$$

We need to find P when $Q = 64$. Substitute $Q = 64$ into the specific variation equation:

$$P = \frac{1}{64}(64)^2$$ $$P = \frac{1}{64} \times 64 \times 64$$ $$P = 64$$

The value of P is $\boxed{\text{64}}$.

:

The electrical resistance ($R$) of a wire varies inversely as the square of its radius ($r$). This can be written as:

$$R = \frac{k}{r^2}$$

where $k$ is the constant of proportionality.

We are given that the resistance is $0.80 \, \Omega$ when the radius is $0.4 \, \text{cm}$. Substitute these values into the equation:

$$0.80 = \frac{k}{(0.4)^2}$$ $$0.80 = \frac{k}{0.16}$$

Multiply both sides by 0.16 to find $k$:

$$k = 0.80 \times 0.16$$ $$k = 0.128$$

Substitute the value of $k$ back into the general equation:

$$R = \frac{0.128}{r^2}$$

We need to find the resistance when the radius is $0.7 \, \text{cm}$. Substitute $r = 0.7$ into the specific variation equation:

$$R = \frac{0.128}{(0.7)^2}$$ $$R = \frac{0.128}{0.49}$$ $$R \approx 0.26122$$

Rounding to two significant figures, the resistance is approximately:

$$R \approx 0.26 \, \Omega$$

The resistance is approximately $\boxed{\text{0.26 \Omega}}$.